Question 12 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^5}\text{dx}$
Answer$\int\text{x}^{-5}\text{dx}$
$=\frac{\text{x}^{-5+1}}{-5+1}+\text{c}$
$=-\frac{1}{4}\text{x}^{-4}+\text{c}$
$=-\frac{1}{4\text{x}^4}+\text{c}$
View full question & answer→Question 22 Marks
Evaluate:
$\int\frac{\text{e}^{6\log_\text{e}\text{x}}-\text{e}^{5\log_\text{e}\text{x}}}{\text{e}^{4\log_\text{e}\text{x}}-\text{e}^{3\log_\text{e}\text{x}}}\text{dx}$
Answer$\int\frac{\text{e}^{6\log_\text{e}\text{x}}-\text{e}^{5\log_\text{e}\text{x}}}{\text{e}^{4\log_\text{e}\text{x}}-\text{e}^{3\log_\text{e}\text{x}}}\text{dx}=\int\frac{\text{x}^6-\text{x}^5}{\text{x}^4-\text{x}^3}\text{dx}$
$=\int\frac{\text{x}^5(\text{x}-1)}{\text{x}^3(\text{x}-1)}\text{dx}$
$=\int\text{x}^2\text{dx}$
$=\frac{\text{x}^3}{3}+\text{C}$
View full question & answer→Question 32 Marks
Evaluate the following integrals:
$\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$
Answer$\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$
$=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
View full question & answer→Question 42 Marks
Evaluate $\int\frac{\text{x}^3-\text{x}^2+\text{x}+1}{\text{x}-1}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3-\text{x}^2+\text{x}+1}{\text{x}-1}\text{ dx}$
$=\int\frac{\text{x}^2(\text{x}-1)+(\text{x}-1)}{(\text{x}-1)}\text{ dx}$
$=\int\text{x}^2\text{dx}+\int\text{dx}$
$=\frac{\text{x}^3}{3}+\text{x}+\text{C}$
View full question & answer→Question 52 Marks
Evaluate the following integrals:
$\int(3\text{x}+4)^2\text{dx}$
Answer$\int(3\text{x}+4)^2\text{dx}$
$=\int(9\text{x}^2+2\times3\text{x}\times4+16)\text{dx}$
$=9\int\text{x}^2\text{dx}+24\int\text{x dx}+16\int\text{dx}$
$=9\Big[\frac{\text{x}^3}{3}\Big]+24\Big[\frac{\text{x}^2}{2}\Big]+16\text{x}+\text{C}$
$=3\text{x}^3+12\text{x}^2+16\text{x}+\text{C}$
View full question & answer→Question 62 Marks
Evaluate:
$\int\Big(\frac{2\cos^2\text{x}-\cos2\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
Answer$\int\Big(\frac{2\cos^2\text{x}-\cos2\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{2\cos^2\text{x}-(2\cos^2\text{x}-1)}{\cos^2\text{x}}\Big)\text{dx}$ $[\because\cos2\text{x}=2\cos^2\text{x}-1]$
$=\int\text{sec}^2\text{x}\text{ dx}$
$=\tan\text{x}+\text{c}$
View full question & answer→Question 72 Marks
Evaluate the following integrals:
$\int\frac{\sin^2\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\frac{\sin^2\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{(1-\cos^2\text{x})}{(1+\cos\text{x})}\text{dx}$
$=\int\frac{(1-\cos\text{x})(1+\cos\text{x})}{(1+\cos\text{x})}\text{dx}$
$=\int(1-\cos\text{x})\text{dx}$
$=\text{x}-\sin\text{x}+\text{C}$
View full question & answer→Question 82 Marks
Evaluate the following integrals:
$\int\frac{\sin^3\text{x}-\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$
Answer$\int\frac{\sin^3\text{x}-\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$
$=\int\bigg(\frac{\sin^3\text{x}}{\sin\text{x}^2\cos^2\text{x}}-\frac{\cos^3\text{x}}{\sin\text{x}^2\cos^2\text{x}}\bigg)\text{dx}$
$=\int(\sin\text{x}\sec^2\text{x}-\cos\text{x }\text{cosec}^2\text{x})\text{dx}$
$=\int(\tan\text{x}\sec\text{x}-\cot\text{x }\text{cosec x})\text{dx}$
$=\sec\text{x}+\text{cosec x}+\text{C}$
View full question & answer→Question 92 Marks
Evaluate the following integrals:$\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}$
Let $\tan\text{e}^{\text{x}}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\text{dx = dt}$
So, $\text{I}=\int\frac{\text{dt}}{1+\text{t}^{2}}$
$=\tan^{-1}(\text{t})+\text{C}$ $\big[\text{Since,}\int\frac{1}{1+\text{x}^2}\text{dx}=\tan^{-1}\text{x+c}\big]$
$\text{I}=\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
