Question 13 Marks
Evaluate the following integrals:
$\int(\tan\text{x}+\cot\text{x})^2\text{dx}$
Answer$\int(\tan\text{x}+\cot\text{x})^2\text{dx}$
$=\int(\tan^2\text{x}+\cot^2\text{x}+2\tan\text{x}\cot\text{x})\text{dx}$
$=\int(\tan^2\text{x}+\cot^2\text{x}+2)\text{dx}$
$=\int\big[(\sec^2\text{x}-1)+(\text{cosec}^2\text{x}-1)+2\big]\text{dx}$
$=\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
View full question & answer→Question 23 Marks
$\int\sin\text{x}\sqrt{1+\cos2\text{x}}\text{ dx}$
Answer$\int\sin\text{x}\sqrt{1+\cos2\text{x}}\text{ dx}$
$=\int\sin\text{x}.\sqrt{2\cos^2\text{x}}\text{ dx}$ $[\therefore1+\cos2\text{x}=2\cos^2\text{x}]$
$=\sqrt{2}\int\sin\text{x}\cos\text{x dx}$
$=\frac{\sqrt{2}}{2}\int2\sin\text{x}\cos\text{x dx}$
$=\frac{1}{\sqrt{2}}\int\sin2\text{x dx}$
$=\frac{1}{\sqrt{2}}\Big[\frac{-\cos2\text{x}}{2}\Big]+\text{c}$
$=\frac{-1}{2\sqrt{2}}\cos2\text{x}+\text{c}$
View full question & answer→Question 33 Marks
If $f'(x) = 8x^3 - 2x, f'(2) = 8,$ find $f'(x)$.
AnswerWe have,
$\text{f}'\text{(x)}=8\text{x}^3-2\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\int\text{f}'\text{(x)dx}=\int(8\text{x}^3-2\text{x})\text{dx}$
$\Rightarrow\text{f}'\text{(x)}=\int(8\text{x}^3-2\text{x})\text{dx}$
$=\int8\text{x}^3\text{dx}-\int2\text{ x dx}$
$=\frac{8\text{x}^4}{4}-\frac{2\text{x}^2}{2}+\text{C}$
$=2\text{x}^4-\text{x}^2+\text{C}$
$\Rightarrow\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2+\text{C}\ \dots(1)$
Since, $\text{f}'(2)=8$
$\therefore\ \text{f}'(2)=2(2)^4-(2)^2+\text{C}=8$
$\Rightarrow32-4+\text{C}=8$
$\Rightarrow28+\text{C}=8$
$\Rightarrow\text{C}=-20$
Putting C = -20 in eq. (1), we get
$\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2-20$
Hence, $\text{f}'\text{(x)}=2\text{x}^4-\text{x}^2-20$
View full question & answer→Question 43 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
Let $\text{x}\text{e}^\text{x}=\text{t}$
$\Rightarrow\big(1.\text{e}^\text{x}+\text{x}\text{e}^\text{x}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
$=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\sec^2\text{t}\text{ dt}$
$=\tan(\text{t})+\text{C}$
$=\tan\big(\text{xe}^\text{x}\big)+\text{C}$
View full question & answer→Question 53 Marks
Evaluate the following integrals:
$\int\text{x}^2\cos2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\cos2\text{x}\text{ dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\cos2\text{x dx}-\int(2\text{x}\int\cos2\text{x dx})\text{dx}$
$=\text{x}^2\frac{\sin2\text{x}}{2}-2\int\text{x}\Big(\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-\int\text{x}\sin2\text{x dx}$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-[\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}]$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-\bigg[\text{x}\Big(\frac{-\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}\bigg]$
$=\frac{1}{2}\text{x}^2\sin2\text{x}+\frac{\text{x}}{2}\cos2\text{x}-\frac{1}{2}\int(\cos2\text{x})\text{dx}$
$\text{I}=\frac{1}{2}\text{x}^2\sin2\text{x}+\frac{\text{x}}{2}\cos2\text{x}-\frac{1}{4}\sin2\text{x}+\text{C}$
View full question & answer→Question 63 Marks
Evaluate the following integrals:
$\int\frac{\sin(\log\text{x})}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin(\log\text{x})}{\text{x}}\text{ dx}\ ....(1)$ Let $\log\text{x}=\text{t}$ then, $\text{d}(\log\text{x})=\text{dt}$ $\Rightarrow\frac{1}{\text{x}}\text{ dx}=\text{dt}$ Putting $\log\text{x}=\text{t}$ and $\frac{1}{\text{x}}\text{ dx}=\text{dt}$ in equation (1), We get,$\text{I}=\int\sin\text{t dt}$
$=-\cos\text{t}+\text{C}$
$=-\cos\big(\log\text{x}\big)+\text{C}$
View full question & answer→Question 73 Marks
Evaluate the following integrals:
$\int\sqrt{\tan\text{x}}\sec^4\text{x}\text{ dx}$
Answer$\int\sqrt{\tan\text{x}}\sec^4\text{x}\text{ dx}$
$=\int\sqrt{\tan\text{x}}\cdot\sec^2\text{x}\cdot\sec^2\text{x}\text{ dx}$
$=\int\sqrt{\tan\text{x}}\cdot(1+\tan^2\text{x})\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\text{ dx}=\text{ dt}$
Now, $\int\sqrt{\tan\text{x}}\cdot(1+\tan^2\text{x})\sec^2\text{x}\text{ dx}$
$=\int\sqrt{\text{t}}(1+\text{t}^2)\text{dt}$
$=\int\Big(\sqrt{\text{t}}+\text{t}^{\frac{5}{2}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{1}{2}}+\text{t}^{\frac{5}{2}}\Big)\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\frac{2}{7}\text{t}^{\frac{7}{2}}+\text{C}$
$=\frac{2}{3}\tan^{\frac{3}{2}}\text{x}+\frac{2}{7}\tan^{\frac{7}{2}}\text{x}+\text{C}$
View full question & answer→Question 83 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^3-3\text{x}^2+5\text{x}-7+\text{x}^2\text{a}^\text{x}}{2\text{x}^2}\text{dx}$
Answer$\int\frac{\text{x}^3-3\text{x}^2+5\text{x}-7+\text{x}^2\text{a}^\text{x}}{2\text{x}^2}\text{dx}$
$=\frac{1}{2}\int\frac{\text{x}^3}{\text{x}^2}\text{dx}-\frac{3}{2}\int\frac{\text{x}^2}{\text{x}^2}\text{dx}+\frac{5}{2}\int\text{x}\frac{\text{x}}{\text{x}^2}\text{dx}-\frac{7}{2}\int\text{x}^{-2}\text{dx}+\frac{1}{2}\int\frac{\text{x}^2\text{a}^\text{x}}{\text{x}^2}\text{dx}$
$=\frac{1}{2}\times\frac{\text{x}^2}{2}-\frac{3}{2}\text{x}+\frac{5}{2}\log\text{x}-\frac{7}{2}\text{x}^{-1}+\frac{1}{2}\frac{\text{a}^\text{x}}{\log\text{a}}+\text{C}$
$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}-3\text{x}+5\log\text{x}-\frac{7}{\text{x}}+\frac{\text{a}^\text{x}}{\log\text{a}}\Big]+\text{C}$
$\therefore\ \int\frac{\text{x}^3-3\text{x}^2+5\text{x}-7+\text{x}^2\text{a}^\text{x}}{2\text{x}^2}\text{dx}$$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}-3\text{x}+5\log\text{x}-\frac{7}{\text{x}}+\frac{\text{a}^\text{x}}{\log\text{a}}\Big]+\text{C}$
View full question & answer→Question 93 Marks
Evalute the following integrals:
$\int\frac{\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}-2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}-2}\text{dx}$
Putting $e^{2x} = t$
$\Rightarrow2\text{e}^{2\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{2\text{x}}\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{\text{t}-2}\text{dt}$
$=\frac{1}{2}\text{ ln}|\text{t}-2|+\text{C}$
$=\frac{1}{2}\text{ ln}\big|\text{e}^{2\text{x}}-2\big|+\text{C }\big[\text{t}=\text{e}^{2\text{x}}\big]$
View full question & answer→Question 103 Marks
Evaluate the following integrals:$\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin2\text{x}}}\text{ dx}$
AnswerTo evaluate the following integral follow the steps:
$\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2-1}}\text{ dx}$
Let $\sin\text{x}+\cos\text{x}=\text{t}$ therefore $(\cos\text{x}-\sin\text{x})\text{ dx}=\text{dt}$
Now,
$\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2-1}}\text{ dx}=-\int\frac{\text{dt}}{\sqrt{\text{t}^2-1}}$
$=-\ln\Big|\text{t}+\sqrt{\text{t}^2-1}\Big|+\text{C}$
$=-\ln\Big|\sin\text{x}+\cos\text{x}+\sqrt{\sin2\text{x}}\Big|+\text{C}$
View full question & answer→Question 113 Marks
Evaluate the following integrals:$\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^3}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
Let the first function be $\cos^{-1}\text{x}$ and second function be $\frac{\text{x}}{\sqrt{1-\text{x}^2}}\cdot$
First we find the integral of the second function, i.e, $\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}.$
Put $\text{t}=1-\text{x}^2.$.Then $\text{dt}=-2\text{x dx}$
Therefore,
$\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=-\frac{1}{2}\int\frac{1}{\sqrt{\text{t}}}\text{dt}$
$=-\sqrt{\text{t}}$
$=-\sqrt{1-\text{x}^2}$
Hence, using integration by parts, we get
$\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=(\cos^{-1}\text{x})\int\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}-\int\bigg[\bigg(\frac{\text{d}(\cos^{-1}\text{x})}{\text{dx}}\bigg)\int\bigg(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\bigg)\bigg]\text{dx}$
$=(\cos^{-1})\big(-\sqrt{1-\text{x}^2}\big)-\int\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)\big(-\sqrt{1-\text{x}^2}\big)\text{dx}$
$=-\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\text{x+C}$
Hence, $\int\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}=-\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\text{x+C}$
View full question & answer→Question 123 Marks
Evaluate the following integrals:
$\int\text{x}^3\sin\big(\text{x}^4+1\big)\text{dx}$
Answer$\int\text{x}^3.\sin\big(\text{x}^4+1\big)\text{dx}$
Let $\text{x}^4+1=\text{t}$
$\Rightarrow4\text{x}^3=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x}^3\text{ dx}=\frac{\text{dt}}{4}$
Now, $\int\text{x}^3.\sin\big(\text{x}^4+1\big)\text{dx}$
$=\frac{1}{4}\int\sin(\text{t})\text{dt}$
$=\frac{1}{4}[-\cos\text{t}]+\text{C}$
$=\frac{1}{4}[-\cos(\text{x}^4+1)]+\text{C}$
View full question & answer→Question 133 Marks
Evaluate the following integrals:
$\int\text{x}\sin2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}$
$=\text{x}\Big(-\frac{\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}$
$=-\frac{\text{x}}{2}\cos2\text{x}+\frac{1}{2}\int\cos2\text{x dx}$
$=-\frac{\text{x}}{2}\cos2\text{x}+\frac{1}{2}\frac{\sin2\text{x}}{2}+\text{C}$
$\text{I}=-\frac{\text{x}}{2}\cos2\text{x}+\frac{1}{4}\sin2\text{x}+\text{C}$
View full question & answer→Question 143 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\big(\cot\text{x}-\text{cosec}^2\text{x}\big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\big(\cot\text{x}-\text{cosec}^2\text{x}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\cot\text{x dx}-\int\text{e}^{\text{x}}\text{cosec}^2\text{x dx}$
Integration by parts
$=\text{e}^{\text{x}}\cot\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\cot\text{x}\Big)\text{dx}-\int\text{e}^{\text{x}}\text{cosec}^2\text{x dx}$
$=\text{e}^{\text{x}}\cot\text{x}+\int\text{e}^{\text{x}}\text{cosec}^2\text{x dx}-\int\text{e}^{\text{x}}\text{cosec}^{2}\text{x dx}$
$=\text{e}^{\text{x}}\cot\text{x+C}$
$\int\text{e}^{\text{x}}\big(\cot\text{x}-\text{cosec}^2\text{x}\big)\text{dx}=\text{e}^\text{x}\cot\text{x}+\text{C}$
View full question & answer→Question 153 Marks
Evaluate the following integrals:
$\int\sec\text{x}.\text{log} (\sec\text{x}+\tan\text{x})\text{dx}$
Answer$\int\sec\text{x}.\text{log} (\sec\text{x}+\tan\text{x})\text{dx}$
$\text{Let }\text{ log}(\sec\text{x}+ \tan\text{x})=\text{t}$
$\Rightarrow \frac{(\sec\text{x} \tan\text{x}+\sec^{2}\text{x})} {(\sec\text{x} +\tan\text{x})}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \frac{\sec\text{x} (\sec\text{x}+\tan\text{x})} {(\sec\text{x} +\tan\text{x})}\text{dx}=\text{dt}$
$\text{Now},\int\sec \text{x}.\text{log}(\sec\text{x}+\tan\text{x})\text{dx}$
$=\int \text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{[\text{log}(\sec\text{x}+\tan\text{x})]^2}{2}+\text{C}$
View full question & answer→Question 163 Marks
Evaluate the following integrals:
$\int\text{x}^2\text{e}^{\text{x}^3}\cos\big(\text{e}^{\text{x}^3}\big)\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\cos\big(\text{e}^{\text{x}^3}\big)\text{dx}\ ....(1)$ Let $\text{e}^{\text{x}^3}=\text{t}$ then, $\text{d}\big(\text{e}^{\text{x}^3}\big)=\text{dt}$ $\Rightarrow3\text{x}^2\text{e}^{\text{x}^3}\text{dx}=\text{dt}$ $\Rightarrow\text{x}^2\text{e}^{\text{x}^3}\text{dx}=\frac{\text{dt}}{3}$Putting $\text{e}^{\text{x}^3}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{3}$ in equation (1), we get
$\text{I}=\int\cos\text{t}\frac{\text{dt}}{3}$ $=\frac{\sin\text{t}}{3}+\text{C}$ $=\frac{\sin\big(\text{e}^{\text{x}^3}\big)}{3}+\text{C}$ $\text{I}=\frac{1}{3}\sin\big(\text{e}^{\text{x}^3}\big)+\text{C}$
View full question & answer→Question 173 Marks
Evaluate the following integrals:$\int\frac{\sec^2\text{x}}{1-\tan^2\text{x}}\text{dx}$
Answer$\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
Let $\tan\text{x = t}$
$\Rightarrow\sec^2\text{x dx = dt}$
Now, $\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
$=\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{2}\log\bigg|\frac{1+\text{t}}{1-\text{t}}\bigg|+\text{C}$
$=\frac{1}{2}\log\bigg|\frac{1+\tan\text{x}}{1-\tan\text{x}}\bigg|+\text{C}$
View full question & answer→Question 183 Marks
Evaluate the following integrals:
$\int\frac{\big\{\text{e}^{\sin^{-1}\text{x}}\big\}^2}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet $\text{e}^{\sin^{-1}\text{z}}=\text{t}$
Differentiating both sides w.r.t. x,
$\text{e}^{\sin^{-1}\text{x}}\times \frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
Now, $\int \frac{(\text{e}^{\sin-1_\text{x}})^2}{\sqrt{1-\text{x}^2}}\text{dx}$
$\int \text{e}^{\sin^{-1}\text{z}}\times \frac{\text{e}^{\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{dx}$
$\int \text{t}.\text{dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$=\frac{(\text{e}^{\sin^{-1}\text{x}})^2}{2}+\text{C}$
View full question & answer→Question 193 Marks
Write a value of $\int\frac{(\log\text{x})^{\text{n}}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{(\log\text{x})^{\text{n}}}{\text{x}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
$\text{dx}=\text{xdt}$
$\therefore\ \int\frac{(\log\text{x})^{\text{n}}}{\text{x}}\text{ dx}=\int(\text{t})^{\text{n}}\text{dt}$
$=\frac{\text{t}^{\text{n}+1}}{\text{n}+1}+\text{C}$
$=\frac{(\log\text{x})^{\text{n}+1}}{(\text{n}+1)}+\text{C}$
View full question & answer→Question 203 Marks
Evaluate the following integrals:
$\int3\sqrt{\cos^2\text{x}}\sin\text{x }\text{dx}$
AnswerLet I $=\int3\sqrt{\cos^2\text{x}}\sin\text{x }\text{dx}\ ....(1)$
Let $\cos\text{x}=\text{t}$ then,
$\text{d}(\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{dx}=-\frac{\text{dt}}{\sin\text{x}}$
Putting $\cos\text{x}=\text{t}$ and $\text{dx}=-\frac{\text{dt}}{\sin\text{x}}$ in equation (1), we get
$\text{I}=\int3\sqrt{\text{t}^2}\sin\text{x}\times\frac{-\text{dt}}{\sin\text{x}}$
$=-\int\text{t}^\frac{2}{3}\sin\text{x}\frac{\text{dt}}{\sin\text{x}}$
$=-\int\text{t}^\frac{2}{3}\text{dt}$
$=-\frac{3}{5}\times\text{t}^\frac{5}{3}+\text{C}$
$=-\frac{3}{5}(\cos\text{x})^\frac{5}{3}+\text{C}$
$\therefore\text{I}=-\frac{3}{5}(\cos\text{x})^\frac{5}{3}+\text{C}$
View full question & answer→Question 213 Marks
Evaluate the following integrals:
$\int \frac{1}{(\text{x}-1)\sqrt{2\text{x}+3}}\text{ dx}$
AnswerLet $\text{I}=\int \frac{1}{(\text{x}-1)\sqrt{2\text{x}+3}}\text{ dx}$
Let $2\text{x}+3=\text{t}^2$
$2\text{dx}=2\text{tdt}$
$\therefore\ \text{I}=\int\frac{\text{t dt}}{\big(\frac{\text{t}^2-3}{2}-1\big)\text{t}}$
$=2\int\frac{\text{dt}}{\text{t}^2-5}$
$=\frac{2}{2\sqrt{5}}\log\Big|\frac{\text{t}-\sqrt{5}}{\text{t}+\sqrt{5}}\Big|+\text{C}$
Thus, $\text{I}=\frac{1}{\sqrt{5}}\log\bigg|\frac{\sqrt{2\text{x}+3}-\sqrt{5}}{\sqrt{2\text{x}+3}+\sqrt{5}}\bigg|+\text{C}$
View full question & answer→Question 223 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)(\text{x}-2)}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(\text{x}+1)(\text{x}-2)}{\sqrt{\text{x}}}\text{dx}$
$=\int\bigg(\frac{\text{x}^2-2\text{x}+\text{x}-2}{\sqrt{\text{x}}}\bigg)\text{dx}$
$=\bigg(\frac{\text{x}^2-\text{x}-2}{\sqrt{\text{x}}}\bigg)\text{dx}$
$=\int\Big(\text{x}^\frac{3}{2}-\text{x}^\frac{1}{2}-2\text{x}^\frac{-1}{2}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]-\Bigg[\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]-2\Bigg[\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{x}^\frac{5}{2}-\frac{2}{3}\text{x}^\frac{3}{2}-4\text{x}^\frac{1}{2}+\text{C}$
View full question & answer→Question 233 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{1-\cos\text{x}}\text{dx}\text{ or }\int\frac{\cot\text{x}}{\text{cosec x}-\cot\text{x}}\text{dx}$
Answer$\int\frac{\cot\text{x}}{\text{cosec x}-\cot\text{x}}\text{dx}$
$=\int\frac{\frac{\cos\text{x}}{\sin\text{x}}}{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{1-\cos\text{x}}\Big)\times\frac{(1+\cos\text{x})}{(1+\cos\text{x})}\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{1-\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\big[(\cot\text{x}\text{ cosec x})+\cot^2\text{x}\big]\text{dx}$
$=\int\big[\cot\text{x}\text{ cosec x}+\text{cosec}^2\text{x}-1\big]\text{dx}$
$=-\text{cosec x}-\cot\text{x}-\text{x}+\text{C}$
View full question & answer→Question 243 Marks
$\int\frac{\text{x}^3}{\text{x}-2}\text{dx}$
Answer$\text{Let I}=\int\frac{\text{x}^3}{\text{x}-2}\text{dx}$
Using long division method, we have
$\frac{\text{x}^3}{\text{x}-2}=\text{x}^2+2\text{x}+4+\frac{8}{\text{x}-2}$
$\text{I}=\int\Big(\text{x}^2+2\text{x}+4+\frac{8}{\text{x}-2}\Big)\text{dx}$
$=\int\text{x}^2\text{dx}+2\int\text{xdx}+4\int\text{dx}+8\int\frac{1}{\text{x}-2}\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{2\text{x}^2}{2}+4\text{x}+8\log|\text{x}-2|+\text{c}$
$=\frac{\text{x}^3}{3}+\text{x}^2+4\text{x}+8\log|\text{x}-2|+\text{c}$
$\therefore\text{I}=\frac{\text{x}^3}{3}+\text{x}^2+4\text{x}+8\log|\text{x}-2|+\text{c}$
View full question & answer→Question 253 Marks
Evaluate the following integrals:
$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
Answer$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
$=\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\times\frac{\text{cosec x}+\cot\text{x}}{\text{cosec x}+\cot\text{x}}\times\text{dx}$
$=\int\frac{\text{cosec x}(\text{cosec x}+\cot\text{x})}{\text{cosec}^2\text{x}-\cot^2\text{x}}\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec x}\cot\text{x})\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 263 Marks
Write a value of $\int\text{x}^2\sin\text{x}^3\text{ dx}$
AnswerLet $\text{I}=\int\text{x}^2\sin\text{x}^3\text{ dx}$
Let $\text{x}^3=\text{t}$
$=3\text{x}^2\text{dx}=\text{dt}$
$=\text{x}^2\text{dx}=\frac{1}{3}\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}(-\cos\text{t})+\text{C}$
Hence, $\text{I}=\frac{-1}{3}\cos\text{x}^3+\text{C}$
View full question & answer→Question 273 Marks
$\int(\text{e}^{\text{x}}+1)^2\text{e}^{\text{x}}\text{ dx}$
AnswerConsider $\text{I}=\int(\text{e}^{\text{x}}+1)^2\text{e}^{\text{x}}\text{dx}$
