Question 13 Marks
Evaluate:
$\int\text{x log 2x dx}$.
Answer$\text{I}=\int\log2\text{x}\cdot\text{x dx}=\log2\text{x}\cdot\frac{\text{x}^{2}}{2}-\int\frac{1}{\text{x}}\cdot\frac{\text{x}^{2}}{2}\text{dx}+\text{c}_{1}$
= $\frac{\text{x}^{2}}{2}\cdot\log\text{2x}-\frac{1}{2}\int\text{x dx + c}_{1}=\frac{\text{x}^{2}}{2}\cdot\log\text{ 2x}-\frac{\text{x}^{2}}{4}+\text{c}.$
View full question & answer→Question 23 Marks
Evaluate:
$\int\sin\text{4x}\cos\text{3x dx}$.
AnswerWriting I = $\frac{1}{2}\int2\sin\text{ 4x }\cos\text{ 3x dx}=\frac{1}{2}\int(\sin\text{7x}+\sin\text{x) dx}$
$=\frac{1}{2}\Bigg[\frac{-\cos\text{ 7x}}{7}-\cos\text{x}\Bigg]+\text{c }\text{ OR }-\frac{1}{14}\cos\text{ 7x}-\frac{1}{2}\cos\text{ x + c}$.
View full question & answer→Question 33 Marks
Evaluate:
$\int\frac{\text{2x.tan-1(x}^{2})}{\text{1 + x}^{4}}\text{dx}. $
Answer$\int\frac{\text{2x.tan}^{-1}\text{(x}^{2})}{\text{1 + x}^{4}}\text{dx}=\int\text{t dt }\text{ where tan}^{-1}\text{(x}^{2})=\text{t }\text{ }\therefore\frac{\text{2x}}{\text{1 + x}^{4}}\text{dx}=\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{c}=\frac{1}{2}\Big[\tan^{-1}\text{(x}^{2})\Big]^{2}+\text{c}.$
View full question & answer→Question 43 Marks
Evaluate:$\int \frac{1 + x^{2}}{1 + x^{4}} \text{dx}$
Answer$\text{I} \int \frac{1 + x^{2}}{1 + x^{4}} \text{dx} = \int \frac{1 +\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}}\text{dx} = \int \frac{1 +\frac{1}{x^{2}}}{\bigg( x - \frac{1}{x}\bigg)^{2} + \bigg(\sqrt{2}\bigg)^{2}} \text{dx}$$= \frac{1}{\sqrt{2}} \tan^{-1} \bigg(x-\frac{\frac{1}{x}}{\sqrt{2}}\bigg) + c $
$= \frac{1}{\sqrt{2}} \tan^{-1}\bigg(\frac{x^{2} -1}{\sqrt{2x}}\bigg) + c $
View full question & answer→Question 53 Marks
Evaluate: $\int \cos \text{4 x} \cos 3\text{x dx}$
Answer$\text{I} = \int \cos \text{4 x} \cos 3\text{x dx} = \frac{1}{2} \int(\cos 7x + \cos x) \text{dx}$$= \frac{1}{2} \bigg[ \frac{\sin 7x}{7} + \sin x\bigg] + c$or $\frac{1}{14} \sin 7 x + \frac{1}{2} \sin x + c$
View full question & answer→Question 63 Marks
Evaluate:$\int{\frac{dx} {\sqrt{x^{2} - 3x + 2}}}$
Answer$\int{\frac{dx} {\sqrt{x^{2} - 3x + 2}}}=\int \frac{dx}{\sqrt{\bigg(x - \frac{3}{2}\bigg)^{2} - \bigg(\frac{1}{2}\bigg)^{2}}}$$= \log\bigg| \bigg(x - \frac{3}{2}\bigg) + \sqrt{x^{2} - 3x + 2}\bigg| + c$
View full question & answer→Question 73 Marks
Evaluate:$\int \frac{\sin (\text{x} - \alpha)}{\sin (\text{x} + \alpha)} \text{dx}$
Answer$I = \int \frac{\sin(x + a)-2\alpha)}{\sin(x + \alpha)}$$= \int \frac{\sin(x + \alpha). \cos 2\alpha - \cos (x + \alpha) .2 \alpha}{\sin(x + \alpha)} dx$
$= \cos 2 \alpha \int dx - \sin 2 \alpha \int \frac{\cos(x + \alpha)}{\sin(x + \alpha)} dx$
$= x \cos 2 \alpha - \sin 2 \alpha \log |\sin (x + \alpha)|+c$
View full question & answer→Question 83 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 93 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 103 Marks
Evaluate the following:
$\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{dx}$ $=\int\frac{(\sin\text{x}+\cos\text{x})}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{dx}$ $=\int1\text{dx}=\text{x}+\text{C}$