View full question & answer→Question 102 Marks
Write a value of $\int\sqrt{9+\text{x}^2}\text{ dx}$
Answer$\int\sqrt{9+\text{x}^2}\text{ dx}$
$=\int\sqrt{3^2+\text{x}^2}\text{ dx}$ $\Big(\because\sqrt{\text{a}^2+\text{x}^2}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\ln\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Big)$
$=\frac{\text{x}}{2}\sqrt{9+\text{x}^2}+\frac{9}{2}\ln\Big|\text{x}+\sqrt{9+\text{x}^2}\Big|+\text{C}$
View full question & answer→Question 112 Marks
Write a value of $\int\cos^4\text{x }\sin\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\cos^4\text{x }\sin\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\text{I}=-\int\text{t}^{4}\text{ dt}$
$=-\frac{\text{t}^5}{5}+\text{C}$
$\text{I}=\frac{\cos^5\text{x}}{5}+\text{C}$
View full question & answer→Question 122 Marks
$\int\cos^2\frac{\text{x}}{2}\text{dx}$
Answer$\int\cos^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\frac{1+\cos\text{x}}{2}\Big)\text{dx}$ $\Big[\therefore\cos^2\frac{\text{x}}{2}=\frac{1+\cos\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1+\cos\text{x})\text{dx}$
$=\frac{1}{2}[\text{x}+\sin\text{x}]+\text{C}$
View full question & answer→Question 132 Marks
Evalute the following integrals:
$\int\frac{\text{e}^{3\text{x}}}{\text{e}^{3\text{x}}+1}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{3\text{x}}}{\text{e}^{3\text{x}}+1}\text{dx}$ then,
Putting $e^{3x} + 1 = t$
$\Rightarrow3\text{e}^{3\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3\text{e}^{3\text{x}}}$
$\therefore\text{I}=\int\frac{\text{e}^{3\text{x}}}{3\text{t}(\text{e}^{3\text{x}})}\text{dt}$
$=\frac{1}{3}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{\text{In}|\text{t}|}{3}+\text{C}$
$=\frac{\text{In}|\text{e}^{3\text{x}}+1|}{3}+\text{C}$
View full question & answer→Question 142 Marks
Evaluate the following integrals:$\int\cos\sqrt{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos\sqrt{\text{x}}\text{dx}$
$\sqrt{\text{x}}=\text{t}$
$\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$=\int2\text{t}\cos\text{t dt}$
$\text{I}=2\int\text{t}\cos\text{t dt}$
$\text{I}=2[\text{t}\int\cos\text{t dt}-\int(1\int\cos\text{t dt})\text{dt}]$
$=2[\text{t}\sin\text{t}-\int\sin\text{t dt}]$
$=2[\text{t}\sin\text{t}+\cos\text{t}]+\text{C}$
$\text{I}=2[\sqrt{\text{x}}\sin\sqrt{\text{x}}+\cos\sqrt{\text{x}}]+\text{C}$
View full question & answer→Question 152 Marks
Evaluate the following integrals:
$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$
Answer$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$ Let $\tan^{-1}\text{x}=\text{t}$ $\Rightarrow\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$ Now, $\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$$=\int\sin\text{t dt}$
$=-\cos(\text{t})+\text{C}$
$=-\cos\big(\tan^{-1}\text{x}\big)+\text{C}$
View full question & answer→Question 162 Marks
Evaluate the following integrals:
$\int\sin^5\text{x}\cos\text{x dx}$
Answer$\int\sin^5\text{x}\cos\text{x dx}$
$\text{Let }\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\text{Now,}\int\sin^5\text{x}\cos\text{x dx}$
$=\int\text{t}^5\text{dt}$
$=\frac{\text{t}^6}{6}+\text{C}$
$=\frac{1}{6}\sin^6\text{x}+\text{C}$
View full question & answer→Question 172 Marks
Evaluate the following integrals:
$\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
Answer$\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
$\text{Let }1+\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\text{Now,}\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
$=\int\sqrt{\text{t}}\text{ dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{2}{3}(1+\text{e}^\text{x})^\frac{3}{2}+\text{C}$