Let $(\text{e}^{\text{x}}+1)=\text{t}\rightarrow\text{e}^{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int(\text{e}^\text{x}+1)^2\text{e}^{\text{x}}\text{dx}$
$=\int(\text{t})^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{c}$
$=\frac{(\text{e}^{\text{x}}+1)^3}{3}+\text{c}$
View full question & answer→Question 283 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2+4)\sqrt{\text{x}^2+1}}\text{ dx}$
Answer$\text{I}=\int\frac{1}{(\text{x}^2+4)\sqrt{\text{x}^2+1}}\text{ dx}$
Let $\text{x}^2+1=\text{u}^2$
$2\text{xdx}=2\text{udu}$
$\therefore\ \text{I}=\int\frac{\text{u}}{(\text{u}^2+3)\text{u}}\text{ du}$
$=\int\frac{1}{\text{u}^2+3}\text{ du}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{u}}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\bigg(\sqrt{\frac{\text{x}^2+1}{3}}\bigg)+\text{C}$
View full question & answer→Question 293 Marks
Evaluate $\int\frac{1}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
AnswerConsider the integral
$\text{I}=\int\frac{1}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\big(\sec^2\text{x}+\text{cosec}^2\text{x}\big)\text{dx}$
$=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
$=\frac{\sin\text{x}}{\cos\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+\text{C}$
$=\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin\text{x}\cos\text{x}}+\text{C}$
$=-\frac{2\cos2\text{x}}{\sin2\text{x}}+\text{C}$
$=-2\cot2\text{x}+\text{C}$
View full question & answer→Question 303 Marks
$\int\cos^42\text{x dx}$
Answer$\int\cos^42\text{x dx}$
$=\int(\cos^22\text{x})^2\text{dx}$
$=\int\Big(\frac{1+\cos4\text{x}}{2}\Big)^2\text{dx}$ $\Big[\therefore\cos^2\text{x}=\frac{1+\cos2\text{x}}{2}\Big]$
$=\frac{1}{4}\int(1+\cos4\text{x})^2\text{dx}$
$=\frac{1}{4}\int(1+\cos^24\text{x}+2\cos4\text{x})\text{dx}$
$=\frac{1}{4}\Big[1+\Big(\frac{1+\cos8\text{x}}{2}\Big)+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\int\Big(\frac{3}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\Big[\frac{3\text{x}}{2}+\frac{\sin8\text{x}}{16}+\frac{2\sin4\text{x}}{4}\Big]+\text{C}$
$=\frac{3\text{x}}{8}+\frac{\sin8\text{x}}{64}+\frac{\sin4\text{x}}{8}+\text{C}$
View full question & answer→Question 313 Marks
Evaluate the following integrals:$\int\frac{1}{\text{x}\sqrt{4-9(\log\text{x})^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\text{x}\sqrt{4-9(\log\text{x})^2}}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{ dx}=\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}\sqrt{4-9(\log\text{x})^2}}$
$=\int\frac{\text{dt}}{\sqrt{4-9\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{2^2-(3\text{t})^2}}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\text{t}}{2}\Big)+\text{C}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\log\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 323 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos2\text{x}+3\sin^2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos2\text{x}+3\sin^2\text{x}}\ \text{dx}$
$=\int\frac{1}{2\cos^2\text{x}-1+3\sin^2\text{x}}\ \text{dx}$
$\text{I}=\frac{1}{\sqrt{2}}\int\frac{1}{1+\text{t}^2}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{2-\sec^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{2-(1+\tan^2\text{x})^2+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{2-1-\tan^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\text{dt}}{1+2\tan^2\text{x}}$
Let $\sqrt{2}\tan\text{x}=\text{t}$
$\sqrt{2}\sec^2\text{x dx}=\text{dt}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\text{t}+\text{C}$
$\text{I}=\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan\text{x})+\text{C}$
View full question & answer→Question 333 Marks
Evaluate the following integrals:
$\int\frac{1+\cos\text{x}}{(\text{x}+\sin\text{x})^3}\text{dx}$
Answer$\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$\text{Let x}+\sin\text{x}=\text{t}$
$\Rightarrow(1+\cos\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\cos\text{x})\text{dx}={\text{dt}}$
$\text{Now,}\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^3}$
$=\int\text{t}^{-3}\text{dt}$
$=\frac{\text{t}^{-3+1}}{-3+1}+\text{C}$
$=\frac{-1}{2\text{t}^2}+\text{C}$
$=\frac{-1}{2(\text{x}+\sin\text{x})^2}+\text{C}$
View full question & answer→Question 343 Marks
Evaluate the following intregals:
$\int\frac{1}{\sin^2\text{x}-\sin2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin^2\text{x}-\sin(2\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin^2\text{x}+2\sin\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+2\text{t}}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1}$
$=\ln\frac{\text{dt}}{(\text{t}+1)^2-(-1)^2}$
$=\frac{1}{2}\ln\Big|\frac{\text{t}}{\text{t}+2}\Big|+\text{C}$
$=\frac{1}{2}\ln\Big|\frac{\tan\text{x}}{\tan\text{x}+2}\Big|+\text{C}$
View full question & answer→Question 353 Marks
Evaluate the following integrals:$\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\Big[\frac{\text{x}}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big]\text{dx}$
$=\int\bigg[\frac{\text{x}}{2\cos^{2}\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg]\text{dx}$
$=\frac{1}{2}\int\text{x}\cdot\sec^2\frac{\text{x}}{2}\text{dx}+\int\tan\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\bigg[\text{x}\cdot\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}-\int1\times2\tan\big(\frac{\text{x}}{2}\big)\text{dx}\bigg]+\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)-\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\log\frac{\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 363 Marks
Evaluate:
$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}$
Answer$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}=\int\frac{\sqrt{\text{x}}}{\text{x}}\text{dx}$
$=\int\text{x}^\frac{1}{2}\times\text{x}^{-1}\text{dx}$
$=\int\text{x}^{\frac{1}{2}-1}\text{dx}$
$=\int\frac{\text{x}^{\frac{-1}{2}+1}+\text{c}}{\frac{-1}{2}+1}$
$=\frac{\text{x}^\frac{1}{2}}{\frac{1}{2}}$
$=\sqrt{\text{x}}+\text{c}$
View full question & answer→Question 373 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{2\text{x}}}{1+\text{e}^\text{x}}\text{ dx}$
Answer$\int\frac{\text{e}^{2\text{x}}}{1+\text{e}^\text{x}}\text{ dx}$ $\Rightarrow\int\frac{\text{e}^\text{x}.\text{e}^\text{x}}{1+\text{e}^\text{x}}\text{ dx}$ then, Let $1+\text{e}^\text{x}=\text{t}$ $\Rightarrow\text{e}^\text{x}=\text{t}-1$ $\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{\text{e}^\text{x}.\text{e}^\text{x}}{1+\text{e}^\text{x}}\text{ dx}$$=\int\frac{(\text{t}-1)\text{dt}}{\text{t}}$
$=\Big(1-\frac{1}{\text{t}}\Big)\text{dt}$
$=\text{t}-\log|\text{t}|+\text{C}$
$=\big(1+\text{e}^\text{x}\big)-\log\big(1+\text{e}^\text{x}\big)+\text{C}$
Let $\text{C}+1=\text{C}^\text{n}$ $=\text{e}^\text{x}-\log\big(1+\text{e}^\text{x}\big)+\text{C}^\text{n}$
View full question & answer→Question 383 Marks
Evaluate the following integrals:
$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
Answer$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
$=\int(\text{a}^2\tan^2\text{x}+\text{b}^2\cot^2\text{x}+2\text{ab}\tan\text{x}\cot\text{x})\text{dx}$
$=\text{a}^2\int\tan^2\text{x dx}+\text{b}^2\int\cot^2\text{x dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2\int(\sec^2\text{x}-1)\text{dx}+\text{b}^2\int(\text{cosec}^2\text{x}-1)\text{dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2[\tan\text{x}-\text{x}]+\text{b}^2[-\cot\text{x}-\text{x}]+2\text{ab}\text{x}+\text{C}$
$=\text{a}^2\tan\text{x}-\text{b}^2\cot\text{x}-(\text{a}^2+\text{b}^2-2\text{ab})\text{x}+\text{C}$
View full question & answer→Question 393 Marks
Evaluate the following integrals:
$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
Answer$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
$=\int\sin^{-1}(\sin2\text{x} )\text{dx}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$
$=\int2\text{ x dx}$
$=2\int\text{x dx}$
$=\frac{2\text{x}^2}{2}+\text{C}$
$=\text{x}^2+\text{C}$
$\therefore\ \int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}=\text{x}^2+\text{C}$
View full question & answer→Question 403 Marks
Evaluate the following integrals:
$\int\frac{3\text{x}^5}{1+\text{x}^{12}}\text{dx}$
AnswerLet $\text{I}=\int\frac{3\text{x}^5}{1+\text{x}^{12}}\text{dx}$
$=\int\frac{3\text{x}^5}{1+(\text{x}^6)^2}\text{dx}$
Let $\text{x}^6=\text{t}$
$\Rightarrow6\text{x}^5\text{dx = dt}$
$\Rightarrow\text{x}^5\text{dx}=\frac{\text{dt}}{6}$
$\text{I}=\frac{3}{6}\int\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\tan^{-1}(\text{t})+\text{C}$ $\Big[\text{Since}\int\frac{1}{\text{x}^2+1}\text{dx}=\tan^{-1}\text{x+C}\Big]$
$\text{I}=\frac{1}{2}\tan^{-1}(\text{x}^6)+\text{C}$
View full question & answer→Question 413 Marks
Evaluate the following intregals:
$\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\sqrt{\frac{(1-\text{x})(1-\text{x})}{(1+\text{x})(1-\text{x})}}\text{dx}$
$=\int\Big(\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}-\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}$
Putting $1-\text{x}^2=\text{t}$
$\Rightarrow-2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{x dx}=-\frac{\text{dt}}{2}$
Then
$\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\sin^{-1}(\text{x})+\frac{1}{2}\times2\sqrt{\text{t}}+\text{C}$
$=\sin^{-1}(\text{x})+\sqrt{1-\text{x}^2}+\text{C}$
View full question & answer→Question 423 Marks
$\int\frac{1}{\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3}}\text{dx}$
Answer$\int\frac{\text{dx}}{(\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3})}$
Rationalise the denominator
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}}-3)}{(\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3})(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}\text{dx}$
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}{(2\text{x}+3)-(2\text{x}-3)}\text{dx}$
$=\frac{1}{6}\int(2\text{x}+3)^{\frac{1}{2}}\text{dx}-\frac{1}{6}\int(2\text{x}-3)^{\frac{1}{2}}\text{dx}$
$=\frac{1}{6}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]-\frac{1}{6}\Bigg[\frac{(2\text{x}-3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{1}{18}\big\{(2\text{x}+3)^{\frac{3}{2}}-(2\text{x}-3)^{\frac{3}{2}}\big\}+\text{c}$
View full question & answer→Question 433 Marks
Evaluate the following integrals:
$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
Answer$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
$=\int\frac{2\sin^2\text{x}}{2\cos^2\text{x}}\text{dx}$
$=\int\tan^2\text{x dx}$
$=\int(\sec^2\text{x}-1)\text{dx}$
$=\int\sec^2\text{x dx}-\int\text{dx}$
$=\tan\text{x}-\text{x}+\text{C}$
$\therefore\ \int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}=\tan\text{x}-\text{x}+\text{C}$
View full question & answer→Question 443 Marks
Evaluate the following integrals:
$\int\sin^5\text{x}\cos\text{x}\text{ dx}$
Answer$\int\sin^5\text{x}\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\sin^5\text{x}\cos\text{x}\text{ dx}$
$=\int\text{t}^5\text{dt}$
$=\frac{\sin^6\text{x}}{6}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}$
View full question & answer→Question 453 Marks
Evalute the following integrals:
$\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
Putting x - a = t
⇒ x = a + t
⇒ dx = dt
$\therefore\text{I}=\int\frac{\cos(\text{a}+\text{t})\text{dt}}{\cos\text{t}}$
$=\frac{\cos\text{a}\cos\text{t}}{\cos\text{t}}-\frac{\sin\text{a}\sin\text{t}}{\cos\text{t}}\text{dt}$
$=\int\big(\cos\text{a}-\sin\text{a}\tan\text{t}\big)\text{dt}$
$=\text{t}\cos\text{a}-\sin\text{a ln}|\sec\text{t}|+\text{C}$
$=(\text{x}-\text{a})\cos\text{a}-\sin\text{a ln}|\sec(\text{a}-\text{a})|+\text{C}\ \big[\text{t}=\text{x}-\text{a}\big]$
View full question & answer→Question 463 Marks
$\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x}}{(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{(1-\tan\text{x})^2}$
$-\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\bigg[\frac{\text{t}^{-2+1}}{-2+1}\bigg]+\text{c}$
$=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1-\tan\text{x}}+\text{c}$
View full question & answer→Question 473 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2+6\text{x}+13}\text{dx}$
AnswerWe have $\text{x}^2+6\text{x}+13=\text{x}^2+6\text{x}+3^2-3^2+13=(\text{x}+3)^2+4$
Sol, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{1}{(\text{x}+3)^2+2^2}\text{dx}$
Let $\text{x}+3=\text{t}$ Then $\text{dx = dt}$
Therefore, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{\text{dt}}{\text{t}^2+2^2}=\frac{1}{2}\tan^{-1}\frac{\text{t}}{2}+\text{c}$ [by 7.4 (3)]
$=\frac{1}{2}\tan^{-1}\frac{\text{x}+3}{2}+\text{C}$
View full question & answer→Question 483 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx},$ then
$\text{I}=\int\frac{\frac{1}{\cos\text{x}}}{\frac{1}{\cos2\text{x}}}\text{dx}$
$=\int\frac{\cos2\text{x}}{\cos\text{x}}\text{dx}$
$=\int\frac{2\cos^2\text{x}-1}{\cos\text{x}}\text{dx}$
$=\int2\cos\text{ x dx}-\int\frac{1}{\cos\text{x}}\text{dx}$
$=2\int\cos\text{dx}-\int\sec\text{x dx}$
$=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
$\because\text{I}=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
View full question & answer→Question 493 Marks
$\int\frac{1+\cos\text{x}}{1-\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
$=\int\bigg(\frac{2\cos^2\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$ $\big[\therefore1+\cos\text{x}=2\cos^2\frac{\text{x}}{2} \&1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\big]$
$=\int\cot^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\text{cosec}^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{-\cot\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}-\text{x+c}$
$=-2\cot\big(\frac{\text{x}}{2}\big)-\text{x+c}$
View full question & answer→Question 503 Marks
Evaluate the following integrals:$\int\text{e}^{\sqrt{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\sqrt{\text{x}}}\text{dx}$
Let $\sqrt{\text{x}}=\text{t}$
$\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\text{I}=2\int\text{e}^{\text{t}}\text{t dt}$
$\text{I}=2[\text{t}\int\text{e}^{\text{t}}\text{dt}-\int(1\int\text{e}^{\text{t}}\text{dt})\text{dt}$
$\text{I}=2[\text{te}^\text{t}-\int\text{e}^{\text{t}}\text{dt}]$
$=2[\text{te}^{\text{t}}-\text{e}^{\text{t}}]+\text{C}$
$=2\text{e}^{\text{t}}(\text{t}-1)+\text{C}$
$\text{I}=2\text{e}^{\sqrt{\text{x}}}(\sqrt{\text{x}}-1)+\text{C}$
View full question & answer→Question 513 Marks
Evaluate the following integrals:$\int\text{x}^2\text{e}^{-\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\text{e}^{-\text{x}}\text{dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\text{e}^{-\text{x}}\text{dx}-\int(2\text{x}\int\text{e}^{-\text{x}}\text{dx})$
$=-\text{x}^2\text{e}^{-\text{x}}-\int(2\text{x})(-\text{e}^{-\text{x}})$
$=-\text{x}^2\text{e}^{-\text{x}}+2\int\text{xe}^{-\text{x}}\text{dx}$
$=-\text{x}^2\text{e}^{-\text{x}}+2[\text{x}\int\text{e}^{-\text{x}}\text{dx}-\int(1\times\int\text{e}^{-\text{x}}\text{dx})\text{dx}]$
$=-\text{x}^2\text{e}^{-\text{x}}+2[\text{x}(-\text{e}^{-\text{x}})-\int(-\text{e}^{-\text{x}})\text{dx}]$
$=-\text{x}^2\text{e}^{-\text{x}}-2\text{xe}^{-\text{x}}+2\int\text{e}^{-\text{x}}\text{dx}$
$\text{I}=-\text{x}^2\text{e}^{-\text{x}}-2\text{xe}^{-\text{x}}-2\text{e}^{-\text{x}}+\text{C}$
$\text{I}=-\text{e}^{-\text{x}}(\text{x}^2+2\text{x}+2)+\text{C}$
View full question & answer→Question 523 Marks
Evaluate the following integrals:$\int\log_{10}\text{x dx}$
AnswerLet $\text{I}=\int\log_{10}\text{x dx}$
$=\int\frac{\log\text{x}}{\log10}\text{dx}$
$=\frac{1}{\log10}\int1\times\log\text{x dx}$
Using integration by parts,
$=\frac{1}{\log10}\Big[\log\text{x}\int\text{dx}-\int\Big(\frac{1}{\text{x}}\int\text{dx}\Big)\text{dx}\Big]$
$=\frac{1}{\log10}\Big[\text{x}\log\text{x}-\int\big(\frac{\text{x}}{\text{x}}\big)\text{dx}\Big]$
$=\frac{1}{\log10}[\text{x}\log\text{x}-\text{x}]$
$\text{I}=\frac{\text{x}}{\log10}(\log\text{x}-1)$
View full question & answer→Question 533 Marks
Evaluate the following integrals:$\int\text{x}^2\sin^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\sin^2\text{x dx}$
$=\int\text{x}^2\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\frac{\text{x}^2}{2}\text{dx}-\int\Big(\frac{\text{x}^2\cos2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{6}-\frac{1}{2}[\int\text{x}^2\cos2\text{x dx}]$
$=\frac{\text{x}^3}{6}-\frac{1}{2}[\text{x}^2\int\cos2\text{x dx}-\int(2\text{x}\int\cos2\text{x dx})\text{dx]}$
$=\frac{\text{x}^3}{6}-\frac{1}{2}\Big(\text{x}^2\frac{\sin2\text{x}}{2}\Big)+\frac{1}{2}\times2\int\Big(\text{x}\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}+\frac{1}{2}[\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}]$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}+\frac{1}{2}\Big[\text{x}\Big(-\frac{\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}\Big]$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{4}\frac{\sin2\text{x}}{2}+\text{C}$
$\text{I}=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{8}\sin2\text{x}+\text{C}$
View full question & answer→Question 543 Marks
$\int\frac{\text{x}+3}{(\text{x}+1)^4}\text{dx}$
Answer$\int\bigg[\frac{\text{x}+3}{(\text{x}+1)^4}\bigg]\text{dx}$
$=\int\bigg[\frac{\text{x}+1+2}{(\text{x}+1)^4}\bigg]\text{dx}$
$=\int\bigg[\frac{(\text{x}+1)}{(\text{x}+1)^4}+\frac{2}{(\text{x}+1)^4}\bigg]\text{dx}$
$=\int\frac{\text{dx}}{(\text{x}+1)^3}+2\int\frac{\text{dx}}{(\text{x}+1)^4}$
$=\int(\text{x}+1)^{-3}\text{dx}+2\int(\text{x}+1)^{-4}\text{dx}$
$=\bigg[\frac{(\text{x}+1)^{-3+1}}{-3+1}\bigg]+2\bigg[\frac{(\text{x}+1)^{-4+1}}{-4+1}\bigg]+\text{c}$
$=-\frac{1}{2}(\text{x}+1)^{-2}-\frac{2}{3}(\text{x}+1)^{-3}+\text{c}$
$=-\frac{1}{2(\text{x}+1)^2}-\frac{2}{3(\text{x}+1)^3}+\text{c}$
View full question & answer→Question 553 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{16-6\text{x}-\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{16-6\text{x}-\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x})}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x}+3^2-3^2)}}$