View full question & answer→Question 113 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{0}^{\frac{\pi}{2}}\cos^{2}\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\cos^{2}\text{x}\ \text{dx}$$\int\cos^{2}\text{x}\ \text{dx}=\int\bigg(\frac{1+\cos2\text{x}}{2}\bigg)\text{dx}=\frac{\text{x}}{2}+\frac{\sin2\text{x}}{4}=\frac{1}{2}\bigg(\text{x}+\frac{\sin2\text{x}}{2}\bigg)=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtin $\text{I}=\bigg[\text{F}\bigg(\frac{\pi}{2}\bigg)-\text{F}(0)\bigg]$$=\frac{1}{2}\Bigg[\bigg(\frac{\pi}{2}-\frac{\sin\pi}{2}\bigg)-\bigg(0+\frac{\sin0}{2}\bigg)\Bigg]$
$=\frac{1}{2}\bigg[\frac{\pi}{2}+0-0-0\bigg]$
$=\frac{\pi}{4}$
View full question & answer→Question 123 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 133 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 143 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 153 Marks
Evaluate the following:
$\int\frac{\cos\text{x}-\cos2\text{x}}{1-\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}-\cos2\text{x}}{1-\cos\text{x}}\text{dx}$
$=\int\frac{2\sin\frac{3\text{x}}{2}\cdot\sin\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\text{dx}$
$=\int\frac{\sin\frac{3\text{x}}{2}}{\sin\frac{\text{x}}{2}}\text{dx}$
$=\int\frac{3\sin\frac{\text{x}}{2}-4\sin^3\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\text{dx}$ $\big[\because\sin3\text{x}=3\sin\text{x}-4\sin^3\text{x}\big]$
$=3\int\text{dx}-4\int\sin^2\frac{\text{x}}{2}\text{dx}$
$=3\int\text{dx}-4\int\frac{1-\cos\text{x}}{2}\text{dx}$ $=\int\text{dx}+2\int\cos\text{xdx}=\text{x}+2\sin\text{x}+\text{C}$
View full question & answer→Question 163 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 173 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 183 Marks
Evaluate the following integrals:
$\int^\limits1_{-1}5\text{x}^4\sqrt{\text{x}^5+1}\text{ dx}$
AnswerLet $\text{I}=\int^\limits1_{-1}5\text{x}^4\sqrt{\text{x}^5+1}\text{ dx}$ Then,
Let $\text{x}^5+1=\text{t}$ Then, $5\text{x}^4\text{ dx}=\text{dt}$
When $\text{x}=-1,\text{t}=0$ and $\text{x}=1,\text{t}=2$
$\therefore\ \text{I}=\int\limits^2_0\sqrt{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{2}{3}\text{t}^{\frac{3}{2}}\Big]^6_0$
$\Rightarrow\text{I}=\frac{2}{3}\sqrt{8}$
$\Rightarrow\text{I}=\frac{4\sqrt{2}}{3}$
View full question & answer→Question 193 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 203 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{1}{2}}\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{1}{2}}\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Put $\text{x}=\sin\theta$
$\therefore\text{ dx}=\cos\theta\text{ d}\theta$
When $\text{x}\rightarrow0,\theta\rightarrow0$
When $\text{x}\rightarrow\frac{1}{2},\theta\rightarrow\frac{\pi}{6}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{6}}\frac{\sin\theta\sin^{-1}(\sin\theta)}{\cos\theta}\cos\theta\text{ d}\theta$
$=\int_{0}^\limits{\frac{\pi}{6}}\theta\sin\theta\text{ d}\theta$
Applying integration by parts, we have
$\text{I}=\big[\theta\big(-\cos\theta\big)\big]^{\frac{\pi}{6}}_0-\int_{0}^\limits{\frac{\pi}{6}}1\times\big(-\cos\theta\big)\text{d}\theta$
$=-\Big(\frac{\pi}{6}\cos\frac{\pi}{6}-0\Big)+\int_{0}^\limits{\frac{\pi}{6}}\cos\theta\text{ d}\theta$
$=-\frac{\pi}{6}\times\frac{\sqrt{3}}{2}+\big[\sin\theta\big]^{\frac{\pi}{6}}_0$
$=-\frac{\pi}{4\sqrt{3}}+\Big(\sin\frac{\pi}{6}-\sin0\Big)$
$=-\frac{\pi}{4\sqrt{3}}+\Big(\frac{1}{2}-0\Big)$
$=\frac{1}{2}-\frac{\pi}{4\sqrt{3}}$
