Question 11 Mark
If $\int_{-2}^3 x^2 d x=k \int_0^2 x^2 d x+\int_2^3 x^2 d x$, then the value of $k$ is
AnswerGiven, $\int_{-2}^3 x^2 d x=k \int_0^2 x^2 d x+\int_2^3 x^2 d x$
$\Rightarrow\left[\frac{x^3}{3}\right]_{-2}^3=k\left[\frac{x^3}{3}\right]_0^2+\left[\frac{x^3}{3}\right]_2^3$
$\Rightarrow \frac{27}{3}+\frac{8}{3}=k\left(\frac{8}{3}\right)+\frac{27}{3}-\frac{8}{3} $
$\Rightarrow \frac{8 k}{3}=\frac{16}{3}$
$ \Rightarrow k=2$
View full question & answer→Question 21 Mark
The value of $\int_0^3 \frac{d x}{\sqrt{9-x^2}}$ is:
AnswerWe have, $\int_0^3 \frac{d x}{\sqrt{9-x^2}}=\left[\sin ^{-1} \frac{x}{3}\right]_0^3=\sin ^{-1} 1-\sin ^{-1} 0=\frac{\pi}{2}$
View full question & answer→Question 31 Mark
The value of $\int_1^e \log x d x$ is
AnswerLet $I=\int_1^e \log x d x$
Using integration by parts, we get $I=[x \log x]_1^e-\int_1^e \frac{1}{x} \cdot x d x$
$I=[x \log x]_1^e-[x]_1^e=e \log e-\log 1-e+1$
$=e-0-e+1=1 \quad[\because \log e=1 \text { and } \log 1=0]$
View full question & answer→Question 41 Mark
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ is equal to :
AnswerLet $I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\sin (\pi / 2-x)-\cos (\pi / 2-x)}{1+\sin (\pi / 2-x) \cos (\pi / 2-x)} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \cdot \sin x} d x$
Adding $(i)$ and $(ii)$, we get
$2 I=\int_0^{\pi / 2} 0 d x=0$
$\Rightarrow I=0$
View full question & answer→Question 51 Mark
Which of these is equal to $\int e^{(x \log 5)} e^x d x$, where $C$ is the constant of integration?
Answer$\frac{(5 e)^x}{\log 5 e}+C$
View full question & answer→Question 61 Mark
For any integer $n$, the value of $\int_{-\pi}^\pi e^{\cos ^2 x} \sin ^3(2 n+1) x d x$ is
Answer$\text {Let } f(x)=e^{\cos ^2 x} \sin ^3(2 n+1) x$
$f(-x)=e^{\cos ^2(-x)} \sin ^3(2 n+1)(-x)$
$=-e^{\cos ^2 x} \sin ^3(2 n+1) x=-f(x)$
$\therefore \int_{-\pi}^\pi e^{\cos ^2 x} \sin ^3(2 n+1) x d x=0 (\because f(-x)=-f(x))$
View full question & answer→Question 71 Mark
If $\frac{d}{d x}(f(x))=\log x$, then $f(x)$ equals:
AnswerWe have, $\frac{d}{d x}(f(x))=\log x$
On integrating both sides, we get
$\int \frac{d}{d x}(f(x))=\int 1 \cdot \log x d x$
$\Rightarrow f(x)=\log x \int 1 \cdot d x-\int\left(\frac{d}{d x}(\log x) \int 1 \cdot d x\right) d x$
${[\text { Integrating by parts }]}$
$\Rightarrow f(x)=x \cdot \log x-\int \frac{1}{x} x x d x \Rightarrow f(x)=x \log x-x+C$
$\Rightarrow f(x)=x(\log x-1)+C$
View full question & answer→Question 81 Mark
$\int_{-1}^1 \frac{|x-2|}{x-2} d x, x \neq 2$ is equal to
AnswerLet $I=\int_{-1}^1 \frac{|x-2|}{x-2} d x$
$
=\int_{-1}^1 \frac{-(x-2)}{x-2} d x=\int_{-1}^1-1 \cdot d x=[-x]_{-1}^1=-[1-(-1)]=-2
$
View full question & answer→Question 91 Mark
$\int_0^{\frac{\pi}{6}} \sec ^2\left(x-\frac{\pi}{6}\right) d x$ is equal to :
Answer$\text {We have, } \int_0^{\pi / 6} \sec ^2\left(x-\frac{\pi}{6}\right) d x$
$=\left[\tan \left(x-\frac{\pi}{6}\right)\right]_0^{\pi / 6}$
$=\tan \left(\frac{\pi}{6}-\frac{\pi}{6}\right)-\tan \left(0-\frac{\pi}{6}\right)$
$=\tan 0^{\circ}-\tan \left(-\frac{\pi}{6}\right) \quad (\because \tan 0^{\circ}=0$ and $\tan (-\theta)=-\tan \theta)$
$=0+\tan (\pi / 6)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$
View full question & answer→Question 101 Mark
$\int_0^4\left(e^{2 x}+x\right) d x$ is equal to
Answer$\text {Let } I=\int_0^4\left(e^{2 x}+x\right) d x=\left[\frac{e^{2 x}}{2}+\frac{x^2}{2}\right]_0^4$
$=\frac{e^8}{2}+\frac{16}{2}-\frac{e^0}{2}-0=\frac{e^8}{2}+\frac{16}{2}-\frac{1}{2}=\frac{e^8+15}{2}$
View full question & answer→Question 111 Mark
The value of $\int_0^{\pi / 6} \sin 3 x\ d x$ is:
AnswerLet $ I=\int_0^{\pi / 6} \sin \ 3 x\ d x$
$=\frac{-1}{3}[\cos 3 x]_0^{\pi / 6}$
$=\frac{-1}{3}\left[\cos \frac{\pi}{2}-\cos 0\right]$
$=\frac{-1}{3}(0-1)=\frac{1}{3}$
View full question & answer→Question 121 Mark