$=\int\frac{\text{dx}}{\sqrt{16+9-(\text{x}+3)^2}}$
$=\int\frac{\text{dx}}{\sqrt{5^2-(\text{x}+3)^2}}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{5}\Big)+\text{C}$
View full question & answer→Question 563 Marks
Evaluate the following integrals:$\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}^2+21\text{x}+4\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}(\text{x}+7)+4(\text{x}+7)}\text{ dx}$ $=\int\frac{\text{x}+7}{(3\text{x}+4)(\text{x}+7)}\text{ dx}$ $=\int\frac{1}{(3\text{x}+4)}\text{ dx}$$=\frac{1}{3}\ln|3\text{x}+4|+\text{C}$
View full question & answer→Question 573 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\cos\text{x}}\text{dx}$
$=\int\frac{1}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{1-\cos^2\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{\sin^2\text{x}}\times\text{dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{dx}+\int\frac{\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\cot\text{x}\times\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{1}{1-\cos\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 583 Marks
Evaluate the following integrals:
$\int\text{x}^2\sqrt{\text{a}^6-\text{x}^6}\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\sqrt{\text{a}^6-\text{x}^6}\text{dx}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\sqrt{\text{a}^6-\text{t}^2}\text{dt}$
$=\frac{1}{3}\begin{Bmatrix}\frac{\text{t}}{2}\sqrt{\text{a}^6-\text{t}^2}+\frac{\text{a}^6}{2}\sin^{-1}\Big(\frac{\text{t}}{\text{a}^3}\Big)\end{Bmatrix}+\text{C}$
$\therefore\ \text{I}=\frac{\text{x}^3}{6}\sqrt{\text{a}^6-\text{x}^6}+\frac{\text{a}^6}{6}\sin^{-1}\Big(\frac{\text{x}^3}{\text{a}^3}\Big)+\text{C}$
View full question & answer→Question 593 Marks
$\int\frac{1}{(7\text{x}-5)^2}+\frac{1}{\sqrt{5\text{x}-4}}\text{dx}$
AnswerLet $\text{I}=\int\bigg[\frac{1}{(7\text{x}-5)^3}+\frac{1}{\sqrt{5\text{x}-4}}\bigg]\text{dx}.$ Then,
$\text{I}=\int(7\text{x}-5)^{-3}\text{dx}+\int(5\text{x}-4)^{\frac{-1}{2}}\text{dx}$
$=\frac{(7\text{x}-5)^{-2}}{7\times(-2)}+\frac{(5\text{x}-4)^{\frac{1}{2}}}{5\times\frac{1}{2}}+\text{c}$
$=-\frac{(7\text{x}-5)^{6-2}}{14}+\frac{2}{5}\sqrt{(5\text{x}-4)}+\text{c}$
$\text{I}=\frac{-1}{14}(7\text{x}-5)^{-2}+\frac{2}{5}\times\sqrt{5\text{x}-4}+\text{c}.$
View full question & answer→Question 603 Marks
Evaluate the following integrals:
$\int\Big(\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}$
Here, $\text{f(x)}=\tan^{-1}\text{x}$ and $\text{f}'\text{(x)}=\frac{1}{1+\text{x}^2}$
and we know thet,
$\int\text{e}^{\text{ax}}(\text{af}(\text{x})+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)+C}$
$\therefore\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
Thus,
$\text{I}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
View full question & answer→Question 613 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\tan^{-1}\text{x}}.(1+\text{x}^2)}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$\text{Let }\tan^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan^{-1}\text{x}}+\text{C}$
View full question & answer→Question 623 Marks
Evaluate the following integrals:
$\int\cot^3\text{x }\text{cosec}^2\text{x}\text{ dx}$
Answer$\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}$
$\text{Let},\cot\text{x}=\text{t}$
$\Rightarrow-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
$\text{Now},\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}$
$=\int\text{t}^3(-\text{dt})$
$=\frac{-\text{t}^4}{4}+\text{C}$
$=\frac{-\cot^4\text{x}}{4}+\text{C}$
View full question & answer→Question 633 Marks
Evaluate the following integrals:
$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}=\int\sqrt{4-(\text{x}-1)^2}\text{dx}$ Let X - 1 = t, so that dx = dt Thus,$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}=\int\sqrt{4-\text{t}^2}\text{dt}$ $=\frac{1}{2}\text{t}\sqrt{4-\text{t}^2}+\frac{4}{2}\sin^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$$=\frac{1}{2}(\text{x}-1)\sqrt{3+2\text{x}-\text{x}^2}+2\sin^{-1}\Big(\frac{\text{x}-1}{2}\Big)+\text{C}$
View full question & answer→Question 643 Marks
Evaluate the following integrals:
$\int2\text{x}\sec^3\big(\text{x}^2+3\big)\tan\big(\text{x}^2+3\big)\text{dx}$
View full question & answer→Question 653 Marks
Evaluate the following integrals:
$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
Answer$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$\text{Let},\text{x}-\cos\text{x}=\text{t}$
$\Rightarrow(1+\sin\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\sin\text{x})\text{dx}=\text{dt}$
$\text{Now,}\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\text{x}-\cos\text{x}}+\text{C}$
View full question & answer→Question 663 Marks
Evaluate the following integrals:
$\int\frac{\log\text{x}^2}{\text{x}}\text{dx}$
Answer$\int\frac{\log\text{x}^2\text{dx}}{\text{x}}$
$=\int\frac{2\log\text{x}}{\text{x}}\text{dx}$
$=2\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$\text{Let }\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{Now, }2\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$=2\int\text{t dt}$
$2\Big[\frac{\text{t}^2}{2}\Big]+\text{C}$
$=\text{t}^2+\text{C}$
$=(\log\text{x})^2+\text{C}$
View full question & answer→Question 673 Marks
Evaluate the following integrals:
$\int\sec^42\text{x}\text{ dx}$
Answer$\int\sec^42\text{x}\text{ dx}$
$=\int\sec^22\text{x}.\sec^22\text{x}\text{ dx}$
$=\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
Let $\tan2\text{x}=\text{t}$
$\sec^22\text{x}.2\text{dx}=\text{dt}$
$\sec^22\text{x}.\text{dx}=\frac{\text{dt}}{2}$
Now, $\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
$=\frac{1}{2}\int(1+\text{t}^2)\text{dt}$
$=\frac{1}{2}\Big[\text{t}+\frac{\text{t}^3}{3}\Big]+\text{C}$
$=\frac{\text{t}}{2}+\frac{\text{t}^3}{6}+\text{C}$
$=\frac{\tan(2\text{x})}{2}+\frac{\tan^3(2\text{x})}{6}+\text{C}$
View full question & answer→Question 683 Marks
Write a value of $\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
$\because\ \int\text{e}^{\text{x}}\big(\text{f(x})+\text{f}'(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
Here, $\text{f(x)}=\sin\text{x}$ and $\text{f}'(\text{x})=\cos\text{x}$
$\therefore\ \text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}=\text{e}^{\text{x}}\sin\text{x}+\text{C}$
View full question & answer→Question 693 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\big[\cot\text{x}-\text{cosec}^2\text{x}\big]\text{dx}$
Here, $\text{f(x)}=\cot\text{x}$
$\Rightarrow\text{f}'\text{(x)}=-\text{cosec}^2\text{x}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\cot\text{x}=\text{t}$
Diff both sides w.r.t x
$\text{e}^{\text{x}}(\cot\text{x}-\text{cosec}^2\text{x})\text{dx = dt}$
$\therefore\text{I}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\cot\text{x + C}$
View full question & answer→Question 703 Marks
$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
Answer$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
$=\int(\text{e}^{2\text{x}}+\frac{1}{\text{e}^{2\text{x}}}2\text{e}^\text{x}\times\frac{1}{\text{e}^{\text{x}}})\text{dx}$
$=\int(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}+2)\text{dx}$
$=\frac{\text{e}^{2\text{x}}}{2}+\frac{\text{e}^{-2\text{x}}}{-2}+2\text{x}+\text{c}$
$=\frac{\text{e}^{2\text{x}}}{2}-\frac{\text{e}^{-2\text{x}}}{2}+2\text{x}+\text{c}$
View full question & answer→Question 713 Marks
Evalute the following integrals:
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}$
Answer$\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}=\frac{2\cos\text{x}-3\sin\text{x}}{2(3\cos\text{x}+2\sin\text{x})}$
Let $3\cos\text{x}+2\sin\text{x}=\text{t}$
$(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}=\int\frac{\text{dt}}{2\text{t}}$
$=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\log|\text{t}|+\text{C}$
$=\frac{1}{2}\log|2\sin\text{x}+3\cos\text{x}|=\text{C}$
View full question & answer→Question 723 Marks
Evaluate the following integrals:$\int\frac{\text{x}}{\sqrt{4-\text{x}^4}}\text{ dx}$
Answer$\int\frac{\text{x}\text{ dx}}{\sqrt{4-\text{x}^4}}$
$\Rightarrow\int\frac{\text{x}\text{ dx}}{\sqrt{2^2-(\text{x}^2)^2}}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x}\text{ dx}=\text{dt}$
$\text{x}\text{ dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x}\text{ dx}}{\sqrt{2^2-(\text{x}^2)^2}}$
$\frac{1}{2}\int\frac{\text{dt}}{\sqrt{2^2-\text{t}^2}}$
$=\frac{1}{2}\times\sin^{-1}\Big(\frac{1}{2}\Big)+\text{C}$
$=\frac{1}{2}\times\sin^{-1}\Big(\frac{\text{x}^2}{2}\Big)+\text{C}$
View full question & answer→Question 733 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
Let $\text{x}+1=\frac{1}{\text{t}}$
$\text{dt}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\frac{1}{\text{t}}\sqrt{\Big(\frac{1}{\text{t}^2}+\frac{1}{\text{t}}-1\Big)}}$
$=-\int\frac{\text{dt}}{\sqrt{1+\text{t}-\text{t}^2}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\frac{1}{4}-\text{t}+\text{t}^2\big)}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\text{t}-\frac{1}{2}\big)^2}}$
$=-\sin^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{5}}{2}}\Bigg)+\text{C}$
$\therefore\ \text{I}=-\sin^{-1}\Big(\frac{2\text{t}-1}{\sqrt{5}}\Big)+\text{C}$ $\Big[\text{When}\text{t}=\frac{1}{\text{x}+1}\Big]$
View full question & answer→Question 743 Marks
Evaluate the following integrals:
$\int\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\text{dx}$
Answer$\int\bigg(\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\bigg(\frac{5\cos^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}+\frac{6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\Big(\frac{5}{2}\frac{\cos\text{x}}{\sin^2\text{x}}+3\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\frac{5}{2}\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}\Big)\text{dx}+3\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}$
$=\frac{5}{2}\int(\text{cosec x}\cot\text{x})\text{dx}+3\int\sec\text{x}\tan\text{x dx}$
$=\frac{5}{2}(-\text{cosec x})+3\sec\text{x}+\text{C}$
$=-\frac{5}{2}\text{cosec x}+3\sec\text{x}+\text{C}$
View full question & answer→Question 753 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{16+\text{t}^2}\text{dt}$
$=\int\sqrt{4^2+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{16+\text{t}^2}+\frac{16}{2}\log\big|\text{t}+\sqrt{16+\text{t}^2}\big|+\text{C}$
$\therefore\ \text{I}=\frac{\log\text{x}}{2}\sqrt{16+(\log\text{x})^2}\\+8\log\Big|\log\text{x}+\sqrt{16+(\log\text{x})^2}\Big|+\text{C}$
View full question & answer→Question 763 Marks
Evaluate the following integrals:
$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
Answer$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^4+7\text{x}^3+5\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^3+7\text{x}^2+5\text{x}^2+7\text{x}}{\text{x}+1}\text{dx}$
$=\int\frac{5\text{x}^2(\text{x}+1)+7\text{x}(\text{x}+1)}{\text{x}+1}\text{dx}$
$=\int(5\text{x}^2+7\text{x})\text{dx}$
$=\frac{5\text{x}^3}{3}+\frac{7\text{x}^2}{2}+\text{C}$
View full question & answer→Question 773 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\sin^2\frac{\pi}{6}}\text{dx}$
$\big[\because \sin(\text{A}+\text{B})\sin(\text{A}-\text{B})=\sin^2\text{A}-\sin^2\text{B}\big]$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\frac{1}{4}}\text{dx}$
Putting $\sin^2\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow2\sin\text{x}\cos\text{ x dx}=\text{dt}$
$\Rightarrow\sin2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}\big|\sin^2\text{x}-\frac{1}{4}\big|+\text{C}\ \Big[\because\text{t}=\sin^2\text{x}-\frac{1}{4}\Big]$
View full question & answer→Question 783 Marks
Write a value of $\int\frac{1}{1+2\text{e}^{\text{x}}}\text{dx}$
AnswerLet $\int\frac{1}{1+2\text{e}^{\text{x}}}\text{dx}$
Dividing and multiplying by $e^x$
$\text{I}=\int\frac{\frac{1}{\text{e}^{\text{x}}}\text{dx}}{\frac{1}{\text{e}^{\text{x}}}+2}$
$=\int\frac{\text{e}^{-\text{x}}\text{dx}}{\text{e}^{-\text{x}}+2}$
Let $\text{e}^{-\text{x}}+2=\text{t}$
$-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|\text{e}^{-\text{x}}+2|+\text{C}$ $(\because\text{t}=\text{e}^{-\text{x}}+2)$
View full question & answer→Question 793 Marks
Evaluate the following integrals:
$\int\text{x}^3\cos\text{x}^4\text{dx}$
Answer$\text{I}=\int\text{x}^3.\cos\big(\text{x}^4\big)\text{dx}$
Let $\text{x}^4=\text{t}$ then,
$\Rightarrow4\text{x}^3\text{ dx}=\text{dt}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
$\Rightarrow\text{x}^3\text{dx}= \frac{\text{dt}}{\text{4} }$
Now, $\int\text{x}^3.\cos\big(\text{x}^4\big)\text{dx}$
$=\frac{1}{4}\int\cos(\text{t})\text{dt}$
$=\frac{1}{4}[\sin(\text{t})]+\text{C}$
$=\frac{1}{4}\big[\sin\text{x}^4\big]+\text{C}$
View full question & answer→Question 803 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}\text{dx}$
Answer$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx = dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Now, $\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}^2-(\text{a}^3)^2}$
$=\frac{1}{3}\times\frac{1}{2\text{a}^3}\log\bigg|\frac{\text{t}-\text{a}^3}{\text{t}+\text{a}^3}\Big|+\text{C}$
$=\frac{1}{6\text{a}^3}\log\Big|\frac{\text{x}^3-\text{a}^3}{\text{x}^3+\text{a}^3}\Big|+\text{C}$
View full question & answer→Question 813 Marks
Evaluate the following intregals:
$\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
We express
$\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+3}$
$\Rightarrow3\text{x}-2=\text{A}(\text{x}+1)(\text{x}+3)+\text{B}(\text{x}+3)+\text{C}(\text{x}+1)^2$
Equating the coefficient of $x^2, x$ and constants, we get
$0 = A + C$ and $3 = 4A + B + 2C$ and $-2 = 3A + 3B + C$
$\text{or }\text{A}=\frac{11}{4}\text{ and }\text{B}=-\frac{5}{2}\text{ and }\text{C}=-\frac{11}{4}$
$\therefore\text{I}=\int\bigg(\frac{\frac{11}{4}}{\text{x}+1}+\frac{-\frac{5}{2}}{(\text{x}+1)^2}+\frac{-\frac{11}{4}}{\text{x}+3}\bigg)\ \text{dx}$
$=\frac{11}{4}\int\frac{1}{\text{x}+1}\text{ dx }-\frac{5}{2}\int\frac{1}{(\text{x}+1)^2}\text{ dx }-\frac{11}{4}\int\frac{1}{\text{x}+3}\text{ dx}$
$=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
Hence,
$\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
View full question & answer→Question 823 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
AnswerWe have
$\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Putting $\text{x}-1=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\big(\frac{1}{\text{t}}\big)\sqrt{\big(1+\frac{1}{\text{t}}}\big)^2+1}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{1+\frac{1}{\text{t}^2}+\frac{2}{\text{t}}+1}}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{\frac{\text{t}^2+1+2\text{t}+\text{t}^2}{\text{t}}}}$
$=\int\frac{-\text{dt}}{\sqrt{2\text{t}^2+2\text{t}+1}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\big(\text{t}+\frac{1}{2}\big)^2+\big(\frac{1}{2}\big)^2}$
$=-\frac{1}{\sqrt{2}}\log\begin{vmatrix}\text{t}+\frac{1}{2}+\sqrt{\Big(\text{t}+\frac{1}{2}\Big)^2+\frac{1}{4}}\end{vmatrix}+\text{C}$ where $\text{t}=\frac{1}{\text{x}-1}$
View full question & answer→Question 833 Marks
Evaluate the following integrals:
$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
Answer$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
Let $\text{e}^\text{x}-1=\text{t}^2$
$\Rightarrow\text{e}^\text{x}=\text{t}^2+1$
$\text{e}^\text{x}=2\text{t}\frac{\text{dt}}{\text{dx}}$
$\text{dx}=\frac{2\text{t dt}}{\text{e}^\text{x}}$
$\text{dx}=\frac{2\text{t dt}}{\text{t}^2+1}$
Now, $\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
$=\int\frac{\text{t. 2t dt}}{\text{t}^2+1}$
$=2\int\frac{\text{t}^2\text{ dt}}{\text{t}^2+1}$
$=2\int\Big(\frac{\text{t}^2+1-1}{\text{t}^2+1}\Big)\text{ dt}$
$=2\int\text{dt}-2\int\frac{\text{dt}}{\text{t}^2+1}$
$=2\text{t}-2\tan^{-1}(\text{t})+\text{C}$
$=2\sqrt{\text{e}^\text{x}-1}-2\tan^{-1}\big(\sqrt{\text{e}^\text{x}-1}\big)+\text{C}$
View full question & answer→Question 843 Marks
Write a value of $\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
Let $\log\text{x}^{\text{n}}=\text{t}$
$\frac{\text{nx}^{\text{n}-1}}{\text{x}^{\text{n}}}\text{ dx}=\text{dt}$
$\frac{\text{n}}{\text{x}}=\text{ dt}$
$\therefore\ \text{I}=\text{n}\int\text{t}\text{ dt}$
$=\text{n}\Big(\frac{\text{t}^2}{2}\Big)+\text{C}$
Putting the value of t
$\text{I}=\frac{\text{n}(\log\text{x}^{\text{n}})^2}{2}+\text{C}$
View full question & answer→Question 853 Marks
Evaluate the following integrals:
$\int\frac{2\text{x}^4+7\text{x}^3+6\text{x}^2}{\text{x}^2+2\text{x}}\text{dx}$
Answer$\int\bigg(\frac{2\text{x}^4+7\text{x}^3+6\text{x}^2}{\text{x}^2+2\text{x}}\bigg)\text{dx}$
$=\int\frac{\text{x}^2(2\text{x}^2+7\text{x}+6}{\text{x}(\text{x}+2)}$
$=\int\frac{\text{x}\big[2\text{x}^2+4\text{x}+3\text{x}+6\big]}{(\text{x}+2)}\text{dx}$
$=\int\frac{\text{x}(2\text{x}(\text{x}+2)+3(\text{x}+2))}{(\text{x}+2)}\text{dx}$
$=\int\frac{\text{x}(2\text{x}+3)(\text{x}+2)}{(\text{x}+2)}\text{dx}$
$=\int(2\text{x}^2+3\text{x})\text{dx}$
$=2\int\text{x}^2\text{dx}+3\int\text{x }\text{dx}$
$=2\Big[\frac{\text{x}^3}{3}\Big]+3\Big[\frac{\text{x}^2}{2}\Big]+\text{C}$
$=\frac{2}{3}\text{x}^3+\frac{3}{2}\text{x}^2+\text{C}$
View full question & answer→Question 863 Marks
$\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\text{dx}.$ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\times\frac{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{\sqrt{\text{x}+1}-\sqrt{\text{x}}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+1}}-\sqrt{\text{x}}}{(\sqrt{\text{x}+1})^2-(\sqrt{\text{x}})^2}\times\text{dx}$
$=\int\frac{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{\text{x}+1-\text{x}}\times\text{dx}$
$=\int(\sqrt{\text{x}+1}-\sqrt{\text{x}})\times\text{dx}$
$=\int(\text{x}+1)^{\frac{1}{2}}\text{dx}-\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\frac{2}{3}(\text{x}+1)^{\frac{3}{2}}-\frac{2}{3}\text{x}^{\frac{3}{2}}+\text{c}$
$\therefore\text{I}=\frac{2}{3}(\text{x}+1)^{\frac{3}{2}}-\frac{2}{3}\text{x}^{\frac{3}{2}}+\text{c}$.