View full question & answer→Question 182 Marks
Evaluate the following integrals:
$\int\bigg (2^{\text{x}}+\frac{5}{\text{x}}-\frac{1}{\text{x}^{\frac{1}{3}}}\bigg)\text{dx}$
Answer $\int\bigg(2^{\text{x}}+\frac{5}{\text{x}}-\frac{1}{\text{x}^{\frac{1}{3}}}\bigg)\text{dx}$
$=\int2^{\text{x}}\text{dx}+5\int\frac{1}{\text{x}}\text{dx}-\int\frac{1}{\text{x}^{\frac{1}{3}}}\text{dx}$
$=\frac{2^{\text{x}}}{\log2}+5\log\text{x}-\frac{3}{2}\text{x}^{\frac{2}{3}}+\text{C}$
View full question & answer→Question 192 Marks
Write tha value of $\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{dx}$
Answer$\int\sec\text{x}(\sec\text{x}+\tan\text{x})\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 202 Marks
Write a value of $\int\tan\text{x}\sec^3\text{x dx}$
AnswerLet $\text{I}=\int\tan\text{x}\sec^3\text{x dx}$
Let $\sec\text{x}=\text{t}$
$\sec\text{x}\tan\text{x dx}=\text{dt}$
$\text{dx}=\frac{\text{dt}}{\sec\text{x}\tan\text{x}}$
$\therefore\ \text{I}=\int\sec^2\text{x}\tan\text{x dx}$
$=\int\text{t}^2\text{ dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$=\frac{\sec^3\text{x}}{3}+\text{C}$
$\therefore\ \text{I}=\frac{\sec^3\text{x}}{3}+\text{C}$
View full question & answer→Question 212 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\text{dx}$
Answer$\int\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\text{dx}$
$=\int\sqrt{\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}}\text{dx}$
$\Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2} \ \&\ 1+\cos\text{x}=2\cos^2\frac{\text{x}}{2}\Big]$
$=\int\tan\frac{\text{x}}{2}\text{dx}$
$=-2\text{ln}\Big|\cos\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 222 Marks
Evaluate the following integral:
$\int\frac{1}{\text{a}^2\text{x}^2+\text{b}^2}\text{ dx}$
Answer$\int\frac{1}{\text{a}^2\text{x}^2+\text{b}^2}\text{ dx}$
$=\frac{1}{\text{a}^2}\int\frac{\text{dx}}{\text{x}^2+\big(\frac{\text{b}}{\text{a}}\big)^2}$
$=\frac{1}{\text{a}^2}\times\frac{\text{a}}{\text{b}}\tan^{-1}\bigg(\frac{\text{x}}{\frac{\text{b}}{\text{a}}}\bigg)+\text{C}$ $\Big[\therefore\int\frac{\text{dx}}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$=\frac{1}{\text{ab}}\tan^{-1}\Big(\frac{\text{ax}}{\text{b}}\Big)+\text{C}$
View full question & answer→Question 232 Marks
Evaluate $\int\frac{\text{x}^3-1}{\text{x}^2}\text{ dx}$
Answer$\int\Big(\frac{\text{x}^3-1}{\text{x}^2}\Big)\text{ dx}$
$=\int\Big(\frac{\text{x}^3}{\text{x}^2}-\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\int(\text{x}-\text{x}^{-2})\text{dx}$
$=\frac{\text{x}^2}{2}-\frac{\text{x}^{-2+1}}{-2+1}+\text{C}$
$=\frac{\text{x}^2}{2}+\frac{1}{\text{x}}+\text{C}$
View full question & answer→Question 242 Marks
Evaluate the following integrals:
$\int\tan^3\text{x}\sec^2\text{x}\text{dx}$
Answer$\int\tan^3\text{x}\sec^2\text{x}\text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\text{dx}=\text{dt}$
Now, $\int\tan^3\text{x}\sec^2\text{x}\text{dx}$
$=\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^4\text{x}}{4}+\text{C}$
View full question & answer→Question 252 Marks
Evaluate $\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Let $\sqrt{\text{x}}=\text{t}$
$\frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=2\int\sin\text{t dt}$
$=-2\cos\text{t}+\text{C}$
$\text{I}=-2\cos\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 262 Marks
Evaluate $\int\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta$
$\therefore\ \text{I}=\int\frac{\cos\theta}{\cos\theta}\text{ d}\theta$
$=\int\text{d}\theta$
$=\theta+\text{C}$
$=\sin^{-1}\text{x}+\text{C}$ $(\because\text{ x}=\sin\theta)$
View full question & answer→Question 272 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{4-\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{2^2-\text{t}^2}}$