$\int \frac{\sec x}{\sec x-\tan x} d x \text { equals }$
Answer$\text {Let } I=\int \frac{\sec x}{\sec x-\tan x} d x$
$=\int \frac{\sec x(\sec x+\tan x)}{(\sec x-\tan x)(\sec x+\tan x)} d x=\int\left(\frac{\sec ^2 x+\sec x \tan x}{\sec ^2 x-\tan ^2 x}\right) d x$
$=\int \sec ^2 x d x+\int \sec x \tan x d x \quad\left[\because \sec ^2 x-\tan ^2 x=1\right]$
$=\tan x+ \sec x+c$
View full question & answer→Question 131 Mark
$\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$ is equal to
Answer$\text {Let, } I=\int \frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x} d x$
$=\int\left(\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cdot \cos ^2 x}\right) d x \quad\left[\because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta\right]$
$=\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x=\int \operatorname{cosec}^2 x d x-\int \sec ^2 x d x$
$=-\cot x-\tan x+C$
View full question & answer→Question 141 Mark
$\int e^{5 \log x} d x$ is equal to
Answer$\text {Let } I=\int e^{5 \log x} d x$
$=\int e^{\log x^5} d x=\int x^5 d x \quad\left[\because e^{\log x}=x\right]$
$=\frac{x^6}{6}+C$
View full question & answer→Question 151 Mark
$\int_0^{\pi / 8} \tan ^2(2 x) d x$ is equal to
Answer$\text {Let } I=\int_0^{\pi / 8} \tan ^2(2 x) d x=\int_0^{\pi / 8}\left(\sec ^2(2 x)-1\right) d x$
$=\left(\frac{1}{2} \tan 2 x-x\right)_0^{\pi / 8}=\frac{1}{2} \tan 2\left(\frac{\pi}{8}\right)-\frac{\pi}{8}=\frac{1}{2} \tan \frac{\pi}{4}-\frac{\pi}{8}$
$=\frac{1}{2}-\frac{\pi}{8}=\frac{4-\pi}{8}$
View full question & answer→Question 161 Mark
$\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x$ equals
Answer$\text { (d) : Let } I=\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x$
$=\int e^x\left[\frac{1}{x+1}+\log (x+1)\right] d x$
It is of the form $\int e^x\left[f(x)+f^{\prime}(x) d x\right]$,
where $f(x)=\log (x+1)$ and $f^{\prime}(x)=\frac{1}{x+1}$
So, $I=e^x \log (x+1)+C$
View full question & answer→Question 171 Mark
$\int x^2 e^{x^3} d x$ equals
Answer$\text {Let } I=\int x^2 e^{x^3} d x$
$\text { Put } x^3=t$
$\Rightarrow 3 x^2 d x=d t$
$\therefore I=\int e^t \frac{d t}{3}=\frac{1}{3} e^t+C=\frac{1}{3} e^{x^3}+C$
View full question & answer→Question 181 Mark
$\int e^x\left(\frac{x \log x+1}{x}\right) d x$ is equal to
Answer$\text {Let } I=\int e^x\left(\log x+\frac{1}{x}\right) d x$
$\Rightarrow I=e^x \log x+c \quad\left(\because \int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\right)$
View full question & answer→Question 191 Mark
$\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x$ is equal to
Answer$\text {Let } I=\int_{-\pi / 4}^{\pi / 4} \sec ^2 x d x=[\tan x]_{-\pi / 4}^{\pi / 4}$
$=\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)=1+1=2$
View full question & answer→Question 201 Mark
$\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$ is equal to
Answer$\text {Let } I=\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$
$\text { Put } x e^x=t$
$\Rightarrow\left(x e^x+e^x\right) d x=d t$
$\Rightarrow e^x(x+1) d x=d t$
$\therefore I=\int \frac{d t}{\cos ^2 t}=\int \sec ^2 t d t=\tan t+c=\tan \left(x e^x\right)+c$
View full question & answer→Question 211 Mark
Evaluate: $\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x$
Answer(b) : Let $I=\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x$ Put $10^x+x^{10}=t$
$
\begin{aligned}
\Rightarrow \quad & \left(10^x \log _e 10+10 x^9\right) d x=d t \\
\therefore \quad & I=\int \frac{10 x^9+10^x \log _e 10}{10^x+x^{10}} d x=\int \frac{d t}{t} \\
& \quad=\log _e t+C=\log _e\left(10^x+x^{10}\right)+C
\end{aligned}
$
View full question & answer→Question 221 Mark
Ramkali is trying to find the solution of the following definite integrals :
(i) $\int_0^{2 \pi} \frac{d x}{e^{\sin x}+1}$
(ii) $\int_0^1 x^2 d x$
(iii) $\int_0^1 e^x d x$
Which of the above integrals solved by using of definite integral properties?