View full question & answer→Question 873 Marks
Evalute the following integrals:
$\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
Dividing numerator and denomimator by $e^x$
$\Rightarrow\text{I}=\int\frac{\text{ae}^{-\text{x}}}{\text{be}^{-\text{x}}+\text{c}}\text{dx}$
Putting $e^{-x} = t$
$\Rightarrow-\text{e}^{-\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{a}}{\text{bt}+\text{c}}\text{dt}$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{bt}+\text{c}|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{be}^{-\text{x}}+\text{c}|+\text{C}\ \big[\because\text{t}=\text{e}^{-\text{x}}+\text{C}\big]$
View full question & answer→Question 883 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}$
AnswerLet I $=\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}\ .....(1)$
Let $\log\text{x}=\text{t}$ then,
$\text{d}(\log\text{x})=\text{dt}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
Putting $\log\text{x}=\text{t}$ and $\frac{1}{\text{x}}\text{dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$=\frac{(\log\text{x})^3}{3}+\text{C}$
$\therefore\text{I}=\frac{1}{3}(\log\text{x})^3+\text{C}$
View full question & answer→Question 893 Marks
Evaluate the following integrals:
$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
Answer$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
$\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
Let $\sec\text{x}=\text{t}$
$\sec\text{x}\tan\text{x}\text{ dx}=\text{dt}$
Now, $\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
$=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\sec^6\text{x}}{6}+\text{C}$
View full question & answer→Question 903 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
Answer$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\sqrt{\text{x}}}+\frac{1}{\sqrt{\text{x}}\text{x}}\Big)\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{-3}{2}}\text{dx}$
$=2\text{x}^{\frac{1}{2}}-2\text{x}^{\frac{-1}{2}}+\text{C}$
$=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
$\therefore\ \int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
View full question & answer→Question 913 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$ Let $\text{xe}^\text{x}=\text{t}$ $\Rightarrow(1.\text{e}^\text{x}+\text{xe}^\text{x})=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2({\text{xe}^\text{x}})}=\text{dx}$$=\int\frac{\text{dt}}{\sin^2\text{t}}$
$=\int\text{cosec}^2\text{t}\text{ dt}$
$=-\cot(\text{t})+\text{C}$
$=-\cot(\text{xe}^\text{x})+\text{C}$
View full question & answer→Question 923 Marks
Evaluate the following integrals:
$\int\text{x e}^{\text{x}^2}=\text{dx}$
AnswerLet $\text{I}=\int\text{x e}^{\text{x}^2}=\text{dx}\ ....(1)$ Let $\text{x}^2=\text{t}$ then, $\text{d}(\text{x}^2)=\text{dt}$ $\Rightarrow2\text{x dx}=\text{dt}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Putting $\text{x}^2=\text{t}$ and $\text{x dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\text{e}^\text{t}+\text{C}$
$=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
$\text{I}=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
View full question & answer→Question 933 Marks
Evaluate the following integrals:
$\int\frac{(1+\text{x})^3}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(1+\text{x})^3}{\sqrt{\text{x}}}\text{dx}$
$=\int\frac{1}{\sqrt{\text{x}}}\text{dx}+\int\frac{\text{x}^3}{\sqrt{\text{x}}}\text{dx}+\int\frac{3\text{x}^2}{\sqrt{\text{x}}}\text{dx}+\int\frac{3\text{x}}{\sqrt{\text{x}}}\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}\text{dx}+\int\text{x}^{\frac{5}{2}}\text{dx}+3\int\text{x}^{\frac{3}{2}}\text{dx}+3\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+\frac{3\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{3\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+\frac{\text{3x}^{\frac{5}{2}}}{\frac{5}{2}}+\frac{\text{3x}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=2\text{x}^{\frac{1}{2}}+\frac{2}{7}\text{x}^{\frac{7}{2}}+\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{6}{3}\text{x}^{\frac{3}{2}}+\text{C}$
$=2\text{x}^{\frac{1}{2}}+\frac{2}{7}\text{x}^{\frac{7}{2}}+\frac{6}{5}\text{x}^{\frac{5}{2}}+2\text{x}^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 943 Marks
Write a value of $\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
We know that,
$\int\text{e}^{\text{x}}\int\text{f}(\text{x})+\text{f}'(\text{x})=\text{e}^{\text{x}}\text{f}(\text{x})+\text{C}$
Hence, $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Then, $\int\text{e}^{\text{ax}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
$\therefore\ \text{I}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
View full question & answer→Question 953 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^6+1}{\text{x}^2+1}\text{dx}$
Answer$\int\Big(\frac{\text{x}^6+1}{\text{x}^2+1}\Big)\text{dx}$
$=\int\bigg[\frac{(\text{x}^2)^3+1^3}{\text{x}^2+1}\bigg]\text{dx}$ $[\text{A}^3+\text{B}^3=(\text{A+B})(\text{A}^2-\text{AB}+\text{B}^2)]$
$=\int\frac{(\text{x}^2+1)(\text{x}^4-\text{x}^2+1)}{(\text{x}^2+1)}\text{dx}$
$=\int(\text{x}^4-\text{x}^2+1)\text{dx}$
$=\int\text{x}^4\text{dx}+\int\text{x}^2\text{dx}+\int1\text{dx}$
$=\frac{\text{x}^{4+1}}{4+1}-\frac{\text{x}^{2+1}}{2+1}+\text{x}+\text{C}$
$=\frac{\text{x}^5}{5}-\frac{\text{x}^3}{3}+\text{x}+\text{C}$
View full question & answer→Question 963 Marks
Integrate the following integrals:
$\int\sin4\text{x}\cos7\text{x dx}$
Answer$\int\sin4\text{x}\cos7\text{x dx}$
$=\frac{1}{2}\int2\cos7\text{x}\sin4\text{x dx}$
$=\frac{1}{2}\int\big[\sin(7\text{x}+4\text{x})-\sin(7\text{x}-4\text{x})\big]\text{dx}$ $[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\big(\sin(11\text{x})-\sin(3\text{x})\big)\text{dx}$
$=\frac{1}{2}\Big[-\frac{\cos(11\text{x})}{11}+\frac{\cos(3\text{x})}{3}\Big]+\text{c}$
$=-\frac{\cos(11\text{x})}{22}+\frac{\cos(3\text{x})}{6}$
View full question & answer→Question 973 Marks
Write a value of $\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
$\text{dx}=\text{xdt}$
$\therefore\ \text{I}=\int\frac{1}{\text{t}^{\text{n}}}\text{ dt}$
$=\frac{\text{t}^{-\text{n}+1}}{-\text{n}+1}+\text{C}$
$\text{I}=\frac{(\log\text{x})^{1-\text{n}}}{1-\text{n}}+\text{C}$
View full question & answer→Question 983 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{2\text{x}-\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2-1+1}}$
$=\int\frac{\text{dx}}{\sqrt{1-(\text{x}^2-2\text{x}+1)}}$
$=\int\frac{\text{dx}}{1-(\text{x}-1)^2}$
$=\sin(\text{x}-1)+\text{C}$ $\Big[\because\ \int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
View full question & answer→Question 993 Marks
Evaluate the following integrals:
$\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
Answer$\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$ Let $\text{x}+\log\text{x}=\text{t}$ $\Rightarrow\Big(1+\frac{1}{\text{x}}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{dx}=\text{dt}$Now, $\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{(\text{x}+\log\text{x})^3}{3}+\text{C}$
View full question & answer→Question 1003 Marks
Evaluate the following integrals:$\int\text{x}^2\cos\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\cos\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\cos\text{x dx}-\int(2\text{x}\int\cos\text{x dx})\text{dx}$
$=\text{x}^2\sin\text{x}-2\int\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2[\text{x}\int\sin\text{x dx}-\int(1\int\sin\text{x dx})\text{dx}]$
$=\text{x}^2\sin\text{x}-2[\text{x}(-\cos\text{x})-\int(-\cos\text{x})\text{dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int(\cos\text{x})\text{dx}$
$\text{I}=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x}+\text{C}$
View full question & answer→Question 1013 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$
Answer$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$$\int\frac{1}{\sqrt{2\sin^2\text{x}}}\text{dx}\ \big[\because 1-\cos 2\text{x}=2\sin^2\text{x}\big]$
$=\frac{1}{\sqrt{2}}\int\text{cosec x dx}$
$=\frac{1}{\sqrt{2}}\text{ln}|\text{cosec x}-\cot\text{x}|=\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big|+\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\bigg|\frac{2\sin^2\frac{\text{x}}{2}}{\sin\text{x}}\bigg|+\text{C} \Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Bigg|\frac{2\sin^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\Bigg|+\text{C}\ \Big[\because\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 1023 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{5\text{x}^2-2\text{x}}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{5\text{x}^2-2\text{x}}}$
$=\int\frac{\text{dx}}{\sqrt{5\big(\text{x}^2-\frac{2}{5}\text{x}\big)}}$
$=\frac{1}{\sqrt5}\int\frac{\text{dx}}{\sqrt{\text{x}^2-\frac{2}{5}\text{x}+\big(\frac{1}{5}\big)^2-\big(\frac{1}{5}\big)^2}}$
$=\frac{1}{\sqrt5}\int\frac{\text{dx}}{\big(\text{x}-\frac{1}{5}\big)^2-\big(\frac{1}{5}\big)^2}$
$=\frac{1}{\sqrt5}\log\bigg|\text{x}-\frac{1}{5}+\sqrt{\big(\text{x}-\frac{1}{5}\big)^2+\big(\frac{1}{5}\big)^2}\bigg|+\text{C}$
$=\frac{1}{\sqrt5}\log\Big|\frac{5\text{x}-1}{5}+\frac{\sqrt{5\text{x}^2-2\text{x}}}{\sqrt5}\Big|+\text{C}$
View full question & answer→Question 1033 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\sin\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\sin\text{x}}\text{dx}$$=\int\frac{(1+\sin\text{x})}{(1-\sin\text{x})\times(1+\sin\text{x})}\text{dx}$
$=\Big(\frac{1+\sin\text{x}}{1-\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1+\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\Big)\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$
View full question & answer→Question 1043 Marks
Write a value of $\int\frac{\sin\text{x}}{\cos^3\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}}{\cos^3\text{x}}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\sin\text{x dx}=-\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}^3}$
$=-\int\text{t}^{-3}\text{dt}$
$=-\Big[\frac{\text{t}^{-3+1}}{-3+1}\Big]+\text{C}$
$=\frac{1}{2\text{t}^2}+\text{C}$
$=\frac{1}{2\cos^2\text{x}}+\text{C}$ $(\because\text{t}=\cos\text{x})$
$=\frac{1}{2}\sec^2\text{x}+\text{C}$
View full question & answer→Question 1053 Marks
Evaluate the following integrals:
$\int\sin^4\text{x}\cos^3\text{x}\text{ dx}$
Answer$\int\sin^4\text{x}\cos^3\text{x}\text{ dx}$
$=\int\sin^4\text{x}\cdot\cos^2\text{x }\cos\text{x}\text{ dx}$
$=\int\sin^4\text{x}\big(1-\sin^2\text{x}\big)\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\sin^4\text{x}\big(1-\sin^2\text{x}\big)\cos\text{x}\text{ dx}$
$=\int\text{t}^4(1-\text{t}^2)\text{dt}$
$=\int(\text{t}^4+\text{t}^6)=\text{dt}$
$=\frac{\text{t}^5}{5}-\frac{\text{t}^7}{7}+\text{C}$
$=\frac{\sin^5\text{x}}{5}-\frac{\sin^7\text{x}}{7}+\text{C}$
View full question & answer→Question 1063 Marks
Evaluate the following integrals:$\int\frac{1}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-4}}\text{ dx}$
View full question & answer→Question 1073 Marks
Evalute the following integrals:
$\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}$
AnswerLet $\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}\ .....\text{(i)}$
Let $\tan\text{x}+2=\text{t}$ then,
$\text{d}(\tan\text{x}+2)=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{1}{\sec^2\text{x}}\text{dt}$
Putting $\tan\text{x}+2=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$ In equation (i), we get,
$\text{I}=\int\frac{\sec^2\text{x}}{\text{t}}\times\frac{1}{\sec^2\text{x}}\text{dt}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\tan\text{x}+2|+\text{C}$
$\Rightarrow\text{I}=\log|\tan\text{x}+2|+\text{C}$
View full question & answer→Question 1083 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
Answer$\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
$\Rightarrow\int\frac{\text{x}^2\text{dx}}{(\text{x}^3)^2+(\text{a}^3)^2}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^3\text{dx = dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Now $\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}^2+(a^3)^2}$
$=\frac{1}{3\text{a}^3}\tan^{-1}\Big(\frac{\text{t}}{\text{a}^3}\Big)+\text{C}$
$=\frac{1}{3\text{a}^3}\tan^{-1}\Big(\frac{\text{x}^3}{\text{a}^3}\Big)+\text{C}$
View full question & answer→Question 1093 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$$=\int\text{e}^{\text{x}}\tan\text{x dx}-\int\text{e}^{\text{x}}\log\cos\text{x dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\Big\{\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\log\cos\text{x}\Big)\text{dx}\Big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\big\{\text{e}^{\text{x}}\log\cos\text{x}+\int\text{e}^{\text{x}}\tan\text{x dx}\big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\tan\text{x dx}+\text{C}$
$=-\text{e}^{\text{x}}\log\cos\text{x}+\text{C}$
$=\text{e}^{\text{x}}\log\sec\text{x}+\text{C}$ $\big[\because\log\sec\text{x}=-\log\cos\text{x}\big]$
View full question & answer→Question 1103 Marks
Evaluate the following integrals:$\int\frac{\log\text{x}}{\text{x}^{\text{n}}}\text{dx}$
Answer$\int\frac{\log\text{x}}{\text{x}^{\text{n}}}\text{dx}=\int(\log\text{x})\Big(\frac{1}{\text{x}^{n}}\Big)\text{dx}$
by integration by parts
$\int(\log\text{x})\Big(\frac{1}{\text{x}^{\text{n}}}\Big)\text{dx}=\log\text{x}\int\Big(\frac{1}{\text{x}^{\text{n}}}\Big)\text{dx}-\int\Big(\frac{\text{d}(\log\text{x})}{\text{dx}}\Big)\Big(\int\Big(\frac{1}{\text{x}^{n}}\Big)\text{dx}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)\text{dx}=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\int\Big(\frac{\text{x}^{-\text{n}}}{1-\text{n}}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\Big(\frac{1}{1-\text{n}}\Big)\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\bigg(\frac{\text{x}^{1-\text{n}}}{[1-\text{n}]^2}\bigg)+\text{C}$
View full question & answer→Question 1113 Marks
Evaluate the following integrals:
$\int\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\text{dx}$
Answer$\int\Big(\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\Big)\text{x}$
$=\int\frac{\cos(2\text{x})}{\sqrt{2\cos^2(2\text{x})}}\text{dx}$ $\Big[\therefore1+\cos\text{A}=2\cos^2\Big(\frac{\text{A}}{2}\Big)\text{ and }\cos^2\text{A}-\sin^2\text{A}=\cos\text{2A}\Big]$
$=\frac{1}{\sqrt2}\int\Big(\frac{\cos2\text{x}}{\cos\text{2x}}\Big)\text{dx}$
$=\frac{1}{\sqrt{2}}[\text{x}]+\text{C}$
$=\frac{\text{x}}{\sqrt{2}}+\text{C}$
View full question & answer→Question 1123 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
Here, $\text{f(x)}=\log\sin\text{x}$ Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'\text{(x)}=\cot\text{x}$
let $\text{e}^{\text{x}}\log\sin\text{x = t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\times\frac{1}{\sin\text{x}}\times\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big[\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\cot\text{x}\big]\text{dx = dt}$
$\Rightarrow\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx = dt}$
$\therefore \int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx} =\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\log\sin\text{x}+\text{C}$
View full question & answer→Question 1133 Marks
Evaluate the following integrals:
$\int\tan^32\text{x}\sec2\text{x dx}$
Answer$\int\tan^32\text{x}\sec2\text{x}=\tan^22\text{x}\tan2\text{x}\sec2\text{x}$
$=\big(\sec^22\text{x}-1\big)\tan2\text{x}\sec2\text{x}$
$=\sec^22\text{x}\tan2\text{x}\sec2\text{x}-\tan2\text{x}\sec2\text{x}$
$\therefore\ \int\tan^32\text{x}\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x}\tan2\text{x}\sec2\text{x dx}-\int\tan2\text{x}\sec2\text{x dx}$
$=\int\sec^22\text{x}\tan2\text{x}\sec2\text{x dx}-\frac{\sec2\text{x}}{2}+\text{C}$
Let $2\text{x}=\text{t}$
$\therefore\ 2\sec2\text{x}\tan2\text{x dx}=\text{dt}$
$\therefore\ \int\tan^32\text{x}\sec2\text{x dx}=\frac{1}{2}\int\text{t}^2\text{dt}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{\text{t}^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{(\sec2\text{x})^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
View full question & answer→Question 1143 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
Answer$\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-\frac{2}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-{2}{\text{x}^{-\frac{1}{2}}}\Big)\text{dx}$
$=\frac{\text{x}^{\frac{7}{2}+1}}{\frac{7}{2}+1}-2\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\text{x}^{\frac{1}{2}}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 1153 Marks
Evaluate the following integrals:
$\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
AnswerLet $\text{I}=\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Let $\text{x}+2=\text{t}^2$
$\therefore\ \text{I}=\int\frac{2\text{tdt}}{(\text{t}^2-3)\text{t}}$
$=2\int\frac{\text{dt}}{\text{t}^2-3}$
$=\frac{2}{\sqrt{3}}\log\Big|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\Big|+\text{C}$
Thus, $\text{I}=\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}-2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
View full question & answer→Question 1163 Marks
Evaluate the following integrals:
$\int\text{x}^2\tan^{-1}\text{x dx} $
AnswerLet $\text{I}=\int\text{x}^2\tan^{-1}\text{x dx}$
$=\tan^{-1}\text{x}\int\text{x}^2\text{dx}-\int\Big(\frac{1}{1+\text{x}^2}\int\text{x}^2\text{dx}\Big)$
$=\tan^{-1}\text{x}\Big(\frac{\text{x}^3}{3}\Big)-\frac{1}{3}\frac{\text{x}^3}{1+\text{x}^2}\text{dx}$
$=\frac{1}3{\text{x}^3}\tan^{-1}\text{x}-\frac{1}{3}\int\Big(\text{x}-\frac{\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{3}\times\frac{\text{x}^2}{2}+\frac{1}3{}\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\text{I}=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{6}\text{x}^2+\frac{1}{6}\log\big|1+\text{x}^2\big|+\text{C}$
View full question & answer→Question 1173 Marks
Evaluate the following integrals:$\int\frac{\sin8\text{x}}{\sqrt{9+\sin^44\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin8\text{x}}{\sqrt{9+(\sin4\text{x})^4}}\text{ dx}$
Let $\sin^24\text{x}=\text{t}$
$\Rightarrow2\sin4\text{x}.\cos4\text{x}(4)\text{dx}=\text{dt}$
$\Rightarrow4\sin8\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin8\text{x}\text{ dx}=\frac{\text{dt}}{4}$
$\text{I}=\frac{1}{4}\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\text{I}=\frac{1}{4}\log\Big|\text{t}+\sqrt{(3)^2+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\frac{1}{4}\log\Big|\sin^24\text{x}+\sqrt{9+\sin^44\text{x}}\Big|+\text{C}$
View full question & answer→Question 1183 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
Putting $\log(\log\text{x})=\text{t}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\log\text{x})|+\text{C}\ \big[\because\text{t}=\log(\log\text{x})\big]$
View full question & answer→Question 1193 Marks
Evaluate the following integrals:
$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$\text{Let a}+\text{b}\cos2\text{x}=\text{t}$
$\Rightarrow-\text{b}\sin(2\text{x})\text{dx}\times2=\text{dt}$
$\Rightarrow\sin(2\text{x})\text{dx}=\frac{-\text{dt}}{2\text{b}}$
$\text{Now,}\int\frac{\sin(2\text{x})}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$=-\frac{1}{2\text{b}}\int\frac{\text{dt}}{\text{t}^2}$
$=\frac{-1}{2\text{b}}\int\text{t}^{-2}\text{dt}$
$=\frac{-1}{2\text{b}}\Big[\frac{\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$
$=\frac{1}{2\text{b}}\times\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{2\text{b}(\text{a}+\text{b}\cos2\text{x})}+\text{C}$
View full question & answer→Question 1203 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\int\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})\text{dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}\tan\text{x})-\int\text{e}^{\text{dx}}\Big\{\frac{\text{d}}{\text{dx}}\log(\sec\text{x}+\tan\text{x})\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})-\int\text{e}^{\text{x}}\sec\text{x dx}$
$=\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})+\text{C}$
View full question & answer→Question 1213 Marks
Evaluate the following integrals:
$\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}\ ....(1)$ Let $\tan^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\tan^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow\frac{1\times2\text{x}}{1+(\text{x}^2)^2}\text{ dx}=\text{dt}$ $\Rightarrow\frac{1\times\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ Putting, $\tan^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{t}\frac{\text{dx}}{2}$
$=\frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{\text{t}^2}{4}+\text{C}$
$=\frac{(\tan^{-1}\text{x}^2)^2}{4}+\text{C}$
$\text{I}=\frac{1}{4}\big(\tan^{-1}\text{x}^2\big)^2+\text{C}$
View full question & answer→Question 1223 Marks
$\int\frac{2\text{x}+3}{(\text{x}-1)^2}\text{dx}$
Answer$\int\bigg(\frac{2\text{x}+3}{(\text{x}-1)^2}\bigg)\text{dx}$
$=\int\bigg[\frac{2\text{x}-2+2+3}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg[\frac{2(\text{x}-1)+5}{(\text{x}-1)^2}\bigg]\text{dx}$
$=2\int\frac{\text{dx}}{(\text{x}-1)}+5\int(\text{x}-1)^{-2}\text{dx}$
$=2\text{ln}|\text{x}-1|+5\bigg[\frac{(\text{x}-1)^{-2+1}}{-2+1}\bigg]+\text{C}$
$=2\text{ln}|\text{x}-1|-\frac{5}{\text{x}-1}+\text{c}$
View full question & answer→Question 1233 Marks
Evaluate the following integrals:
$\int\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\cos^2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\cos^2\text{x dx}-\int(1\int\cos^2\text{x dx})\text{dx}$
$=\text{x}\int\Big(\frac{\cos2\text{x}+1}{2}\Big)\text{dx}-\int\bigg(\int\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\bigg)\text{dx}$
$=\frac{\text{x}}{2}\Big[\frac{\sin2\text{x}}{2}+\text{x}\Big]-\frac{1}{2}\int\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{2}-\frac{1}{2}\times\frac{\text{x}^2}{2}-\frac{1}{4}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{4}+\frac{1}{8}\cos2\text{x}+\text{C}$
View full question & answer→Question 1243 Marks
Evaluate the following integrals:
$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
$=\int\frac{1+\text{x}+2\sqrt{\text{x}}}{\text{x}^{\frac{1}{2}}}\text{dx}$
$\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{1}{2}}\text{dx}+2\int\text{dx}$
$=2\sqrt{\text{x}}+\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}+\text{C}$
$\therefore\ \int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}+\text{C}$
View full question & answer→Question 1253 Marks