$=\sin^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 282 Marks
Evaluate the following integrals:
$\int\text{x}\cos\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\cos\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\cos\text{x dx}-\int(1\times\int\cos\text{x dx})\text{dx+C}$
$=\text{x}\sin\text{x}-\int\sin\text{x dx+C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x+C}$
View full question & answer→Question 292 Marks
Write a value of $\int\sin^3\text{x}\cos\text{x dx}$
AnswerLet $\text{I}=\int\sin^3\text{x}\cos\text{x dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^3\text{ dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\sin^4\text{x}}{4}+\text{C}$ $(\because\text{t}=\sin\text{x})$
View full question & answer→Question 302 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{\text{x}}(\sqrt{\text{x}}+1)}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}}(\sqrt{\text{x}}+1)}\text{dx}$
Putting $\sqrt{\text{x}}+1=\text{t}$
$\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{\text{x}}}\text{dx}=2\text{dt}$
$\therefore\text{I}=2\int\frac{1}{\text{t}}\text{dt}$
$=2\text{ In }|\text{t}|+\text{C}$
$=2\text{ In }|\sqrt{\text{x}}+1|+\text{C }\big[\because\text{t}=\sqrt{\text{x}}+1\big]$
View full question & answer→Question 312 Marks
Evalute the following integrals:
$\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$
Putting $\cot\text{x}=\text{t}$
$\Rightarrow-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{cosec}^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{1+\text{t}}$
$=-\text{ln}|1+\text{t}|+\text{C}$
$=-\text{ln}|1+\cot\text{x}|+\text{C}\ \big[\because\text{t}=\cot\text{x}\big]$
View full question & answer→Question 322 Marks
Evaluate the following integrals:
$\int\Big\{3\sin\text{x}-4\cos\text{x}+\frac{5}{\cos^2\text{x}}-\frac{6}{\sin^2\text{x}}+\tan^2\text{x}-\cot^2\text{x}\Big\}\text{dx}$
Answer$\int\Big\{3\sin\text{x}-4\cos\text{x}+\frac{5}{\cos^2\text{x}}-\frac{6}{\sin^2\text{x}}+\tan^2\text{x}-\cot^2\text{x}\Big\}\text{dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+5\int\sec^2\text{dx}\\-6\int\text{cosec}^2\text{x}+\int\tan^2\text{x dx}-\int\cot^2\text{x dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+5\int\sec^2\text{x dx}\\-6\int\text{cosec}^2\text{x}+\int(\sec^2\text{x}-1)\text{dx}-\int(\text{cosec}^2\text{x}-1)\text{dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+6\int\sec^2\text{x dx}-7\int\text{cosec}^2\text{x dx}$
$=-3\cos\text{x}-4\sin\text{x}+6\tan\text{x}+7\cot\text{x}+\text{C}$
View full question & answer→Question 332 Marks
Evaluate the following integrals:
$\int\frac{1}{1+\cos2\text{x}}\text{dx}$
Answer$\int\frac{1}{1+\cos(2\text{x})}\text{dx}$ $\Big[\therefore\ 1+\cos\theta=2\cos^2\Big(\frac{\theta}{2}\Big)\Big]$
$=\int\frac{\text{dx}}{2\cos^2\text{x}}$
$=\frac{1}{2}\int\sec^2\text{x dx}$
$=\frac{1}{2}\tan\text{x}+\text{C}$
View full question & answer→Question 342 Marks
Evaluate the following integral:
$\int\frac{\text{x}^2-1}{\text{x}^2+4}\text{ dx}$
Answer$\int\frac{\text{x}^2-1}{\text{x}^2+4}\text{ dx}$
$=\int\Big(\frac{\text{x}^2+4-4-1}{\text{x}^2+4}\Big)\text{ dx}$
$=\int\Big(\frac{\text{x}^2+4}{\text{x}^2+4}\Big)\text{ dx}-5\int\frac{\text{dx}}{\text{x}^2+2^2}$
$=\int\text{dx}-5\int\frac{\text{dx}}{\text{x}^2+2^2}$
$=\text{x}-\frac{5}{2}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$$\Big[\therefore\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
View full question & answer→Question 352 Marks
$\int\sin^2(2\text{x}+5)\text{dx}$
Answer$\int\sin^2(2\text{x}+5)\text{dx}$
$=\int\Big(\frac{1-\cos(4\text{x}+10)}{2}\Big)\text{dx}$ $\Big[\therefore\sin^2\text{A}=\frac{1-\cos2\text{A}}{2}\Big]$
