Question 13 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 23 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 33 Marks
Evaluate the following intregals:
$\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
Answer Let $\text{I}=\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\sqrt{\frac{(1-\text{x})(1-\text{x})}{(1+\text{x})(1-\text{x})}}\text{dx}$
$=\int\Big(\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}-\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}$
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 43 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 53 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 63 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 73 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 83 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}$$\int\frac{1}{\sqrt{2\sin^2\text{x}}}\text{dx}\ \big[\because 1-\cos 2\text{x}=2\sin^2\text{x}\big]$
$=\frac{1}{\sqrt{2}}\int\text{cosec x dx}$
$=\frac{1}{\sqrt{2}}\text{ln}|\text{cosec x}-\cot\text{x}|=\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big|+\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\bigg|\frac{2\sin^2\frac{\text{x}}{2}}{\sin\text{x}}\bigg|+\text{C} \Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Bigg|\frac{2\sin^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\Bigg|+\text{C}\ \Big[\because\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 93 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\sin\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\sin\text{x}}\text{dx}$$=\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 103 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
Answer Let $\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 113 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 123 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 133 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 143 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 153 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 163 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 173 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 183 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 193 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 203 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 213 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 223 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 233 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 243 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 253 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 263 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 273 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 283 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 293 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 303 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 313 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 323 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 333 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 343 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 353 Marks
Evaluate the following integrals:$\int\frac{1}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-4}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\Big(\text{x}^{\frac{1}{3}}\Big)^2-2^2}}$
Let $\text{x}^{\frac{1}{3}}=\text{t}$
$\Rightarrow\frac{1}{3}\text{x}^{\frac{-2}{3}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{3\text{x}^{\frac{2}{3}}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}^{\frac{2}{3}}}=3\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=3\int\frac{\text{dt}}{\sqrt{\text{t}^2-2^2}}$
$=3\log\Big|\text{t}+\sqrt{\text{t}^2-2^2}\Big|+\text{C}$
$=3\log\Bigg|\text{x}^{\frac{1}{3}}+\sqrt{\text{x}^{\frac{2}{3}}-4}\Bigg|+\text{C}$
View full question & answer→Question 363 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 373 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 383 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 393 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 403 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 413 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 423 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 433 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 443 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 453 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 463 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 473 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 483 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 493 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}$
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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,
$\text{I}=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\int\frac{\text{dt}}{\text{t}}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|\text{t}|+\text{C}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|1+\text{x}^2|+\text{C}$
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