Evaluate the following integrals:$\int(\text{x}+1)\log\text{x dx}$
Answer$\int(\text{x}+1).\log\text{x dx}$
$=\log \text{x}\int(\text{x}+1)\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int(\text{x}+1)\text{dx}\Big\}\text{dx}$
$=\log\text{x}\Big[\frac{\text{x}^2}{2}+\text{x}\Big]-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\int\big(\frac{\text{x}}{2}+1\big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\Big(\frac{\text{x}^2}{4}+\text{x}\Big)+\text{C}$
View full question & answer→Question 1263 Marks
Evaluate the following integrals:
$\int\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet first function be $\sin^{-1}\text{x}$ and second dunction be $\frac{\text{x}}{\sqrt{1-\text{x}^2}}.$
First we find the intergral of the second function, i.e, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}.$
Put $\text{t}=1-\text{x}^{2}.$ Then $\text{dt}=-2\text{x dx}$
Therefore, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}=-\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}=-\sqrt{\text{t}}=-\sqrt{1-\text{x}^2}$
Hence, $\int\frac{\text{x}\sin^{-1}}{\sqrt{1-\text{x}^2}}\text{dx}=(\sin^{-1}\text{x})\Big(-\sqrt{1-\text{x}^2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}\Big(-\sqrt{1-\text{x}^2}\Big)\text{dx}$
$=-\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\text{x + C}=\text{x}-\sqrt{1-\text{x}^2}\sin^{-1}\text{x+C}$
View full question & answer→Question 1273 Marks
Evaluate the following integrals:
$\int \frac{\text{1}}{\sqrt{\text{x}} + \text{x}} \text{ dx}$
Answer$\int \frac{\text{dx}}{\sqrt{\text{x}} + \text{x}}\text{dx}$
$=\int \frac{\text{dx}}{\sqrt{\text{x}}\big(1 + \sqrt{\text{x}}\big)}$
Let $1 + \sqrt{\text{x}} = \text{t}$
$\Rightarrow \frac{1}{2\sqrt{\text{x}}} = \frac{\text{dt}}{\text{dx}}$
$\Rightarrow \frac{\text{dx}}{\sqrt{\text{x}}} = 2\text{dt}$
Now, $\int \frac{\text{dx}}{\sqrt{\text{x}}\big(1 + \sqrt{\text{x}}\big)}$
$= \int \frac{2\text{dt}}{\text{t}}$
$= 2\int\frac{\text{dt}}{\text{t}}$
$= 2\log|\text{t}| + \text{C}$
$= 2\log \big(1 + \sqrt{\text{x}}\big) + \text{C}$
View full question & answer→Question 1283 Marks
Write a value of $\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$ Let $\tan\text{x}=\text{t}$ $\sec^2\text{x dx}=\text{dt}$ $\therefore\ \text{I}=\int\text{t}^3\text{ dt}$ $=\frac{\text{t}^4}{4}+\text{C}$$=\frac{\tan^{4}\text{x}}{4}+\text{C}$ $(\because\text{t}=\tan\text{x})$
View full question & answer→Question 1293 Marks
Evaluate the following integrals:
$\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)(1+\text{xe}^\text{x}-\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{\int(\text{x}+1)(1+\text{xe}^\text{x})}{\text(1+\text{xe}^\text{x})}\ \text{dx}-\int\frac{(\text{x}+1)(\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)}{\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\int\frac{(\text{x}+1)\text{e}^\text{x}}{\text{xe}^\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\log|\text{xe}^\text{x}|-\log|1+\text{xe}^\text{x}|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{xe}^\text{x}}{1+\text{xe}^\text{x}}\Big|+\text{C}$
View full question & answer→Question 1303 Marks
Evaluate the following integrals:
$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
Answer$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$ Let $\text{x}+\tan^{-1}\text{x}=\text{t}$ $\Big(1+\frac{1}{1+\text{x}^2}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}^2-1+1}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$ $\Rightarrow\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$Now, $\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
$=\int5^\text{t}\text{dt}$
$=\frac{5^\text{t}}{\log5}+\text{C}$
$=\frac{5^{\text{x}+\tan^{-1}\text{x}}}{\log5}+\text{C}$
View full question & answer→Question 1313 Marks
Write a value of $\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\int\text{dx}$
$\therefore\ \text{I}=\text{x}+\text{C}$
View full question & answer→Question 1323 Marks
Write a value of $\int\log_\text{e}\text{x}\text{ dx}$
Answer$\int\log_\text{e}\text{x}\text{ dx}$
$=\int1\cdot\log_\text{e}\text{x dx}$
$=\log_\text{e}\text{x}\int1\text{ dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})\int1\text{dx}\Big\}\text{dx}$
$=\log_\text{e}\text{x}\int1\cdot\text{dx}-\int\frac{1}{\text{x}}\cdot\text{x dx}$
$=\log_\text{e}\text{x}\cdot\text{x}-\int\text{dx}$
$=\text{x}\log_\text{e}\text{x}-\text{x}+\text{C}$
$=\text{x}(\log_\text{e}\text{x}-1)+\text{C}$
View full question & answer→Question 1333 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin\text{x}\cos\text{x}+2\cos^2\text{x}}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\ \text{dx}$
Let $2+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$\text{I}=\log|2+\tan\text{x}|+\text{C}$
View full question & answer→Question 1343 Marks
Evaluate the following integrals:
$\int\cos^7\text{x}\text{ dx}$
Answer$\int\cos^7\text{x}\text{ dx}$
$=\int\cos^6\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(\cos^2\text{x})^3\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^3\cdot\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^3\cdot\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^3\text{dt}$
$=\int\big(1-\text{t}^6-3\text{t}^2+3\text{t}^4)\text{dt}$
$=\Big[\text{t}-\frac{\text{t}^7}{\text{7}}-\frac{3\text{t}^3}{3}+\frac{3\text{t}^5}{5}\Big]+\text{C}$
$=\sin\text{x}-\frac{1}{7}\sin^7\text{x}-\sin^3\text{x}+\frac{3}{5}\sin^5\text{x}+\text{C}$
View full question & answer→Question 1353 Marks
Evaluate the following integrals:
$\int (3\text{x}\sqrt{\text{x}}+4\sqrt{\text{x}}+5)\text{dx}$
Answer$\int(3\text{x}\sqrt{5}+4\sqrt{\text{x}}+5)\text{dx}$
$=\int3\text{x}\sqrt{5}\text{dx}+\int4\sqrt{\text{x}}\text{dx}+\int5\text{dx}$
$=\int3\text{x}^{\frac{3}{2}}\text{dx}+4\int\text{x}^{\frac{1}{2}}\text{dx}+5\int\text{dx}$
$=\frac{\text{x}\frac{3}{2}+1}{\frac{3}{2}+1}+\frac{4\text{x}^{\frac{1}{2}}}{\frac{1}{2}+1}+5\text{x}+\text{C}$
$=\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{8}{3}\text{x}^{\frac{3}{2}}+5\text{x}+\text{C}$
View full question & answer→Question 1363 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$
Let $\text{x}^2+9=\text{u}^2$
$2\text{xdx}=2\text{udu}$
$\therefore\ \text{I}=\int\frac{\text{u}}{(\text{u}^2-5)\text{u}}\text{ du}$
$=\int\frac{\text{du}}{\text{u}^2-5}$
$=\frac{1}{2\sqrt{5}}\log\Big(\frac{\text{u}-\sqrt{5}}{\text{u}+\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\log\bigg(\frac{\sqrt{\text{x}^2+9}-\sqrt{5}}{\sqrt{\text{x}^2+9}+\sqrt{5}}\bigg)+\text{C}$
View full question & answer→Question 1373 Marks
Evaluate the following integrals:$\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\frac{1}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\bigg(\frac{1}{2\cos\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\text{e}^{\text{}x}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}$
Putting $\text{e}^{\text{x}}\tan\frac{\text{x}}{2}=\text{t}$
Diff. both sides w.r.t.x
$\text{e}^{\text{x}}.\tan\big(\frac{\text{x}}{2}\big)+\text{e}^{\text{x}}\times\frac{1}{2}\sec^{2}\frac{\text{x}}{2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\big[\tan\frac{\text{x}}{2}+\frac{1}{2}\sec^2\big(\frac{\text{x}}{2}\big)\big]\text{dx}=\text{dt}$
$\therefore\int\text{e}^{\text{x}}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 1383 Marks
Integrate the following integrals:
$\int\cos3\text{x}\cos4\text{x dx}$
AnswerLet I $=\int\cos3\text{x}\cos4\text{x dx}.$ Then,
$\text{I}=\frac{1}{2}\int(2\cos3\text{x}\cos4\text{x})\text{dx}$
$=\frac{1}{2}\int(\cos7\text{x}+\cos(-\text{x}))\text{dx}$
$=\frac{1}{2}\int\cos7\text{x}+\frac{1}{2}\int\cos\text{dx}$ $[\because\cos(-0)=\cos0]$
$=\frac{\sin7\text{x}}{2\times7}+\frac{\sin\text{x}}{2}+\text{C}$
$=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\sin\text{x}+\text{C}$
$\therefore\text{I}=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\times\sin\text{x}+\text{C}$
View full question & answer→Question 1393 Marks
Evaluate the following integrals:
$\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
Answer$\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
Let $5^\text{x}=\text{t}$
$\Rightarrow5^\text{x}\log5=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow5^\text{x}\text{dx}=\frac{\text{dt}}{\log5}$
Now, $\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
$=5\int5^{5^{\text{t}}}.5^\text{t}.\frac{\text{dt}}{\log5}$
Again let $5^\text{t}=\text{p}$
$\Rightarrow5^\text{t}\log5=\frac{\text{dp}}{\text{dt}}$
$\Rightarrow5^\text{t}\text{dt}=\frac{\text{dp}}{\log5}$
Again $\int5^{5^\text{t}}.5^\text{x}.\frac{\text{dt}}{\log5}$
$=\int5^\text{p}.\frac{\text{dp}}{(\log5)^2}$
$=\frac{5^\text{p}}{(\log5)^3}+\text{C}$
$=\frac{5^{5^{5^{\text{x}}}}}{(\log5)^3}+\text{C}$
View full question & answer→Question 1403 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}^3+8)(\text{x}-1)}{\text{x}^2-2\text{x}+4}\text{dx}$
Answer$\int\frac{(\text{x}^3+8)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}^3+2^3)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}+2)(\text{x}^2-2\text{x}+4)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$ $\big[\therefore\ \text{a}^3+\text{b}^3=(\text{a + b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\int(\text{x}+2)(\text{x}-1)\text{dx}$
$=\int(\text{x}^2-\text{x}+2\text{x}-2)\text{dx}$
$=(\text{x}^2+\text{x}-2)\text{dx}$
$=\int\text{x}^2\text{dx}+\int\text{x dx}-2\int1\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{2}-2\text{x}+\text{C}$
View full question & answer→Question 1413 Marks
Evaluate the following integrals:
$\int\sqrt{16\text{x}^2+25}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{16\text{x}^2+25}\text{dx}$
$=\int\sqrt{(4\text{x})^2+5^2}\text{dx}$
$=4\int\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\text{dx}$
$=4\begin{Bmatrix}\frac{\text{x}}{2}\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}+\frac{\big(\frac{5}{4}\big)^2}{2}\log\Bigg|\text{x}+\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\Bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=2\text{x}\sqrt{\text{x}^2+\frac{25}{16}}+\frac{25}{8}\log\bigg|\text{x}+\sqrt{\text{x}^2+\frac{25}{16}}\bigg|+\text{C}$
View full question & answer→Question 1423 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
AnswerWe have,
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
$\int\frac{1}{\sqrt{2\cos^2\frac{\text{x}}{2}}}\text{dx}$
$=\int\frac{1}{\sqrt{2}\cos\frac{\text{x}}{2}}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\sec\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\text{cosec}\Big(\frac{\pi}{2}+\frac{\text{x}}{2}\Big)\text{dx}$
$=\frac{2}{\sqrt{2}}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
$\because\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}=\sqrt{2}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
View full question & answer→Question 1433 Marks
Evaluate the following integrals:
$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ Let $\sqrt{\text{x}}=\text{t}$ $\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$ Now, $\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$$=2\int\sec^2\text{t dt}$
$=2\tan(\text{t})+\text{C}$
$=2\tan\big(\sqrt{\text{x}}\big)+\text{C}$
View full question & answer→Question 1443 Marks
Write a value of $\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
Let $3+\text{a}^{\text{x}}=\text{t}$
$\text{a}^{\text{x}}\log\text{a dx}=\text{dt}$
$\text{a}^{\text{x}}\text{dx}=\frac{\text{dt}}{\log\text{a}}$
$\text{I}=\int\frac{\text{dt}}{\log\text{a}\cdot\text{t}}=\frac{1}{\log\text{a}}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\log\text{a}}\log(3+\text{a}^{\text{x}})+\text{C}$
View full question & answer→Question 1453 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$$=\int\frac{\cos^2\text{x}-\sin^2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
Putting $\cos\text{x}+\sin\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}+\cos\text{x}=\frac{\text{dt}}{\text{dt}}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\cos\text{x}+\sin\text{x}|+\text{C}\ \big[\because\text{t}=\cos\text{x}+\sin\text{x}\big]$
View full question & answer→Question 1463 Marks
Evaluate the following integrals:
$\int\big\{\sqrt{\text{x}}\big(\text{ax}^2+\text{bx}+\text{c}\big)\big\}\text{dx}$
Answer$\int\sqrt{\text{x}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\text{x}^{\frac{1}{2}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\Big(\text{ax}^{2+\frac{1}{2}}+\text{bx}^{\frac{1}{2}+1}+\text{cx}^{\frac{1}{2}}\Big)\text{dx}$
$=\text{a}\int\text{x}^{\frac{5}{2}}\text{dx}+\text{b}\int\text{x}^{\frac{3}{2}}\text{dx}+\text{c}\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\text{a}\begin{bmatrix}\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}\end{bmatrix}+\text{b}\begin{bmatrix}\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\end{bmatrix}+\text{c}\begin{bmatrix}\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}+\text{C}$
$=\frac{2\text{a}}{7}\text{x}^{\frac{7}{2}}+\frac{2\text{b}}{5}\text{x}^{\frac{3}{2}}+\frac{2\text{c}}{3}\text{x}^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 1473 Marks
Evaluate the following integrals:
$\int\tan^5\text{x }\sec^4\text{x}\text{dx}$
Answer$\int\tan^5\text{x }\sec^4\text{x}\text{dx}$
$=\int\tan^5\text{x }\sec^2\text{x}.\sec^2\text{x}\text{dx}$
$=\int\tan^5\text{x}(1+\tan^2\text{x})\sec^2\text{xdx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{xdx}=\text{dt}$
Now, $\int\tan^5\text{x}(1+\tan^2\text{x})\sec^2\text{xdx}$
$=\int\text{t}^5(1+\text{t}^2)\text{dt}$
$=\int(\text{t}^5+\text{t}^7)\text{dt}$
$=\frac{\text{t}^6}{6}+\frac{\text{t}^8}{8}+\text{C}$
$=\frac{\tan^6\text{x}}{6}+\frac{\tan^8\text{x}}{8}+\text{C}$
View full question & answer→Question 1483 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}^2-2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\text{x}^2-2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\int\sqrt{\text{x}^2-2\text{x}+1-1}\text{dx}$
$\Rightarrow\text{I}=\int\sqrt{(\text{x}-1)^2-1^2}\text{dx}$
$\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\ln\big(|\text{x}+\sqrt{\text{x}^2-\text{a}^2}|\big)+\text{C}$
$\therefore\ \text{I}=\frac{(\text{x}-1)}{2}\sqrt{(\text{x}-1)^2-1}-\frac{1}{2}\ln\big|(\text{x}-1)+\sqrt{\text{x}^2-2\text{x}}\big|+\text{C}$
View full question & answer→Question 1493 Marks
Evaluate the following integrals:
$\int\text{e}^\text{x}\sqrt{\text{e}^{2\text{x}}+1}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^\text{x}\sqrt{\text{e}^{2\text{x}}+1}\text{dx}$
Putting $\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\ln\Big|\text{t}+\sqrt{\text{t}^2+1}\Big|+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2+\text{a}^2}+\frac{1}{2}\ln\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C}\Big]$
$=\frac{\text{e}^\text{x}}{2}\sqrt{\text{e}^{2\text{x}}+1}+\frac{1}{2}\ln\Big|\text{e}^\text{x}+\sqrt{\text{e}^{2\text{x}}+1}\Big|+\text{C}\ \big(\because\ \text{t}=\text{e}^\text{x}\big)$
View full question & answer→Question 1503 Marks
Evaluate the following integrals:
$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
Answer$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$\text{Let, }\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$=\int\text{t}^\frac{3}{2}\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=\frac{2}{5}\tan^\frac{5}{2}\text{x}+\text{C}$
View full question & answer→Question 1513 Marks
Evaluate the following integrals:$\int\frac{\cos2\text{x}}{\sqrt{\sin^22\text{x}+8}}\text{ dx}$
Answer$\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
Let $\sin(2\text{x})=\text{t}$
$\Rightarrow\cos(2\text{x})\times2.\text{dx}=\text{dt}$
$\Rightarrow\cos(2\text{x}).\text{dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\big(2\sqrt2\big)^2}}$
$=\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2+8}\Big|+\text{C}$
$=\frac{1}{2}\log\Big|\sin(2\text{x})+\sqrt{\sin^2(2\text{x})+8}\Big|+\text{C}$
View full question & answer→Question 1523 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}$
Answer$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}=\frac{\cos\text{x}-\sin\text{x}}{(\sin^2\text{x}+\cos^2\text{x})+2\sin\text{x}\cos\text{x}}$
$[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\sin\text{x}+\cos\text{x}=\text{t}$
$\therefore(\sin\text{x}+\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^2} $
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
View full question & answer→Question 1533 Marks
$\int\frac{1}{1-\sin\frac{\text{x}}{2}}\text{dx}$
Answer$\int\frac{\text{dx}}{1-\sin\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\big(1+\sin\frac{\text{x}}{2}\big)}{\big(1-\sin\frac{\text{x}}{2}\big)\big(1+\sin\frac{\text{x}}{2}\big)}\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{1-\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\big(\sec^2\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\tan\frac{\text{x}}{2}\big)\text{dx}$
$=\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\frac{\sec\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\text{c}$
$=2\big(\tan\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\big)+\text{c}$
View full question & answer→Question 1543 Marks
Evalute the following integrals:
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
AnswerNote: Here we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
Putting $\log\sin\text{x}=\text{t}$
$\Rightarrow\cot\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log\sin\text{x}|+\text{C}\ \big[\because\text{t}=\log\sin\text{x}\big]$
View full question & answer→Question 1553 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{8+3\text{x}-\text{x}^2}}\text{ dx}$
Answer$8 + 3x - x^2$ can be written as $8-\Big(\text{x}^2-3\text{x}+\frac{9}{4}-\frac{9}{4}\Big).$Therefore,
$8-\Big(\text{x}^2-3\text{x}+\frac{9}{4}-\frac{9}{4}\Big)$
$=\frac{41}{4}-\Big(\text{x}-\frac{3}{2}\Big)^2$
$=\int\frac{1}{8+3\text{x}-\text{x}^2}\text{ dx}$
$=\int\frac{1}{\sqrt{\frac{41}{4}-\big(\text{x}-\frac{3}{2}\big)^2}}\text{ dx}$
Let $\text{x}-\frac{3}{2}=\text{t}$
$\therefore\ \text{dx}=\text{dt}$
$=\int\frac{1}{\sqrt{\frac{41}{4}-\big(\text{x}-\frac{3}{2}\big)^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Big(\frac{\sqrt{41}}{2}\Big)^2-\text{t}^2}}\text{ dt}$
$=\sin^{-1}\bigg(\frac{\text{t}}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}$
$=\sin^{-1}\bigg(\frac{\text{x}-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}$
$=\sin^{-1}\Big(\frac{2\text{x}-3}{\sqrt{41}}\Big)+\text{C}$
View full question & answer→Question 1563 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
$=\int\sqrt{\text{x}^2+\text{x}+\frac{1}{4}+\frac{3}{4}}\text{dx}$
$=\int\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{dx}$
$=\frac{\big(\text{x}+\frac{1}{2}\big)}{2}\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}+\frac{\big(\frac{\sqrt{3}}{2}\big)^2}{2}\times\\\log\Bigg|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\Bigg|+\text{C}$
$=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|\Big(\frac{2\text{x}+1}{2}\Big)+\frac{1}{2}\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$
$\therefore\ \text{I}=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|2\text{x}+1+\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$
View full question & answer→Question 1573 Marks
Evalute the following integrals:
$\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
AnswerLet $\text{I}=\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
Putting $\sin2\text{x}+\tan\text{x}-5=\text{t}$
$\Rightarrow2\cos2\text{x}+\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(2\cos2\text{x}+\sec^2\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\sin2\text{x}+\tan\text{x}-5|+\text{C} $ $\big[\because\text{t}=\sin2\text{x}+\tan\text{x}-5\big]$
View full question & answer→Question 1583 Marks
Evaluate the following integrals:
$\int\sqrt{9-{\text{x}^2}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{3^2-{\text{x}^2}}$
We know that,
$\int\sqrt{\text{a}^2-\text{x}^2}=\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$\therefore\ \text{I}=\frac{\text{x}}{2}\sqrt{9-\text{x}^2}+\frac{9}{2}\sin^{-1}\frac{\text{x}}{3}+\text{C}$
View full question & answer→Question 1593 Marks
Evaluate the following integrals:
$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
Answer$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\int\frac{\text{dx}}{\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}}}$
$=\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
Let $\text{e}^{\text{x}}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\text{ dx = dt}$
Now, $\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
$=\int\frac{\text{dt}}{1+\text{t}^2}$
$=\tan^{-1}(\text{t})+\text{C}$
$=\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
View full question & answer→Question 1603 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
$=\int\sqrt{\text{a}^2+2\text{ax}-\text{x}^2-\text{a}^2}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}^2-2\text{ax}+\text{a}^2)}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}-\text{a})^2}\text{dx}$
$=\Big(\frac{\text{x}-\text{a}}{2}\Big)\sqrt{2\text{ax}-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 1613 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^{\frac{3}{2}}}\text{dx}$
Answer$\int\frac{1}{\text{x}^\frac{3}{2}}\text{dx}=\int\text{x}\frac{-3}{2}\text{dx}$
$=\int\text{x}^\frac{-3}{2}\text{dx}$
$=\frac{\text{x}^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+\text{c}$
$=\frac{\text{x}^\frac{-1}{2}}{\frac{-1}{2}}+\text{c}$
$=-2\text{x}\frac{1}{\sqrt{\text{x}}}+\text{c}$
$=\frac{-2}{\sqrt{\text{x}}}+\text{c}$
View full question & answer→Question 1623 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{(1-\text{x}^2)\big\{9+\big(\sin^{-1}\text{x}\big)^2\big\}}}\text{ dx}.$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(1-\text{x}^2)\Big[9+\big(\sin^{-1}\text{x}\big)^2\Big]}}\text{ dx}$
Let $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\Rightarrow\text{I}=\log\Big|\text{t}+\sqrt{9+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\log\Big|\sin^{-1}\text{x}+\sqrt{9+\big(\sin^{-1}\text{x}\big)^2}\Big|+\text{C}$
View full question & answer→Question 1633 Marks
If $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C},$ then write the value of f(x).