$=\frac{1}{2}\int(1-\cos(4\text{x}+10))\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin(4\text{x}+10)}{4}\Big]+\text{C}$
$=\frac{1}{2}\text{x}-\frac{\sin(4\text{x}+10)}{8}+\text{C}$
View full question & answer→Question 362 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^\text{x}}{(1+\text{e}^\text{x})^2}\text{dx}$
Answer$\int\frac{\text{e}^\text{x}}{(1+\text{e}^\text{x})^2}\text{dx}$
$\text{Let}\ 1+\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\text{Now},\int\frac{\text{e}^\text{x}\text{dx}}{(1+\text{e}^\text{x})^2}$
$=\int\frac{\text{dt}}{\text{t}^2}$
$=\int\text{t}^{-2}\text{dt}$
$=\frac{\text{t}^{-2}+1}{-2+1}+\text{C}$
$=\frac{-1}{\text{t}}+\text{C}$
$=-\frac{1}{1+\text{e}^\text{x}}+\text{C}$
View full question & answer→Question 372 Marks
$\int\tan^2(2\text{x}-3)\text{dx}$
Answer$\int\tan^2(2\text{x}-3)\text{dx}$
$=\int[\sec^2(2\text{x}-3)-1]\text{dx}$
$=\int\sec^2(2\text{x}-3)\text{dx}-\int1\text{dx}$
$=\frac{\tan(2\text{x}-3)}{2}-\text{x}+\text{c}$
View full question & answer→Question 382 Marks
Evaluate the following integrals:$\int\frac{\text{e}^\text{x}}{\sqrt{16-\text{e}^{2\text{x}}}}\text{ dx}$
Answer$\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$ Let $\text{e}^\text{x}=\text{t}$ $\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$$=\int\frac{\text{dt}}{\sqrt{16-\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{4^2-\text{t}^2}}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{e}^\text{x}}{4}\Big)+\text{C}$
View full question & answer→Question 392 Marks
Evaluate the following integrals:$\int\frac{\sec^2\text{x}}{\sqrt{4+\tan^2\text{x}}}\text{ dx}$
AnswerLet $\tan\text{x}=\text{t}$ $\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{ dx}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+2^2}}$$=\log\Big|\text{t}+\sqrt{\text{t}^2+4}\Big|+\text{C}$
$=\log\Big|\tan+\sqrt{\tan^2\text{x}+4}\Big|+\text{C}$
View full question & answer→Question 402 Marks
$\int\sin^2\frac{\text{x}}{2}\text{dx}$
AnswerLet I $=\int\sin^2\frac{\text{x}}{2}\text{dx}.$ Then,
$\text{I}=\frac{1}{2}\int2\sin^2\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int(1-\cos\text{x})\text{dx}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=\frac{1}{2}\int\text{dx}-\frac{1}{2}\int\cos\text{xdx}$
$=\frac{1}{2}\times\text{x}-\frac{1}{2}\times\sin\text{x}+\text{C}$
$=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$
$\therefore\text{I}=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$
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Evalute the following integrals:
$\int\frac{\text{e}^\text{x}+1}{\text{e}^\text{x}+\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^\text{x}+1}{\text{e}^\text{x}+\text{x}}\text{dx}\ .....(\text{i})$
let $e^x + x = t$ then,
$d(e^x + x) = dt$
$\Rightarrow (e^x + x)dx = dt$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{e}^\text{x}+1}$
Putting $e^x + x = t$ and $\text{dx}=\frac{\text{dt}}{\text{e}^\text{x}+1}$ in equation (i), we get,
$\text{I}=\int\frac{\text{e}^\text{x}+1}{\text{e}}\times\frac{\text{dt}}{\text{e}^\text{x}+1}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{e}^\text{x}+\text{x}|+\text{C}$
$\therefore\text{I}=\log|\text{e}^\text{x}+\text{x}|+\text{C}$
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Write a value of $\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
Let $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{t}^3}{1}\text{dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$\text{I}=\frac{(\tan^{-1}\text{x})^4}{4}+\text{C}$
View full question & answer→Question 432 Marks
Evaluate $\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Let $\sqrt{\text{x}}=\text{t}$