View full question & answer→Question 213 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 223 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 233 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 243 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 253 Marks
Integrate the function in Exercise:
$\frac{\sin\text{x}}{\sin(\text{x}-\text{a)}}$
Answer$\frac{\sin\text{x}}{\sin(\text{x}-\text{a)}}$
$\text{Let}\ \text{x}-\text{a}=\text{t}\Rightarrow\text{dx}=\text{dt}$
$\int\frac{\sin\text{x}}{\sin(\text{x}-\text{a)}}\text{dx}=\int\frac{\sin(\text{t}+\text{a})}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin\text{t}\cos\text{a}+\cos\text{t}\sin\text{a}}{\sin\text{t}}\text{dt}$
$=\int(\cos\text{a}+\cot\text{t}\sin\text{a)}\text{dt}$
$=\text{t}\cos\text{a}+\sin\text{a}\log|\sin\text{t|}+\text{C}_{1}$
$=(\text{x}-\text{a)}\cos\text{a}+\sin\text{a}\log|\sin(\text{x}-\text{a)}|+\text{C}_{1}$
$=\text{x}\cos\text{a}+\sin\text{a}\log|\sin(\text{x}-\text{a)}|-\text{a}\cos\text{a}+\text{C}_{1}$
$=\sin\text{a}\log|\sin(\text{x}-\text{a)}|+\text{x}\cos\text{a}+\text{C}$
View full question & answer→Question 263 Marks
Evaluate the following integrals:
$\int^\limits{\pi}_{0}5\big(5-4\cos\theta\big)^{\frac{1}{4}}\sin\theta\text{ d}\theta$
AnswerLet $\text{I}=\int^\limits{\pi}_{0}5\big(5-4\cos\theta\big)^{\frac{1}{4}}\sin\theta\text{ d}\theta$
Let $\big(5-4\cos\theta\big)=\text{t}$ Then, $4\sin\theta\text{ d}\theta=\text{dt}$
When $\theta=0,\text{t}=1$ and $\theta=\pi,\text{t}=9$
$\therefore\ \text{I}=\frac{5}{4}\int\limits^9_1\text{t}^{\frac{1}{4}}\text{ dt}$
$\Rightarrow\text{I}=\frac{5}{4}\Bigg[\frac{4\text{t}^{\frac{5}{4}}}{5}\Bigg]^9_1$
$\Rightarrow\text{I}=\big(9\sqrt{3}-1\big)$
View full question & answer→Question 273 Marks
Integrate the rational function in exercise:
$\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}$
[Hint: Put sin x = t]
Answer$\text{I}=\int\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}\text{dx}\dots(\text{i})$
Putting $\sin \text{x = t}$
$\Rightarrow \ \cos\text{x}=\frac{\text{dx}}{\text{dt}}$
$\Rightarrow \cos \text{x dx = dt}$
$\therefore$ From eq. (i),
$\text{I}=\int\frac{1}{(1-\text{t})(2-\text{t})}\text{dt}=\int\frac{(2-\text{t})-(1-\text{t})}{(1-\text{t})(2-\text{t})}\text{dt}$
$=\int\Bigg(\frac{(2-\text{t})}{(1-\text{t})(2-\text{t})}-\frac{(1-\text{t})}{(1-\text{t})(2-\text{t})}\Bigg)\text{dt}=\int\Bigg(\frac{1}{(1-\text{t})}-\frac{1}{(2-\text{t})}\Bigg)\text{dt}$
$=\int\frac{1}{1-\text{t}}\text{dt}-\int\frac{1}{2-\text{t}}\text{dt}=\frac{\text{log}|1-\text{t}|}{-1\rightarrow\text{Coeff. of t}}-\frac{\text{log}|2-\text{t}}{-1}+\text{c}$
$=-\text{log}|1-\text{t}|+\text{log}|2-\text{t}|+\text{c}=\text{log}\Bigg|\frac{2-\text{t}}{1-\text{t}}\Bigg|+\text{c}=\text{log}\Bigg|\frac{2-\sin\text{x}}{1-\sin\text{x}}\Bigg|+\text{c}$
View full question & answer→Question 283 Marks
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\text{a}}_{0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}+\sqrt{\text{a}-\text{x}}}}\text{dx}$