Answer$\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\int\Big(\frac{\text{x}}{\text{x}^2}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
Consider, $\text{f(x)}=\frac{1}{\text{x}},$ then $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big|\text{f}(\text{x})+\text{f}'(\text{x})\big|$
Therefore, $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Hence, $\text{f(x)}=\frac{1}{\text{x}}$
View full question & answer→Question 1643 Marks
$\int\frac{2\text{x}-1}{(\text{x}-1)^2}\text{dx}$
Answer$\int\bigg[\frac{2\text{x}-1}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg[\frac{2\text{x}-2+2-1}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg(\frac{2(\text{x}-1)}{(\text{x}-1)^2}+\frac{1}{(\text{x}-1)^2}\bigg)\text{dx}$
$=2\int\frac{\text{dx}}{\text{x}-1}+\int(\text{x}-1)^{-2}\text{dx}$
$=2\text{ ln}|\text{x}-1|+\frac{(\text{x}-1)^{-2+1}}{-2+1}+\text{C}$
$=2\text{ ln}|\text{x}-1|-\frac{1}{\text{x}-1}+\text{C}$
View full question & answer→Question 1653 Marks
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Answer$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Let $\text{t}=\cos^2\text{x}\rightarrow-\text{dt}=2\cos\text{x}\sin\text{x}\text{ dx}$
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
$=\int\frac{-1}{\sqrt{\text{t}^2-(1-\text{t})+2}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+1}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2+\frac{3}{4}}}\text{ dt}$
$=-\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\text{t}^2+\text{t}+1}\Big|$
$=-\log\Big|\Big(\cos^2\text{x}+\frac{1}{2}\Big)+\sqrt{\cos^4\text{x}+\cos^2\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 1663 Marks
Write a value of $\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Let $\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}=\text{t}$
$\big(2\text{a}^2\sin\text{x}\cos\text{x}-2\text{b}^2\cos\text{x}\sin\text{x}\big)\text{dx}=\text{dt}$
$2(\text{a}^2-\text{b}^2)\sin\text{x}\cos\text{x dx}=\text{dt}$
$(\text{a}^2-\text{b}^2)\sin2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{a}^2-\text{b}^2}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\log\big(\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}\big)+\text{C}$
View full question & answer→Question 1673 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$
AnswerHere, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.Let $\text{I}=\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{dx}}{\text{x}}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{3+\text{t}}$
$=\log|3+\text{t}|+\text{C}$
$=\log|3+\log\text{x}|+\text{C}\ \big[\because\text{t}=\log\text{x}\big]$
View full question & answer→Question 1683 Marks
Evaluate the following integrals:
$\int\tan\text{x }\sec^4\text{x}\text{dx}$
AnswerLet $\text{I}=\int\tan\text{x }\sec^4\text{x}\text{dx}$ Then
$\text{I}=\int\tan\text{x }\sec^2\text{x}\sec^2\text{x}\text{dx}$
$=\int\tan\text{x}(1+\tan^2\text{x})\sec^2\text{x}\text{dx}$
$\text{I}=\int\big(\tan\text{x}+\tan^3\text{x}\big)\sec^2\text{x}\text{dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{xdx}=\text{dt},$ we get
$\text{I}=\int(\text{t}+\text{t}^3)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4}{4}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\times\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$
View full question & answer→Question 1693 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}+1)(\text{x}^2+2\text{x}+2)}\text{ dx}$
Answer$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$ Let $\text{I}=\int\frac{1}{(\text{x}+1)(\text{x}^2+2\text{x}+2)}\text{ dx}$ $=\int\frac{1}{(\text{x}+1)\big((\text{x}+1)^2+\text{1}\big)}\text{ dx}$Let $\text{x}+1=\tan\text{u}$
$\Rightarrow\text{dx}=\sec^2\text{u du}$ $\therefore\ \text{I}=\int\frac{\sec^2\text{u}}{\tan\text{u}(\tan^2\text{u}+1)}\text{ du}$ $=\int\frac{\cos\text{u}}{\sin\text{u}}\text{ du}$ $=\log|\sin\text{u}|+\text{C}$ $=\log\Big|\frac{\tan\text{u}}{\sec^2\text{u}}\Big|+\text{C}$ $=\log\bigg|\frac{\text{x}+1}{\sqrt{\text{x}^2+2\text{x}+2}}\bigg|+\text{C}$
View full question & answer→Question 1703 Marks
Evaluate the following integrals:
$\int\cos^5\text{x}\text{ dx}$
Answer$\int\cos^5\text{x}\text{ dx}$
$=\int\cos^4\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^2\text{ dt}$
$=\int(1+\text{t}^4-2\text{t}^2)\text{dt}$
$=\int\text{dt}+\int\text{t}^4\text{ dt}-2\int\text{t}^2\text{ dt}$
$=\text{t}+\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}+\text{C}$
View full question & answer→Question 1713 Marks
Evaluate the following integrals:
$\int\text{x}^3\sin\text{x}^4\text{dx}$
Answer$\int\text{x}^3.\sin\text{x}^4\text{dx}$
Let $\text{x}^4=\text{t}$
$\Rightarrow4\text{x}^3=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
Now, $\int\text{x}^3.\sin\text{x}^4\text{dx}$
$=\frac{1}{4}\int\sin\text{t}\text{ dt}$
$=\frac{1}{4}[-\cos(\text{t})]+\text{C}$
$=\frac{1}{4}\big[-\cos\text{x}^4\big]+\text{C}$
View full question & answer→Question 1723 Marks
Evaluate the following integrals:$\int\text{x}^3\cos\text{x}^2\text{dx}$
AnswerLet $\text{I}=\int\text{x}^3\cos\text{x}^2\text{dx}$
Let $\text{x}^2=\text{t}$
$2\text{x dx = dt}$
$\text{I}=\frac{1}{2}\int\text{t}\cos\text{t dt}$
Using integration by parts,
$=\frac{1}{2}[\text{t}\int\cos\text{t dt}-\int(1\times\int\cos\text{t dt})\text{dt}]$
$=\frac{1}{2}[\text{t}\times\sin\text{t}-\int\sin\text{t dt}]$
$=\frac{1}{2}[\text{t}\sin\text{t}+\cos\text{t}]+\text{C}$
$\text{I}=\frac{1}{2}[\text{x}^2\sin\text{x}^2+\cos\text{x}^2]+\text{C}$
View full question & answer→Question 1733 Marks
Write a value of $\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{ dx}$
Let $\log\sin\text{x}=\text{t}$
$\cot\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$=\log(\log\sin\text{x})+\text{C}$
View full question & answer→Question 1743 Marks
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\tan^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\sec^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{\tan\frac{\text{x}}{2}}{\frac{1}2{}}-\text{x+c}$
$=2\tan\frac{\text{x}}{2}-\text{x+c}$
View full question & answer→Question 1753 Marks
Write a value of $\int(\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}}\text{ dx}$
$=\int(\text{e}\log\text{a}^{\text{x}}+\text{e}\log\text{x}^{\text{a}})\text{dx}$
$=\int(\text{a}^{\text{x}}+\text{x}^{\text{a}})\text{dx}$
$=\int\text{a}^{\text{x}}\text{dx}+\int\text{x}^{\text{a}}\text{dx}$
$=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
$\therefore\ \text{I}=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
View full question & answer→Question 1763 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{2\tan\text{x}\sec^2\text{x}}{\tan^4\text{x}+1}\ \text{dx}$
Let $\tan^2\text{x}=\text{t}$
$2\tan\text{x}\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+1}$
$=\tan^{-1}\text{t}+\text{C}$
$\text{I}=\tan^{-1}(\tan^2\text{x})+\text{C}$
View full question & answer→Question 1773 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$\text{Let,}\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$=\int\frac{\text{dt}}{\text{t}^2}$
$=\int\text{t}^{-2}\text{dt}$
$=\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=\frac{-1}{\text{t}}+\text{C}$
$=-\frac{1}{\sin^{-1}\text{x}}+\text{C}$
View full question & answer→Question 1783 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}(1-\cos\text{x})}{(1+\cos\text{x})(1-\cos\text{x})}\text{dx}$
$=\int\frac{\cos\text{x}-\cos^2\text{x}}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}-\cos^2\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}}{\sin^2\text{x}}-\frac{\cos^2\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int(\cot\text{x}\text{ cosec x}-\cot^2\text{x})\text{dx}$
$=\int(\cot\text{x}\text{ cosec x}-\text{cosec}^2\text{x}+1)\text{dx}$
$=\int\cot\text{x}\text{ cosec x dx}-\int\text{cosec}^2\text{x dx}+\int1\text{dx}$
$=-\text{cosec x}+\cot\text{x}+\text{x}+\text{C}$
View full question & answer→Question 1793 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{\sin^2\text{x}+4\sin\text{x}+5}\text{dx}$
Answer$\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
Let $\sin\text{x = t}$
$\Rightarrow\cos\text{x dx = dt}$
Now, $\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+4\text{t}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+2\times\text{t}\times2+4+1}$
$=\int\frac{\text{dt}}{(\text{t}+2)^2+1^2}$
$=\frac{1}{1}\tan^{-1}\Big(\frac{\text{t}+2}{1}\Big)+\text{C}$
$=\tan^{-1}(\sin\text{x}+2)+\text{C}$
View full question & answer→Question 1803 Marks
Evaluate the following integrals:
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\frac{(1-\cos\text{x})^2}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{1+\cos^2\text{x}-2\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\sin^2\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}-\frac{2\cos\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int(\text{cosec}^2\text{x}+\cot^2\text{x}-2\cot\text{x}\text{ cosec x})\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int(2\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int2\text{cosec}^2\text{x dx}-\int1\text{dx}-\int2\cot\text{x cosec x }\text{dx}$
$=-2\cot\text{x}-\text{x}+2\text{cosec x}+\text{C}$
$=2(\text{cosec x}-\cot\text{x})-\text{x + C}$
View full question & answer→Question 1813 Marks
Evaluate the following integrals:
$\int\sqrt{4\text{x}^2-5}\text{dx}$
Answer$\text{I}=\int\sqrt{4\text{x}^2-5}\text{dx}$
$=\int\sqrt{4\Big(\text{x}^2-\frac{5}{4}\Big)}\text{dx}$
$=2\int\sqrt{\text{x}^2-\Big(\frac{\sqrt5}{2}}\Big)^2\text{dx}$
$=2\Big[\frac{\text{x}}{2}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{8}\int\Big|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\Big|\Big]+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}\frac{1}{2}\text{a}^2\int\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}\Big]$
$=\text{x}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{4}\int\bigg|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\bigg|+\text{C}$
View full question & answer→Question 1823 Marks
Write the anti-derivative of $\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
AnswerLet $\text{I}=\int\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$\text{I}=3\sqrt{\text{x}}\text{ dx}+\int\frac{\text{dx}}{\sqrt{\text{x}}}$
$=3\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=3\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{C}$
$=2\times3\times\frac{\text{x}^\frac{3}{2}}{3}+2\times\frac{\text{x}^{\frac{1}{2}}}{1}+\text{C}$
$=2\Big(\text{x}^{\frac{3}{2}}+\text{x}^{\frac{1}{2}}\Big)+\text{C}$
View full question & answer→Question 1833 Marks
Evaluate the following integrals:
$\int\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\text{dx}$
Answer$\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$\text{Let }\sqrt{2\text{x}^2+3\text{x}+1}=\text{t}$
$\Rightarrow(4\text{x}+3)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(4\text{x}+3)\text{dx}=\text{dt}$
$\text{Now,}\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{t}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{2\text{x}^2+3\text{x}+1}+\text{C}$
View full question & answer→Question 1843 Marks
Evaluate the following integrals:
$\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$
Here, power of $\sin\text{x}$ is odd, so we substitute
$\cos\text{x}=\text{t}$
$-\sin\text{x}\text{dt}=\text{dt}$
$\therefore\ \text{I}=\int\sin^2\text{x}\text{ t}^6(-\text{dt})$
$=-\int\big(1-\cos^2\text{x}\big)\text{t}^6\text{dt}$
$=-\int\big(\text{t}^6-\text{t}^8\big)\text{dt}$
$=-\frac{\text{t}^7}{7}+\frac{\text{t}^9}{9}+\text{C}$
$\therefore\ \text{I}=\frac{1}{7}\cos^7\text{x}+\frac{1}{9}\cos^9\text{x}+\text{C}$
View full question & answer→Question 1853 Marks
Evalute the following integrals:
$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
$=\int\frac{1}{4\cos^2\text{x}-4\cos\text{x}}\text{dx}\big[\because\cos3\text{x}=\text{4}\cos^3\text{x}\cos\text{x}\big]$
$=\int\frac{1}{4\cos\text{x}(\cos^2\text{x}-1)}\text{dx}$
$=\frac{-1}{4}\int\frac{1}{\cos\text{x}\sin^2\text{x}}\text{dx}$
$=\frac{-1}{4}\int\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\cos\text{x}\sin^2\text{x}}\text{dx}\Big)$
$=\frac{-1}{\text{4}}\Big[\int\sec\text{ x dx}+\int\cot\text{x cosec x dx}$
$=\frac{-1}{4}\big(\text{ln }|\sec\text{x}+\tan\text{x}|-\text{cosec x}\big)+\text{C}$
$=\frac{1}{4}\big(\text{cosec x}-\text{ln}|\sec\text{x}+\tan\text{x}|\big)+\text{C}$
View full question & answer→Question 1863 Marks
If $\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C},$ then write the value of f(x).