$\frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
$\text{dx}=2\sqrt{\text{x}}\text{ dt}$
$\therefore\ \text{I}=2\cos\text{t}+\text{dt}$
$\text{I}=2\sin\sqrt{\text{x}}+\text{C}$
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Evaluate:
$\int\sqrt{\frac{1+\cos\ 2\text{x}}{2}}\text{dx}$
Answer$\int\sqrt{\frac{1+\cos\ 2\text{x}}{2}}\text{dx}$
$\int\sqrt{\frac{2\cos^2\text{x}}{2}}\text{dx}\ \ [\therefore1+\cos2\text{A}=2\cos^2\text{A]}$
$=\int\cos\text{x dx}$
$=\sin\text{x}+\text{c}$
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Write a value of $\int\sqrt{4-\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int\sqrt{4-\text{x}^2}\text{ dx}$
We know that,
$\int\sqrt{\text{a}^2-\text{x}^2}=\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
Thus here a = 2
$\therefore\ \text{I}=\frac{\text{x}}{2}\sqrt{4-\text{x}^2}+2\sin^{-1}\frac{\text{x}}{2}+\text{C}$
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Evalute the following integrals:
$\int\sqrt{\frac{1+\cos2\text{x}}{1-\cos2\text{x}}}\text{dx}$
Answer$\int\sqrt{\frac{1+\cos2\text{x}}{1-\cos2\text{x}}}\text{dx}$
$=\int\sqrt{\frac{2\cos^2\text{x}}{2\sin^2\text{x}}}\text{dx}$
$=\int\cot\text{x dx}$
$=\text{ln}|\sin\text{x}|+\text{C}$
View full question & answer→Question 472 Marks
Write a value of $\int\frac{\sec^2\text{x}}{(5+\tan\text{x})^4}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sec^2\text{x}}{(5+\tan\text{x})^4}\text{ dx}$
Let $5+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}^4}$
$=-\frac{1}{3\text{t}^3}+\text{C}$
$\text{I}=-\frac{1}{3(5+\tan\text{x})^3}+\text{C}$
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Evaluate the following integrals:$\int\text{x}^{\text{n}}.\log\text{x dx}$
Answer$\int\text{x}^{\text{n}}\log\text{x dx}$
Taking $log\ x$ as the first function and $x^n$ as the second function.
$=\log\text{x}\int\text{x}^\text{n}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}\log\text{x}\int\text{x}^\text{n}\text{dx}\Big)\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{1}{\text{x}}\bigg(\frac{\text{x}^{n+1}}{\text{n}+1}\bigg)\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{\text{x}^{\text{n}}}{\text{n}+1}\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{\text{x}^{\text{n}+1}}{(\text{n}+1)^2}+\text{C}$
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Evaluate the following integral:
$\int\frac{1}{\text{a}^2\text{x}^2-\text{b}^2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{a}^2\text{x}^2-\text{b}^2}\text{ dx}$
$=\frac{1}{\text{a}^2}\int\frac{1}{\text{x}^2-\frac{\text{b}^2}{\text{a}^2}}\text{ dx}$
$=\frac{1}{\text{a}^2}\int\frac{1}{\text{x}^2-\big(\frac{\text{b}}{\text{a}}\big)^2}\text{ dx}$
$\text{I}=\frac{1}{\text{a}^2}\times\frac{1}{2\times\big(\frac{\text{b}}{\text{a}}\big)}\times\Bigg|\log\frac{\text{x}-\frac{\text{b}}{\text{a}}}{\text{x}+\frac{\text{b}}{\text{a}}}\Bigg|+\text{C}$ $\Big[$Since $\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$
$\text{I}=\frac{1}{2\text{ab}}\log\Big|\frac{\text{ax}-\text{b}}{\text{ax}+\text{b}}\Big|+\text{C}$
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Evaluate the following integrals:
$\int\text{e}^{\cos^2\text{x}}\sin2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{\cos^2\text{x}}\sin2\text{x}\text{ dx}$
Let $\cos^2\text{x}=\text{t}$
On differentiating both sides, we get
$-2\cos\text{x}\sin\text{x}\text{ dx}=\text{dt}$
$\therefore\text{I}=\int\text{e}^\text{t}2\sin\text{x}\cos\text{x}\frac{\text{dt}}{-2\sin\text{x}\cos\text{x}}$
$=-\int\text{e}^\text{t}\text{dt}$
$=-\text{e}^\text{t}+\text{C}$
$=-\text{e}^{\cos^2\text{x}}+\text{C}$
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