Answer$\text{Let}\ \text{I}= \int^{\text{a}}\limits_{0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}+\sqrt{\text{a}-\text{x}}}}\text{dx}\ \ ...(\text{i})$ $\Rightarrow\ \ \ \text{I}=\int^{\text{a}}\limits_{0}\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{a}-\text{(a}-\text{x)}}}\text{dx}=\int^{\text{a}}\limits_{0}\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\text{dx}\ \ ...(\text{ii})$ Adding eq.(i) and (ii),$ 21=\int^{\text{a}}\limits_{0}\bigg(\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}+\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\bigg)\text{dx}=\int^{\text{a}}\limits_{0}\bigg(\frac{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\bigg)\text{dx}=\int^{\text{a}}\limits_{0}1\text{dx}=\text{(x)}^{\text{a}}_{0}=\text{a}$
$\Rightarrow\ \ \text{I}=\frac{\text{a}}{2}$
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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, $\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
$=\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{2}\log\bigg|\frac{1+\text{t}}{1-\text{t}}\bigg|+\text{C}$
$=\frac{1}{2}\log\bigg|\frac{1+\tan\text{x}}{1-\tan\text{x}}\bigg|+\text{C}$
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Find the integrals of the functions in Exercises:
$\cos^42\text{x}$
Answer$\cos^42\text{x}=(\cos^22\text{x})^2$
$=\bigg(\frac{1+\cos4\text{x}}{2}\bigg)^2$
$=\frac{1}{4}\Big[1+\cos^24\text{x}+2\cos4\text{x}\Big]$
$=\frac{1}{4}\Bigg[1+\bigg(\frac{1+\cos8\text{x}}{2}\bigg)+2\cos4\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[1+\frac{1}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[\frac{3}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Bigg]$
$\therefore\int\cos^42\text{x}\text{ dx}=\int\bigg(\frac{3}{8}+\frac{\cos8\text{x}}{8}+\frac{\cos4\text{x}}{2}\bigg)\text{ dx}$
$=\frac{3}{8}\text{x}+\frac{\sin8\text{x}}{64}+\frac{\sin4\text{x}}{8}+\text{C}$
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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, $\int \frac{(\text{e}^{\sin-1_\text{x}})^2}{\sqrt{1-\text{x}^2}}\text{dx}$
$\int \text{e}^{\sin^{-1}\text{z}}\times \frac{\text{e}^{\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{dx}$
$\int \text{t}.\text{dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$=\frac{(\text{e}^{\sin^{-1}\text{x}})^2}{2}+\text{C}$
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Evaluate the following integrals:
$\int\limits^1_{-1}\text{x|x|}\text{dx}$
Answer$|\text{x}|=\begin{cases}-\text{x},&-1<\text{x}<0\\\text{x},&0<\text{x}<1\end{cases}$
$\therefore\ \text{x}|\text{x}|=\begin{cases}-\text{x}^2,&-1<\text{x}<0\\\text{x}^2,&0<\text{x}<1\end{cases}$
Now, $\int\limits^1_{-1}\text{x|x|}\text{dx}$
$=\int\limits^0_{-1}-\text{x}^2\text{ dx}+\int\limits^1_{0}\text{x}^2\text{ dx}$
$=-\int\limits^0_{-1}\text{x}^2\text{ dx}+\int\limits^1_{0}\text{x}^2\text{ dx}$
$=-\Big[\frac{\text{x}^3}{3}\Big]^0_{-1}+\Big[\frac{\text{x}^3}{3}\Big]^1_0$
$=-\Big(0+\frac{1}{3}\Big)+\Big(\frac{1}{3}-0\Big)$
$=0-\frac{1}{3}+\frac{1}{3}-0$
$=0$
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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
$\text{I}=\int3\sqrt{\text{t}^2}\sin\text{x}\times\frac{-\text{dt}}{\sin\text{x}}$
$=-\int\text{t}^\frac{2}{3}\sin\text{x}\frac{\text{dt}}{\sin\text{x}}$
$=-\int\text{t}^\frac{2}{3}\text{dt}$
$=-\frac{3}{5}\times\text{t}^\frac{5}{3}+\text{C}$
$=-\frac{3}{5}(\cos\text{x})^\frac{5}{3}+\text{C}$
$\therefore\text{I}=-\frac{3}{5}(\cos\text{x})^\frac{5}{3}+\text{C}$
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Find the integrals of the functions in Exercises:
$\frac{\sin^2\text{x}}{1+\cos\text{x}}$
Answer$\frac{\sin^2\text{x}}{1+\cos\text{x}}=\frac{\bigg(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\bigg)^2}{2\cos^2\frac{\text{x}}{2}}$ $\ \ \ \ \ \bigg[\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2};\cos\text{x}=2\cos^2\frac{\text{x}}{2}-1\bigg]$ $=\frac{4\sin^2\frac{\text{x}}{2}\cos^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}$ $=2\sin^2\frac{\text{x}}{2}$ $=1-\cos\text{x}$ $\therefore\int\frac{\sin^2\text{x}}{1+\cos\text{x}}\text{ dx}=\int(1-\cos\text{x})\text{ dx}$ $=\text{x}-\sin\text{x}+\text{C}$
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Evaluate the following:
$\int\frac{\text{e}^{6\log\text{x}}-\text{e}^{5\log\text{x}}}{\text{e}^{4\log\text{x}}-\text{e}^{3\log\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{6\log\text{x}}-\text{e}^{5\log\text{x}}}{\text{e}^{4\log\text{x}}-\text{e}^{3\log\text{x}}}\text{dx}$
$=\int\frac{\text{e}^{\log\text{x}^6}-\text{e}^{\log\text{x}^5}}{\text{e}^{\log\text{x}^4}-\text{e}^{3\log\text{x}^3}}\text{dx}$ $\big[\because\ \text{a}\log\text{b}-\text{b}-\log\text{b}^{\text{a}}\big]$
$=\int\frac{\text{x}^6-\text{x}^5}{\text{x}^4-\text{x}^3}\text{dx}$ $\big[\because\ \text{e}^{\log\text{x}}=\text{x}\big]$
$=\int\frac{\text{x}^3-\text{x}^2}{\text{x}-1}\text{dx}=\int\frac{\text{x}^2(\text{x}-1)}{\text{x}-1}\text{dx}$ $=\int\text{x}^2\text{dx}=\frac{\text{x}^3}{3}+\text{C}$
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Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
Let $\cos\theta=\text{t}$ Then, $-\sin\theta\text{ d}\theta=\text{dt}$
When $\theta=0,\text{t}=1$ and $\theta=\frac{\pi}{2},\text{t}=0$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
$=\int_{1}^\limits{0}\frac{-\text{dt}}{\sqrt{1+\text{t}}}$
$=\int_{0}^\limits{1}\frac{\text{dt}}{\sqrt{1+\text{t}}}$
$=2\big[\sqrt{1+\text{t}}\big]^1_0$
$=2\big(\sqrt{2}-1\big)$
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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$\int\frac{\cot\text{x}}{\text{cosec x}-\cot\text{x}}\text{dx}$
$=\int\frac{\frac{\cos\text{x}}{\sin\text{x}}}{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{1-\cos\text{x}}\Big)\times\frac{(1+\cos\text{x})}{(1+\cos\text{x})}\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{1-\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\big[(\cot\text{x}\text{ cosec x})+\cot^2\text{x}\big]\text{dx}$
$=\int\big[\cot\text{x}\text{ cosec x}+\text{cosec}^2\text{x}-1\big]\text{dx}$
$=-\text{cosec x}-\cot\text{x}-\text{x}+\text{C}$
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Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)\text{dx}$
Here, $\text{f(x)}=\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)$
$\text{f}(-\text{x})=\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)$
$=\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)=-\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)=-\text{f(x)}$
Hence f(x) is an odd function
$\therefore\ \text{I}=0$
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Evaluate the following definite integrals:
$\int_{1}^\limits{2}\log\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\log\text{x}\text{ dx}$ Then,
$\text{I}=\int_{1}^\limits{2}1\log\text{x}\text{ dx}$
Integrating by parts.