AnswerSince, $\int\text{e}^{\text{x}}\big(\text{f(x)}+\text{f}(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
We can write the expression as
$\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$
Comparing with equation $\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$ with standard equation we have
$\text{f(x)}=\sec\text{x},\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Therefore,
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
Thus, $\text{f(x)}=\sec\text{x}$
View full question & answer→Question 1873 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
$=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
Here,$\text{f(x)}=\sec\text{x}$ Put$\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$
Diff. both sides e.r.t.x
$\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x})\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
View full question & answer→Question 1883 Marks
Evalute the following integrals:
$\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
Putting $e^x + x^e = t$
$\Rightarrow\text{e}^\text{x}+\text{ex}^{\text{e}-1}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)\text{dx}=\frac{\text{dt}}{\text{e}}$
$\therefore\text{I}=\frac{1}{\text{e}}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{\text{e}}\text{ ln}|\text{t}|+\text{C}$
$=\frac{1}{\text{e}}\text{ ln}\Big|\text{e}^\text{x}+\text{x}^\text{e}\Big|+\text{C}\big[\because\text{t}=\text{e}^\text{x}+\text{x}^\text{e}\big]$
View full question & answer→Question 1893 Marks
Evaluate the following integrals:
$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$
Answer$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$ Let $2+3\log\text{x}=\text{t}$ $\Rightarrow\frac{3}{\text{x}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\text{x}}=\frac{\text{dt}}{3}$ Now, $\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$$=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}[-\cos\text{t}]+\text{C}$
$=-\frac{1}{3}\cos(2+3\log\text{x})+\text{C}$
View full question & answer→Question 1903 Marks
Evaluate the following integrals:
$\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{I}=\int\sqrt{4-\text{t}^2}\text{dt}$
$=\int\sqrt{2^2-\text{t}^2}\text{dt}$
$=\frac{\text{t}^2}{2}\sqrt{2^2-\text{t}^2}+\frac{4}{2}\sin^{-1}\frac{\text{t}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\sin\text{x}\sqrt{4-\sin^2\text{x}}+2\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 1913 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{-(\text{x}^2-\text{x})}\text{dx}$
$=\int\sqrt{-\bigg\{\text{x}^2-\text{x}+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}\text{dx}$
$=\int\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$=\Big(\frac{\text{x}-\frac{1}{2}}{2}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\bigg)+\text{C}$
$\Big[\because\ \int\sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{a}^2-\text{x}^2}+\frac{1}{2}\text{a}^2\sin^{-1}\frac{\text{x}}{\text{a}}=\text{C}\Big]$
$=\Big(\frac{2\text{x}-1}{4}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}(2\text{x}-1)+\text{C}$
View full question & answer→Question 1923 Marks
Evaluate the following integrals:
$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
Answer$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
$\text{Let}\ \text{log}\sin\text{x} =\text{t}$
$\Rightarrow\frac{1}{\sin\text{x}}\times \cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx} =\text{dt}$
$\text{Now},\int\cot\text{x}. \log\sin\text{x dx} $
$=\int\text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{(\text{log}\begin{vmatrix}\sin\text{x}\end{vmatrix})^{2}}{2}+\text{C}$
View full question & answer→Question 1933 Marks
Write a value of $\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
Let $\text{x}+\log\sin\text{x}=\text{t}$
$(1+\cot\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
Hence,
$\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$
View full question & answer→Question 1943 Marks
Evaluate the following integrals:$\int\frac{\sin\text{x}}{\sqrt{4\cos^2\text{x}-1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}}{\sqrt{4\cos^2\text{x}-1}}\text{ dx}$
Let $2\cos\text{x}=\text{t}$
$\Rightarrow-2\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin\text{x}\text{ dx}=-\frac{\text{dt}}{2}$
$\text{I}=-\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}^2-1}}$
$=-\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2-1}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C}\Big]$
$=-\frac{1}{2}\log\Big|2\cos\text{x}+\sqrt{4\cos^2\text{x}-1}\Big|+\text{C}$
View full question & answer→Question 1953 Marks
Write a value of $\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$ $=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$ Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$ $(\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x})\text{dx}=\text{dt}$ $\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}=\text{dt}$ $\therefore\text{I}=\int\text{dt}$ $=\text{t}+\text{C}$$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$ $(\because\text{t}=\text{e}^{\text{x}}\sec\text{x})$
View full question & answer→Question 1963 Marks
Evalute the following integrals:
$\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
$=\int\bigg(\frac{1+\frac{\sin\text{x}}{\cos\text{x}}}{1-\frac{\sin\text{x}}{\cos\text{x}}}\bigg)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big)\text{dx}$
Putting $\cos\text{x}-\sin\text{x}=\text{t}$
$\Rightarrow(-\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow(\sin\text{x}+\cos\text{x})\text{dx}=-\text{dt}$
$\therefore\text{I}=-\int\frac{1}{\text{t}}\text{dt}$
$=-\text{ln}|\text{t}|+\text{C}$
$=-\text{ln}|\cos\text{x}-\sin\text{x}|+\text{C}\ \big[\because\text{t}\cos\text{x}-\sin\text{x}\big]$
View full question & answer→Question 1973 Marks
Evaluate the following integrals:$\int\text{x}\sin\text{x}\cos\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin\text{x}\cos\text{x dx}$
$=\int\frac{\text{x}}{2}(2\sin\text{x} \cos\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x}\sin2\text{x dx}$
Using integration by parts,
$=\frac{1}{2}[\text{x}\int\sin2\text{x dx}-\int(1\times\int\sin2\text{x dx})\text{dx}]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos2\text{x}}{2}\Big)-\int\Big(\frac{-\cos2\text{x}}{2}\Big)\text{dx}\Big]$
$=-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{4}\int\cos2\text{x dx}$
$\text{I}=-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{8}\sin2\text{x}+\text{C}$
View full question & answer→Question 1983 Marks
Evalute the following integrals:
$\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}$
AnswerLet $\text{I}=\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}\ .....(\text{i})$
Let $10^\text{x}+\text{x}^{10}=\text{t}$ then,
$\text{d}(10^\text{x}+\text{x}^{10})=\text{dt}$
$\Rightarrow(10^\text{x}\log_\text{e}10+10\text{x}^9)\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{(10\text{x}^9+10^\text{x}\log_\text{e}10)}$
Putting $10^x + x^{10} = t$ and $\text{dx}=\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$ in equation (i), we get,
$\text{I}=\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{\text{t}}\times\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
$\therefore\text{I}=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
View full question & answer→Question 1993 Marks
$\int\frac{1}{1+\cos3\text{x}}\text{dx}$
Answer$\int\frac{\text{dx}}{1+\cos3\text{x}}$
$=\int\frac{(1-\cos3\text{x})}{(1+\cos3\text{x})(1-\cos3\text{x})}\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{1-\cos^23\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{\sin^23\text{x}}\Big)\text{dx}$
$=\int\text{cosec}^23\text{x}\text{ dx}-\int\text{cosec}3\text{x}\cot3\text{x dx}$
$=-\frac{\cot3\text{x}}{3}+\frac{\text{cosec}3\text{x}}{3}+\text{c}$
$=\frac{1}{3}[\text{cosec}3\text{x}-\cot3\text{x}]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1}{\sin3\text{x}}-\frac{\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1-\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
View full question & answer→Question 2003 Marks
$\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
Answer$\text{Let I}=\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
$\text{Putting}\ \ 2\text{x}-1=\text{t}$
$\Rightarrow2\text{x}=\text{t}+1$
$\Rightarrow\text{x}=\frac{\text{t}+1}{2}$
$\&\ 2\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int\Big[5\Big(\frac{\text{t}+1}{2}\Big)+3\Big]\times\sqrt{\text{t}}\times\frac{\text{dt}}{2}$
$=\int\Big(\frac{5\text{t}}{5}+\frac{5}{2}+3\Big)\times\frac{\sqrt{\text{t}}\text{ dt}}{2}$
$=\frac{1}{4}\int(5\text{t}+11)\text{t}^\frac{1}{2}\text{ dt}$
$=\frac{1}{4}\int(5\text{t}^\frac{3}{2}+11\text{t}^\frac{1}{2})\text{ dt}$
View full question & answer→Question 2013 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}$
Answer$\text{I}=\int\text{e}^{\text{x}}\big(\text{x}^{-2}-2\text{x}^{-3}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\text{x}^{-2}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
Integration by parts
$=\text{e}^{\text{x}}\text{x}^{-2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\big(\text{x}^{-2}\big)\Big)\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\text{e}^{\text{x}}\text{x}^{-2}+2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{\text{x}^2}+\text{C}$
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}=\frac{\text{e}^\text{x}}{\text{x}^2}+\text{C}$
View full question & answer→Question 2023 Marks
Write a value of $\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
Let $3+2\sin\text{x}=\text{t}$
$2\cos\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{2}\log\text{t}+\text{C}$
$=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
$\therefore\ \int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
View full question & answer→Question 2033 Marks
Write a value of $\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}+\cos\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Let $\sin\text{x}+\cos\text{x}=\text{t}$
$(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log\text{t}+\text{C}$
$\text{I}=\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 2043 Marks
Write a value of $\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$
Dividing and multiplying by $e^x$
$=\frac{\text{e}^{-\text{x}}}{\text{e}^{-\text{x}}+1}\text{dx}$
Let $\text{e}^{-\text{x}}+1=\text{t}$
$-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=-\log|1+\text{e}^{-\text{x}}|+\text{C}$
View full question & answer→Question 2053 Marks
Evaluate the following integrals:
$\int\log(\text{x}+1)\text{dx}$
AnswerLet $\text{I}=\int\log(\text{x}+1)\text{dx}$
$=\int1\times\log(\text{x}+1)\text{dx}$
Using integration by parts,
$\text{I}=\log(\text{x}+1)\int1\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(\frac{\text{x}}{\text{x}+1}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx+C}$
$\text{I}=\text{x}\log(\text{x}+1)-\text{x}+\log(\text{x}+1)+\text{C}$
View full question & answer→Question 2063 Marks
Evaluate the following integrals:
$=\int\frac{\sin\text{x}}{(1+\cos\text{x})^2}\text{dx}$
AnswerLet I $=\int\frac{\sin\text{x}}{(1+\cos\text{x})^2}\text{dx}\ ....(1)$
Let $1+\cos\text{x}=\text{t}$ then,
$\text{d}(1+\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\Rightarrow\sin\text{x dx}=-\text{dt}$
Putting $1+\cos\text{x}=\text{t}$ and $\sin\text{dx}=-\text{dt}$ in equation (2), we get
$\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\big(-1\text{t}^{-1}\big)+\text{C}$
$=\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{1+\cos\text{x}}+\text{C}$
$\therefore\text{I}=\frac{1}{1+\cos\text{x}}+\text{C}$
View full question & answer→Question 2073 Marks
Evaluate the following integrals:
$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
Answer$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(6+4\text{x}-9\text{x}-6\text{x}^2\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2-5\text{x}+6\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2+12\text{x}^3-5\text{x}+10\text{x}^2+6-12\text{x}\big)\text{dx}$
$=\int\big(4\text{x}^2+12\text{x}^3-17\text{x}+6\big)\text{dx}$
$=\int\big(12\text{x}^3+4\text{x}^2-17\text{x}+6\big)\text{dx}$
$=\frac{12}{4}\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
$=3\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
View full question & answer→Question 2083 Marks
Evaluate the following integrals:
$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Answer$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
Now, $\int\cot^{\text{n}}\text{x }\text{cosec}^2\text{x}\text{ dx}$
$=-\int\text{t}^{\text{n}}\text{dt}$
$=\frac{-\text{t}^{\text{n}+1}}{\text{n}+1}+\text{C}$
$=-\frac{\cot^{\text{n}+1}}{\text{n}+1}+\text{C}$
View full question & answer→Question 2093 Marks
Evaluate the following integrals:$\int\frac{\cos\text{x}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}\text{ dx}$
Answer$\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2-2\text{t}-3}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2-2\text{t}+1-1-3}}$
$=\int\frac{\text{dt}}{\sqrt{(\text{t}-1)^2-2^2}}$
$=\log\Big|\text{t}-1+\sqrt{(\text{t}-1)^2-2^2}\Big|+\text{C}$
$=\log\Big|\text{t}-1+\sqrt{\text{t}^2-2\text{t}-3}\Big|+\text{C}$
$=\log\Big|\sin\text{x}-1+\sqrt{\sin^2\text{x}-2\sin\text{x}-3}\Big|+\text{C}$
View full question & answer→Question 2103 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x cosec x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\tan\text{x})|+\text{C}$
View full question & answer→Question 2113 Marks
Evaluate the following integrals:$\int\tan^{-1}\Big(\frac{3\text{x}-\text{x}^3}{1-3\text{x}^2}\Big)\text{dx}$
AnswerLet $\int\tan^{-1}\Big(\frac{3\text{x}-\text{x}^3}{1-3\text{x}^2}\Big)\text{dx}$$=\int3\tan^{-1}(\text{x})\text{dx}$
$=3\int\big[\tan^{-1}(\text{x})\times1\big]\text{dx}$
$=3\Big[\tan^{-1}\text{x}\times\text{x}-\int\frac{1}{1+\text{x}^2}\times\text{x dx}\Big]$
$=3\text{x}\tan^{-1}\text{x}-3\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
Let $1+\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
Then,
$\text{I}=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\int\frac{\text{dt}}{\text{t}}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|\text{t}|+\text{C}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|1+\text{x}^2|+\text{C}$
View full question & answer→Question 2123 Marks
Evaluate the following integrals:$\int\frac{\text{e}^\text{x}}{(1+\text{e}^{\text{x}})(2+\text{e}^\text{x})}\text{dx}$
AnswerTo evaluate the following integral follow tha steps:
Let $\text{e}^\text{x}=\text{t}$ therefore $\text{e}^\text{x}\text{dx = dt}$
Now
$\int\frac{\text{e}^{\text{x}}}{(1+\text{e}^\text{x})(2+\text{e}^\text{x})}\text{dx}=\int\frac{\text{dt}}{(1+\text{t})(2+\text{t})}$
$=\int\frac{\text{dt}}{(1+\text{t})}-\int\frac{\text{dt}}{(2+\text{t})}$
$=\ln|1+\text{t}|-\ln|2+\text{t}|+\text{C}$
$=\ln\bigg|\frac{1+\text{t}}{2+\text{t}}\bigg|+\text{C}$
$=\ln\bigg|\frac{1+\text{e}^{\text{x}}}{2+\text{e}^{x}}\bigg|+\text{C}$
View full question & answer→Question 2133 Marks
$\int\frac{2\text{x}}{(2\text{x}+1)^2}\text{dx}$
Answer$\int\frac{2\text{x}}{(2\text{x}+1)^2}\text{dx}$
$=\int\bigg(\frac{2\text{x}+1-1}{(2\text{x}+1)^2}\bigg)\text{dx}$
$=\int\bigg[\frac{2\text{x}+1}{(2\text{x}+1)^2}-\frac{1}{(2\text{x}+1)^2}\bigg]\text{dx}$
$=\int\frac{\text{dx}}{2\text{x}+1}-\int(2\text{x}+1)^{-2}\text{dx}$
$=\frac{\log(2\text{x}+1)}{2}-\bigg[\frac{(2\text{x}+1)^{-2+1}}{2(-2+1)}\bigg]+\text{c}$
$=\frac{\log(2\text{x}+1)}{2}+\frac{(2\text{x}+1)^{-1}}{2}+\text{c}$
$=\frac{\log(2\text{x}+1)}{2}+\frac{1}{2(2\text{x}+1)}+\text{c}$
View full question & answer→Question 2143 Marks
Evaluate the following integrals:
$\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
Answer$\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
$=\int\cot^5\text{x}\text{ cosec}^2\text{ x}.\text{ cosec}^2\text{x}\text{ dx}$
$=\int\cot^5\text{x}.(1+\cot^2\text{x}).\text{ cosec}^2\text{x}\text{ dx}$
Let $\cot\text{x}=\text{t}$
$=-\text{ cosec}^2\text{x}\text{ dx}=\text{dt}$
$=\text{ cosec}^2\text{x}\text{ dx}=-\text{dt}$
Now, $\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
$=\int\text{t}^5(1+\text{t}^2)\text{dt}$
$=\int(\text{t}^5+\text{t}^7)\text{dt}$
$=-\Big[\frac{\text{t}^6}{6}+\frac{\text{t}^8}{8}\Big]+\text{C}$
$=-\Big[\frac{\cot^6\text{x}}{6}+\frac{\cot^8\text{x}}{8}\Big]+\text{C}$
View full question & answer→Question 2153 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}\sqrt{\text{x}^4-1}}\text{ dx}$
Answer$\text{I}=\int\frac{1}{\text{x}\sqrt{\text{x}^4-1}}\text{ dx}\ ....(1)$
Let $\text{x}^2=\text{t}$ then,
$\text{d}\big(\text{x}^2\big)=\text{dt}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2\text{x}}$
Putting $\text{x}^2=\text{t}$ and $\text{dx}=\frac{\text{dt}}{2\text{x}}$ in equation (1), we get,
$\text{I}=\int\frac{1}{\text{x}\sqrt{\text{t}^2-1}}\times\frac{\text{dt}}{2\text{x}}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2\sqrt{\text{t}^2-1}}\text{ dt}$
$=\frac{1}{2}\int\frac{1}{\text{t}\sqrt{\text{t}^2-1}}\text{ dt}$
$=\frac{1}{2}\sec^{-1}\text{t}+\text{C}$
$=\frac{1}{2}\sec^{-1}\text{x}^2+\text{C}$
$\text{I}=\frac{1}{2}\sec^{-1}\big(\text{x}^2\big)+\text{C}$
View full question & answer→Question 2163 Marks
Write a value of $\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^7}{7}+\text{C}$
Thus, $\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$
View full question & answer→Question 2173 Marks
$\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\text{dx}.$ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\times\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}\text{dx}$
$=\int\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\text{x}+3-\text{x}-2}\text{dx}$
$=\int\Big[(\text{x}+3)^{\frac{1}{2}}+(\text{x}+2)^{\frac{1}{2}}\Big]\text{dx}$
$=\frac{(\text{x}+3)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(\text{x}+2)^{\frac{3}{2}}}{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\times(\text{x}+3)^{\frac{3}{2}}+\frac{2}{3}(\text{x}+2)^{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
$\text{I}=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
View full question & answer→Question 2183 Marks
Evaluate the following integrals:$\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}^2+15\text{x}-2\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}(\text{x}+5)-2(\text{x}+5)}\text{ dx}$
$=\int\frac{\text{x}+5}{(3\text{x}-2)(\text{x}+5)}\text{ dx}$
$=\int\frac{1}{3\text{x}-2}\text{ dx}$
$\therefore\ \text{I}=\frac{1}{3}\int|3\text{x}-2|+\text{C}$
View full question & answer→Question 2193 Marks
Evaluate the following integrals:$\int\frac{\cos\text{x}}{\sqrt{4+\sin^2\text{x}}}\text{ dx}$
Answer$\int\frac{\cos\text{x}\text{ dx}}{\sqrt{4+\sin^2\text{x}}}$ Let $\sin\text{x}=\text{t}$ $\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$ Now, $\int\frac{\cos\text{x}\text{ dx}}{\sqrt{4+\sin^2\text{x}}}$$=\int\frac{\text{dt}}{\sqrt{2^2-\text{t}^2}}$
$=\log\Big|\text{t}+\sqrt{4+\text{t}^2}\Big|+\text{C}$$=\log\Big|\sin\text{x}+\sqrt{4+\sin^2\text{x}}\Big|+\text{C}$
View full question & answer→Question 2203 Marks
Write a value of $\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
Let $3+\text{x}\log\text{x}=\text{t}$
$\Big(\log\text{x}+\text{x}\cdot\frac{1}{\text{x}}\Big)\text{dx}=\text{at}$
$(1+\log\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$\text{I}=\log(3+\text{x}\log\text{x})+\text{C}$
View full question & answer→Question 2213 Marks
Evaluate the following integrals:
$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
Answer$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
$=\int\sin^2\text{x}\cdot\cos^5\text{x}\cdot\sin\text{x}\text{ dx}$
$=\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x}\text{ dx}=\text{dt}$
$\sin\text{x}\text{ dx}=-\text{dt}$
Now, $\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
$=-\int(1-\text{t}^2)\text{t}^5\text{dt}$
$=-(\text{t}^5-\text{t}^7)\text{dt}$
$=-\int(\text{t}^7-\text{t}^5)\text{dt}$
$=\frac{\text{t}^8}{8}-\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\cos^8\text{x}}{8}-\frac{\cos^6\text{6}\text{x}}{6}+\text{C}$
View full question & answer→Question 2223 Marks
Evaluate the following integrals:
$\int\text{x}^2\cos\text{x dx}$
Answer$\int\text{x}^2\cos\text{x dx}$
Taking $x^2$ as the first function and cos x as the second function.
$=\text{x}^2\int\cos\text{x dx}-\int\big(\frac{\text{d}}{\text{dx}}\text{x}^2\int\cos\text{x dx}\big)\text{dx}$
$=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2\big[\text{x}\int\sin\text{x}-\int\big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\big\}\text{dx}\big]$
$=\text{x}^2\sin\text{x}-2[-\text{x}\cos\text{x}+\int\cos\text{x dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x+C}$
View full question & answer→Question 2233 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin\text{x}\cos\text{x}+2\cos^2\text{x}}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\ \text{dx}$
Let $2+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$\text{I}=\log|2+\tan\text{x}|+\text{C}$
View full question & answer→Question 2243 Marks
$\int \text{(2x} - 3)^{5} + \sqrt{3\text{x + 2}}\text{ dx}$
Answer$\int\big[(2\text{x}-3)^5+\sqrt{3\text{x}+2}\big]\text{dx}$
$=\int(2\text{x}-3)^5\text{dx}+\int{(3\text{x}+2)^{\frac{1}{2}}}\text{dx}$
$=\frac{(2\text{x}-3)^{5+1}}{2(5+1)}+\frac{(3\text{x}+2)^{\frac{1}{2}{+1}}}{3\Big(\frac{1}{2}+1\Big)}+\text{c}$
$=\frac{(2\text{x}-3)^6}{12}+\frac{2}{9}(3\text{x}+2)^{\frac{3}{2}}+\text{c}$
View full question & answer→Question 2253 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{\sin(5\text{x}-3\text{x})}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{5\text{x}\cos3\text{x}-\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}$
$=\int\frac{\sin5\text{x}\cos3\text{x}}{\sin5\text{x}\sin3\text{x}}-\frac{\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\big[\cot3\text{x}-\cot5\text{x}\big]\text{dx}$
$=\int\cot3\text{x dx}-\int\cot5\text{x dx}$
$=\frac{1}{3}\text{ln}|\sin3\text{x}|-\frac{1}{5}\text{ln}|\sin5\text{x}|+\text{C}$
View full question & answer→Question 2263 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$ then,
$=\int\sqrt{\frac{1-\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}{1+\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\frac{2\sin^2\Big(\frac{\pi}{4}-\text{x}\Big)}{2\cos^2\Big(\frac{\pi}{4}-\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\tan^2\Big(\frac{\pi}{4}-\text{x}\Big)}\text{dx}$
$=\int\tan\Big(\frac{\pi}{4}-\text{x}\Big)\text{dx}$
$=\log\Big|\cos\Big(\frac{\pi}{4}-\text{x}\Big)\Big|+\text{C}$
View full question & answer→Question 2273 Marks
Evaluate the following integrals:$\int2\text{x}^3\text{e}^{\text{x}^{2}}\text{dx}$
Answer$\int2\text{x}^3\cdot\text{e}^{\text{x}^{2}}\text{dx}$
$=\int\text{x}^2\cdot\big(\text{e}^{\text{x}^2}\big)\cdot2\text{x dx}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$=\int\text{t}\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t}\cdot\text{e}^{\text{t}}-\int1\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t e}^{\text{t}}-\text{e}^{\text{t}}+\text{C}$
$=\text{x}^2\text{e}^{\text{x}^{2}}-\text{e}^{\text{x}^{2}}+\text{C}$
$=\text{e}^{\text{x}^2}(\text{x}^2-1)+\text{C}$
View full question & answer→Question 2283 Marks
Evaluate the following integrals:
$\int\text{x}^3\log\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^3\log\text{x dx}$
Using integration by parts,
$\text{I}=\log\text{x}\int\text{x}^3\text{dx}-\int\Big(\frac{1}{\text{x}}\times\int\text{x}^3\text{dx}\Big)\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\int\frac{\text{x}^4}{4\text{x}}\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\text{x}^3\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\frac{\text{x}^4}{4}\text{dx+C}$
$\text{I}=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{16}\text{x}^4+\text{C}$
View full question & answer→Question 2293 Marks
Evaluate the following integrals:$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Answer$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Taking log log x as the first function and $\frac{1}{\text{x}}$ as the second function.
$=\log \log\text{x}\int\frac{1}{\text{x}}\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\log(\log\text{x})\int\frac{1}{\text{x}}\text{dx}\Big\}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}\log\text{x}}(\log\text{x})\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\log\text{x}+\text{C}$
$=\log\text{x}[\log(\log\text{x})-1]+\text{C}$
View full question & answer→Question 2303 Marks
Write a value of $\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}.$
AnswerLet $\text{I}=\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}$
$=\int\text{e}^{\log\text{x}^3}\cdot\text{x}^{4}\text{ dx}$
$=\int\text{x}^3\cdot\text{x}^4\text{ dx}$ $\big[\because\text{e}^{\log\text{x}}=\text{x}\big]$
$=\int\text{x}^{7}\text{ dx}$
$\therefore\ \text{I}=\frac{\text{x}^{8}}{8}+\text{C}$
View full question & answer→Question 2313 Marks
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
Answer$\int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{2\sin\text{x}\cos\text{x}}{2\cos^2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{\sin\text{x}}{\cos\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}(\tan\text{x})\text{dx}$
$=\int\text{x dx}$
$=\frac{\text{x}^2}{2}+\text{C}$
$\therefore\ \int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}=\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 2323 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
Considering sin x as first function and $e^{2x} as$ second function
$\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\frac{1}{2}\int\cos\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{1}{2}\Big[\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int(-\sin\text{x})\frac{\text{e}^{2\text{x}}}{2}\text{dx}\Big]$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{\cos\text{x }\text{e}^{2\text{x}}}{4}-\frac{1}{2}\int\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}\text{dx}$
$\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{4}-\frac{\text{I}}{4}$
$\Rightarrow5\text{I}=\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{5}+\text{C}$
View full question & answer→Question 2333 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^{\frac{-1}{3}}+\sqrt{\text{x}}+2}{\sqrt[3]{\text{x}}}\text{dx}$
Answer$\int\Bigg(\int\frac{\text{x}^{-\frac{1}{3}}+\sqrt{\text{x}}+2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Bigg(\frac{\text{x}^{-\frac{1}{3}}}{\text{x}^{\frac{1}{3}}}+\frac{\text{x}^{\frac{1}{2}}}{\text{x}^{\frac{1}{3}}}+\frac{2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Big(\text{x}^{-\frac{2}{3}}+\text{x}^{\frac{1}{6}}+2\text{x}^{-\frac{1}{3}}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\frac{\text{x}^{\frac{1}{6}+1}}{\frac{1}{6}+1}+2\frac{\text{x}^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\Bigg]$
$=\Bigg[\frac{\text{x}^{\frac{1}{3}}}{\frac{1}{3}}+\frac{\text{x}^{\frac{7}{6}}}{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}\Bigg]+\text{C}$
$=3\text{x}^{\frac{1}{3}}+\frac{6}{7}\text{x}^{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}+\text{C}$
View full question & answer→Question 2343 Marks
$\int\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\text{dx}$
Answer$\int\Big(\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{2-3\text{x}}+\int(3\text{x}-2)^{-\frac{1}{2}}\text{dx}$
$=\frac{\ln(2-3\text{x})}{-3}+\Bigg[\frac{(3\text{x}-2)^{-\frac1{2}+1}}{3\big(-\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{\ln(2-3\text{x})}{-3}+\frac{2}{3}(3\text{x}-2)^{\frac{1}{2}}+\text{c}$
$=-\frac{1}{3}\ln(2-3\text{x})+\frac{2}{3}\sqrt{3\text{x}-2}+\text{c}$
View full question & answer→Question 2353 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
Putting $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\therefore\text{I}=\frac{1}{2+3\text{t}}\text{dt}$
$=\frac{1}{3}\text{ln}|2+3\text{t}|+\text{C}$
$=\frac{1}{3}\text{ln}|2+3\sin^{-1}\text{tx}|+\text{C }\big[\because\text{t}=\sin^{-1}\text{x}\big]$
View full question & answer→Question 2363 Marks
Evalute the following integrals:
$\int\big\{1+\tan\text{x}\tan(\text{x}+\theta)\big\}\text{dx}$
AnswerSince,
$\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$
$\therefore\tan(\text{x}+\theta-\text{x})=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{1+\tan(\text{x}+\theta)\tan\text{x}}$
$\Rightarrow 1+\tan(\text{x}+\theta)\tan\text{x}=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{\tan\theta}$
$\Rightarrow\int1+\tan(\text{x}+\theta)\tan\text{x dx}$
$=\frac{1}{\tan\theta}\big[\int\tan(\text{x}+\theta)\text{dx}-\int\tan\text{x dx}\big]$
$=\frac{1}{\tan\theta}\big[-\log|\cos(\text{x}+\theta)++\log|\cos\text{x}|\big]+\text{C}$
$=\frac{1}{\tan\theta}\big[\log|\cos\text{x}|-\log|\cos(\text{x}+\theta)|\big]+\text{C}$
$=\frac{1}{\tan\theta}\log\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C}$
View full question & answer→Question 2373 Marks
Evaluate the following integrals:$\int\sec^{-1}\sqrt{\text{x}}\text{dx}$
Answer$\int1.\sec^{-1}\sqrt{\text{x}}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\Big(\sec^{-1}\sqrt{\text{x}}\Big)\int1\text{dx}\Big\}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}.\text{x}-\int\frac{1}{\sqrt{\text{x}}\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\times\text{x dx}$
$=\text{x}\sec^{-1}\sqrt{\text{x}}-\frac{1}{2}\int(1-\text{x})^{-\frac{1}{2}\text{dx}}$
$=\text{x}\sec^{-1}\text{x}-\frac{1}{2}\Bigg[\frac{(1-\text{x})^{-\frac{1}{2}+1}}{\big(-\frac{1}{2}+1\big)(-1)}\Bigg]+\text{C}$
$=\text{x}\sec^{-1}\text{x}+(1-\text{x})^{\frac{1}{2}}+\text{C}$
View full question & answer→Question 2383 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{\text{x}^2-1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+1}{\text{x}^2-1}\ \text{dx}$
$=\int1+\frac{2}{\text{x}^2-1}\ \text{dx}$
$=\int\text{dx}+\int\frac{2\text{dx}}{(\text{x}+1)(\text{x}-1)}$
$=\int\text{dx}+\int\frac{-1}{\text{x}+1}+\frac{1}{\text{x}-1}\ \text{dx}$
$=\text{x}-\log|\text{x}+1|+\log|\text{x}-1|+\text{C}$
$\text{I}=\text{x}+\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 2393 Marks
Write a value of $\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
$=\int\text{e}^{2\text{x}^2}\cdot\text{e}^{\ln{\text{x}}}\text{dx}$
$=\int\text{x}\cdot\text{e}^{2\text{x}^2}\text{dx}$ $\big[\because\text{e}^{\ln\text{x}}=\text{x}\big]$
$=\int\text{x}\cdot\big(\text{e}^{\text{x}^2}\big)\text{dx}$
Let $\text{e}^{\text{x}^2}=\text{t}$
$\text{e}^{\text{x}^2}\cdot2\text{x dx}=\text{dt}$
$\therefore\ \frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\frac{\text{t}^2}{2}+\text{C}$
$=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
View full question & answer→Question 2403 Marks
Evaluate the following integrals:
$\int\frac{\text{x}+\sqrt{\text{x}+1}}{\text{x}+2}\text{ dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{x}+\sqrt{\text{x}+1}}{\text{x}+2}\text{ dx}$
Let $\text{x}+1=\text{t}^2$
Differentiating both sides we get
$\text{dx}=2\text{t dt}$
Now, integration becomes
$\text{I}=\int\frac{(\text{t}^2-1+\text{t})}{\text{t}^2+1}2\text{t dt}$
$=2\int\frac{\text{t}^3+\text{t}^2-\text{t}}{\text{t}^2+1}\text{ dt}$
$=2\int\frac{\text{t}^3+\text{t}-\text{t}+\text{t}^2+1-1-\text{t}}{\text{t}^2+1}\text{ dt}$
$=\int\frac{\text{t}^3+\text{t}+\text{t}^2+1-\text{t}-\text{t}-1}{\text{t}^2+1}\text{ dt}$
$=2\int\frac{\text{t}^3+\text{t}}{\text{t}^2+1}+2\int\frac{\text{t}^2+1}{\text{t}^2+1}+2\int\frac{-2\text{t}-1}{\text{t}^2+1}\text{ dt}$
$=2\int\text{t dt}+2\int\text{dt}-2\int\frac{2\text{t}}{\text{t}^2+1}\text{ dt}-2\int\frac{1}{\text{t}^2+1}\text{ dt}$
$=\text{t}^2+2\text{t}-2\log\big|\text{t}^2+1\big|-2\tan^{-1}\text{t}+\text{C}$
$=(\text{x}+1)+2\sqrt{\text{x}+1}-2\log|\text{x}+2|-2\tan^{-1}\sqrt{\text{x}+1}+\text{C}$
View full question & answer→Question 2413 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
$=\int\text{e}^{\text{x}}\cos\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
Integrating by parts
$=\text{e}^{\text{x}}\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\cos\text{x}\Big)\text{dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\int\text{e}^{\text{x}}\sin\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
$\therefore\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
View full question & answer→Question 2423 Marks
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
Answer$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
$=\int\frac{\tan\text{x}}{(\sec\text{x}+\tan\text{x})}\times\Big(\frac{\sec\text{x}-\tan\text{x}}{\sec\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{\tan\text{x}(\sec\text{x}-\tan\text{x})}{(\sec^2\text{x}-\tan^2\text{x})}\text{dx}$
$=\int\Big(\frac{\sec\text{x}\tan\text{x}-\tan^2\text{x}}{1}\Big)\text{dx}$
$=\int\sec\text{x}\tan\text{x dx}-\int(\sec^2\text{x}-1)\text{dx}$
$=\sec\text{x}-\tan\text{x}+\text{x}+\text{C}$
View full question & answer→Question 2433 Marks
Evaluate the following intregals:
$\int\frac{1}{(\text{x}-1)(\text{x}+1)(\text{x}+2)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{(\text{x}-1)(\text{x}+1)(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}-1)(\text{x}+2)+\text{C}(\text{x}^2-1)$
Put x = 1
$\Rightarrow1=6\text{A}\Rightarrow\text{A}=\frac{1}{6}$
put = -1
$\Rightarrow1=-2\text{B}\Rightarrow\text{B}=-\frac{1}{2}$
put = -2
$\Rightarrow1=3\text{C}\Rightarrow\text{C}=\frac{1}{3}$
So,
$\text{I}=\frac{1}{6}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{3}\int\frac{\text{dx}}{\text{x}+2}$
$\text{I}=\frac{1}{6}\log|\text{x}-1|-\frac{1}{2}\log|\text{x}+1|+\frac{1}{3}\log|\text{x}+2|+\text{C}$
View full question & answer→Question 2443 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
Here, $\text{f(x)}=\log\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\text{x}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\log\text{x}=\text{t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log\text{x}+\text{e}^{\text{x}}\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}(\log\text{x}+\frac{1}{\text{x}})\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}\big[\log\text{x}+\frac{1}{\text{x}}\big]\text{dx}=\int\text{dt}$
$\Rightarrow\text{I}=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\log\text{x}+\text{C}$
View full question & answer→Question 2453 Marks
$\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}$
AnswerLet $\text{l}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}. $ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\times\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{x+a}-\text{x-b}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{a}-\text{b}}\times\text{dx}$
$=\frac{1}{\text{a}-\text{b}}\bigg[\frac{2}{3}(\text{x+a})^{\frac{3}{2}}-\frac{2}{3}(\text{x+b})^{\frac{3}{2}}\bigg]+\text{c}$
$=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
$ \text{I}=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
View full question & answer→Question 2463 Marks
Evaluate $\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Answer$\text{I}=\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Let $(1+\log\text{x})=\text{t}$
Or, $\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$\text{I}=\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=\log|1+\log\text{x}|+\text{C}$
View full question & answer→Question 2473 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$
Putting $\text{x}^2+\sin2\text{x}+2\text{x}=\text{t}$
$\Rightarrow2\text{x}+2\cos2\text{x}+2=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+\cos2\text{x}+1)\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\text{ln}|\text{t}|+\text{C}$
$=\frac{1}{2}\text{ln}|\text{x}^2+\sin2\text{x}+2\text{x}|+\text{C}$
$\big[\because\text{t}=\text{x}^2+\sin2\text{x}+2\text{x}\big]$
View full question & answer→Question 2483 Marks
$\int\sin^3(2\text{x}+1)\text{dx}$
AnswerWe need to evaluate $\int\sin^3(2\text{x}+1)\text{dx}$
By using the formula
$\sin3\theta=-4\sin^3\theta+3\sin\theta$
$\therefore\sin^3(2\text{x}+1)=\frac{3\sin(2\text{x}+1)-\sin3(2\text{x}+1)}{4}$
$\int\sin^3(2\text{x}+1)\text{dx}$
$=\int\frac{3\sin(2\text{x}+1)-\sin3(2\text{x}+1)}{4}\text{dx}$
$=-\frac{3}{8}\cos(2\text{x}+1)+\frac{1}{24}\cos3(2\text{x}+1)+\text{C}$
View full question & answer→Question 2493 Marks
Evaluate the following integrals:$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
Answer$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
$=\int\sqrt{\frac{1}{\sin\text{x}}-1}\text{ dx}$
$=\int\frac{\sqrt{1-\sin\text{x}}}{\sqrt{\sin\text{x}}}\text{ dx}$
$=\int\frac{\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{\sqrt{\sin\text{x}(1+\sin\text{x})}}\text{ dx}$
$=\int\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\int\frac{\text{dt}}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\Big|+\text{C}$
$=\log\Big|\text{t}+\frac{1}{2}+\sqrt{\text{t}^2+\text{t}}\Big|+\text{C}$
$=\log\Big|\sin\text{x}+\frac{1}{2}\sqrt{\sin^2\text{x}+\sin\text{x}}\Big|+\text{C}$
View full question & answer→Question 2503 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{2\big[\frac{5}{2}-2\text{x}-\text{x}^2}\big]}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-2\text{x}-\text{x}^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}(\text{x}^2+2\text{x})}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}^2+2\text{x}+1-1)}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}+1)^2+1}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt3}\Big)^2-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\frac{(\text{x}+1)\sqrt2}{\sqrt7}\Big)+\text{C}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\sqrt{\frac{2}{7}}(\text{x}+1)\Big)+\text{C}$
View full question & answer→Question 2513 Marks
$\int\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\text{dx}$
Answer$\int\Big(\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{(1+\cos4\text{x})}{\big(\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\big)}\text{dx}$
$=\int\frac{2\cos^22\text{x}\times\sin\text{x}\cos\text{x}}{(\cos^2\text{x}-\sin^2\text{x})}\text{dx}$
$=\int\frac{\cos^22\text{x}\times2\sin\text{x}\cos\text{x}}{\cos2\text{x}}\text{dx}$
$=\int\cos2\text{x}\sin2\text{x}\text{ dx}$
$=\frac{1}{2}\int2\sin2\text{x}\cos2\text{x dx}$
$=\frac{1}{2}\int\sin4\text{x dx}$
$=\frac{1}{2}\Big[-\frac{\cos4\text{x}}{4}\Big]+\text{c}$
$=-\frac{1}{8}\cos4\text{x}+\text{c}$
View full question & answer→Question 2523 Marks
Evaluate the following integrals:$\int\frac{\log(\text{x}+2)}{(\text{x}+2)^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\log(\text{x}+2)}{(\text{x}+2)^2}\text{dx}$
Let $\frac{1}{\text{x}+2}=\text{t}$
$-\frac{1}{(\text{x}+2)^2}\text{dx = dt}$
$\text{I}=-\int\log\big(\frac{1}{\text{t}}\big)\text{dt}$
$=-\int\log\text{t}^{-1}\text{dt}$
$=-\int1\times\log\text{t dt}$
Using integration by parts,
$\text{I}=\log\text{t}\int\text{dt}-\int\big(\frac{1}{\text{t}}\int\text{dt}\big)\text{dt}$
$=\text{t}\log\text{t}-\int\Big(\frac{1}{\text{t}}\times\text{t}\Big)\text{dt}$
$=\text{t}\log\text{t}-\int\text{dt}$
$=\text{t}\log\text{t}-\text{t+C}$
$=\frac{1}{\text{x}+2}\big(\log(\text{x}+2)^{-1}-1\big)+\text{C}$
$\text{I}=\frac{-1}{\text{x}+2}-\frac{\log(\text{x}+2)}{\text{x}+2}+\text{C}$
View full question & answer→Question 2533 Marks
Evaluate the following integrals:$\int\frac{\text{x}}{\sqrt{\text{x}^4+\text{a}^4}}\text{ dx}$
Answer$\int\frac{\text{x}\text{ dx}}{\sqrt{\text{x}^4+\text{a}^4}}$ $\int\frac{\text{x}\text{ dx}}{\sqrt{(\text{x}^2)^2+(\text{a}^2)^2}}$ Let $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\text{x}\text{ dx}=\frac{\text{dt}}{2}$Now, $\int\frac{\text{x}\text{ dx}}{\sqrt{(\text{x}^2)^2+(\text{a}^2)^2}}$
$=\frac{1}{2}\int\frac{\text{x}\text{ dx}}{\sqrt{{\text{t}^2+(\text{a}^2)^2}}}$
$=\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2+\text{a}^4}\Big|+\text{C}$
$=\frac{1}{2}\log\Big|\text{x}^2+\sqrt{\text{x}^4+\text{a}^4}\Big|+\text{C}$
View full question & answer→Question 2543 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
Answer$\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
Let $\frac{1}{\text{x}}=\text{t}$
$\Rightarrow-\frac{1}{\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}^2}\text{ dx}=-\text{dt}$
Now, $\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
$=-\int\cos^2\text{t}\ \text{dt}$
$=-\int\Big(\frac{1+\cos2\text{t}}{2}\Big)\text{dt}$
$=-\frac{1}{2}\int(1+\cos2\text{t})\text{dt}$
$=-\frac{1}{2}\Big[\text{t}+\frac{\sin2\text{t}}{2}\Big]+\text{C}$
$=-\frac{1}{2}\Bigg[\frac{1}{\text{x}}+\frac{\sin\big(\frac{2}{\text{x}}\big)}{2}\Bigg]+\text{C}$
$=-\frac{1}{2}\Big(\frac{1}{\text{x}}\Big)-\frac{1}{4}\sin\Big(\frac{2}{\text{x}}\Big)+\text{C}$
View full question & answer→Question 2553 Marks
Evaluate the following integrals:$\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$
AnswerLet $\text{I}=\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta \text{d}\theta$
$=\int\sin^{-1}(3\sin\theta-4\sin^3\theta)\cos\theta\text{d}\theta$
$=\int\sin^{-1}(\sin3\theta)\cos\theta\text{d}\theta$
$=\int3\theta\cos\theta\text{d}\theta$
$=3[\theta\int\cos\theta\text{d}\theta-\int(1\int\cos\theta\text{d}\theta)\text{d}\theta]$
$=3[\theta\sin\theta-\int\sin\theta\text{d}\theta]$
$=3[\theta\sin\theta+\cos\theta]+\text{C}$
$\text{I}=3\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]+\text{C}$
View full question & answer→Question 2563 Marks
Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
Consider,
$\text{X}=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)+\text{B}$
$\text{x}=\text{A}(2\text{x}+1)+\text{B}$
$\Rightarrow\text{x}=(2\text{A})\text{x}+\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{2}$
And
$\text{A}+\text{B}=0$
$\Rightarrow\frac{1}{2}+\text{B}=0$
$\Rightarrow\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\frac{\big(\frac{1}{2}(2\text{x}+1)-\frac{1}{2}\big)}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
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