$\Rightarrow\text{I}=\big[\text{x }\log\text{ x}\big]^2_1-\int_{1}^\limits{2}\frac{1}{\text{x}}\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x }\log\text{ x}\big]^2_1-\int_{1}^\limits{2}\text{dx}$
$\Rightarrow\text{I}=\big[\text{x }\log\text{ x}\big]^2_1-\big[\text{x}\big]^2_1$
$\Rightarrow\text{I}=2\log2-2+1$
$\Rightarrow\text{I}=2\log2-1$
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Evaluate the following integrals:
$\int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}$
AnswerLet $\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=1\Rightarrow\text{t}=1$
$\therefore\ \int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}=\int_{0}^\limits{1}\frac{\text{e}^{\text{t}}\text{ dt}}{2}$
$=\frac{1}{2}\int_{0}^\limits{1}\text{e}^{\text{t}}\text{ dt}$
$=\frac{1}{2}\big[\text{e}^{\text{t}}\big]^1_0$
$=\frac{1}{2}\big[\text{e}^1-\text{e}_0\big]$ $\big[\because\text{e}^0=1\big]$
$=\frac{1}{2}\big(\text{e}-1\big)$
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Evaluate the following integrals:
$\int_{\pi}^\limits{\frac{3\pi}{2}}\sqrt{1-\cos2\text{x}}\text{ dx}$
Answer$\int_{\pi}^\limits{\frac{3\pi}{2}}\sqrt{1-\cos2\text{x}}\text{ dx}$
$=\int_{\pi}^\limits{\frac{3\pi}{2}}\sqrt{2\sin^2\text{x}}\text{ dx}$
$=\sqrt{2}\int_{\pi}^\limits{\frac{3\pi}{2}}\big[\sin\text{x}\big]\text{dx}$
$=\sqrt{2}\int_{\pi}^\limits{\frac{3\pi}{2}}\sin\text{x}\text{ dx}$ $(\sin\text{x}<0\text{ for }\pi\leq\text{x}\leq2\pi)$
$=\sqrt{2}\big[(-\cos\text{x})\big]^{\frac{3\pi}{2}}_\pi$
$=\sqrt{2}\Big(\cos\frac{3\pi}{2}-\cos\pi\Big)$
$=\sqrt{2}\big[0-(-1)\big]$
$=\sqrt{2}\times1$
$=\sqrt{2}$
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Evaluate the following definite integrals:
$\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
AnswerWe have,
$\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
We know that $\int\cot\text{x dx}=\log(\sin\text{x})$
Now, $\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
$=\big[\log(\sin\text{x})\big]^{\frac{\pi}{2}}_\frac{\pi}{4}$
$=\Big[\log\Big(\sin\frac{\pi}{2}\Big)-\log\Big(\sin\frac{\pi}{4}\Big)\Big]$
$=\Big[\log1-\log\frac{1}{\sqrt{2}}\Big]$
$=\big[0-\log\text{2}\big]$
$=\log\sqrt{2}$ $[\because\log\text{a}^{\text{n}}-\text{n}\log\text{a}\big]$
$=\frac{1}{2}\log2$
View full question & answer→Question 433 Marks
Evaluate the following integrals:
$\int\limits^{{\pi}}_0\cos^5\text{x dx}$
AnswerLet $\text{I}=\int\limits^{{\pi}}_0\cos^5\text{x dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(\cos^2\text{x}\big)^2\text{dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(1-\sin^2\text{x}\big)^2\text{dx}$
Let $\sin\text{x}=\text{t},$ then $\cos\text{x dx}=\text{dt}$
When, $\text{x}\rightarrow0;\text{ t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{ t}\rightarrow0$
Therefore,
$\text{I}=\int\limits^0_0\big(1-\text{t}^2\big)^2\text{dt}$
$\text{I}=0$
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Evaluate the following integrals:
$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
Answer$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
$=\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\times\frac{\text{cosec x}+\cot\text{x}}{\text{cosec x}+\cot\text{x}}\times\text{dx}$
$=\int\frac{\text{cosec x}(\text{cosec x}+\cot\text{x})}{\text{cosec}^2\text{x}-\cot^2\text{x}}\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec x}\cot\text{x})\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
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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, $\text{I}=\frac{-1}{3}\cos\text{x}^3+\text{C}$
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$\int(\text{e}^{\text{x}}+1)^2\text{e}^{\text{x}}\text{ dx}$
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 473 Marks
$\int\cos^42\text{x dx}$
Answer$\int\cos^42\text{x dx}$
$=\int(\cos^22\text{x})^2\text{dx}$
$=\int\Big(\frac{1+\cos4\text{x}}{2}\Big)^2\text{dx}$ $\Big[\therefore\cos^2\text{x}=\frac{1+\cos2\text{x}}{2}\Big]$
$=\frac{1}{4}\int(1+\cos4\text{x})^2\text{dx}$
$=\frac{1}{4}\int(1+\cos^24\text{x}+2\cos4\text{x})\text{dx}$
$=\frac{1}{4}\Big[1+\Big(\frac{1+\cos8\text{x}}{2}\Big)+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\int\Big(\frac{3}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\Big[\frac{3\text{x}}{2}+\frac{\sin8\text{x}}{16}+\frac{2\sin4\text{x}}{4}\Big]+\text{C}$
$=\frac{3\text{x}}{8}+\frac{\sin8\text{x}}{64}+\frac{\sin4\text{x}}{8}+\text{C}$
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Evaluate the following integrals:
$\int\limits^8_2|\text{x}-5|\text{dx}$
Answer$\int\limits^8_2|\text{x}-5|\text{dx}$
We know that,
$|\text{x}-5|=\begin{cases}-(\text{x}-5),&2\leq\text{x}\leq5\\\text{x}-5,&5<\text{x}\leq8\end{cases}$
$\therefore\ \text{I}=\int\limits^8_2|\text{x}-5|\text{dx}$
$\Rightarrow\text{I}=\int\limits^5_2-(\text{x}-5)\text{dx}+\int\limits^8_5(\text{x}-5)\text{dx}$
$\Rightarrow\text{I}=-\Big[\frac{\text{x}^2}{2}-5\text{x}\big]^5_2+\Big[\frac{}{}\frac{\text{x}^2}{2}-5\text{x}\Big]^8_5$
$\Rightarrow\text{I}=\frac{-25}{2}+25+2-10+32-40-\frac{25}{2}+25$
$\Rightarrow\text{I}=9$
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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, $\int\frac{\text{dx}}{\text{x}\sqrt{4-9(\log\text{x})^2}}$
$=\int\frac{\text{dt}}{\sqrt{4-9\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{2^2-(3\text{t})^2}}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\text{t}}{2}\Big)+\text{C}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\log\text{x}}{2}\Big)+\text{C}$
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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 $\sqrt{2}\tan\text{x}=\text{t}$
$\sqrt{2}\sec^2\text{x dx}=\text{dt}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\text{t}+\text{C}$
$\text{I}=\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan\text{x})+\text{C}$
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