Answer(a) : (i) Let $I=\int_0^{2 \pi} \frac{d x}{e^{\sin x}+1}$ ....(1)
$\Rightarrow \quad I=\int_0^{2 \pi} \frac{d x}{e^{\sin (2 \pi-x)}+1} \quad\left(\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right)$
$\Rightarrow \quad I=\int_0^{2 \pi} \frac{d x}{e^{-\sin x}+1} \Rightarrow I=\int_0^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x$ ....(2)
Adding (1) and (2), we get
$
2 I=\int_0^{2 \pi} 1 \cdot d x=2 \pi \quad \therefore \quad I=\pi
$
(ii) $\int_0^1 x^2 d x=\left[\frac{x^3}{3}\right]_0^1=\frac{1}{3}$
(iii) $\int_0^1 e^x d x=\left[e^x\right]_0^1=e^1-1$
View full question & answer→Question 231 Mark
Evaluate: $\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x$
Answer$\text { (b) : Let } I=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x$
$\text { Put } x^{3 / 2}=t$
$\Rightarrow \frac{3}{2} x^{1 / 2} d x=d t$
$\therefore I=\frac{2}{3} \int \frac{d t}{\sqrt{a^3-t^2}}=\frac{2}{3} \int \frac{d t}{\sqrt{\left(a^{3 / 2}\right)^2-t^2}}$
$=\frac{2}{3}\left[\sin ^{-1}\left(\frac{t}{a^{3 / 2}}\right)\right]+C=\frac{2}{3}\left[\sin ^{-1}\left(\frac{x^{3 / 2}}{a^{3 / 2}}\right)\right]+C$
$=\frac{2}{3} \sin ^{-1}\left(\frac{x}{a}\right)^{3 / 2}+C$
View full question & answer→Question 241 Mark
Evaluate: $\int \sin ^3 x \cos ^3 x d x$
Answer$\text { (c) : Let } I=\int \sin ^3 x \cos ^3 x d x$
$\Rightarrow I=\frac{1}{8} \int(2 \sin x \cos x)^3 d x$
$\Rightarrow I=\frac{1}{8} \int \sin ^3 2 x d x$
$\Rightarrow I=\frac{1}{8} \int \frac{3 \sin 2 x-\sin 6 x}{4} d x$
$\Rightarrow I=\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
View full question & answer→Question 251 Mark
Evaluate: $\int \frac{1}{\sin x+\sqrt{3} \cos x} d x$
Answer$(a) :$ Let $I=\int \frac{1}{\sin x+\sqrt{3} \cos x} d x$
$=\frac{1}{2} \int \frac{d x}{\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x}$
$\Rightarrow I=\frac{1}{2} \int \frac{1}{\sin \left(x+\frac{\pi}{3}\right)} d x=\frac{1}{2} \int \operatorname{cosec}\left(x+\frac{\pi}{3}\right) d x$
$\Rightarrow I=\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+C$
${\left[\because \int \operatorname{cosec} x d x=\log \left|\tan \frac{x}{2}\right|+C\right]}$
View full question & answer→Question 261 Mark
Evaluate: $\int \frac{x^3-x^2+x-1}{x-1} d x$
Answer$\text { (a) : Let } I=\int \frac{x^3-x^2+x-1}{x-1} d x$
$=\int \frac{x^2(x-1)+1(x-1)}{x-1}=\int \frac{\left(x^2+1\right)(x-1)}{x-1} d x$
$=\int\left(x^2+1\right) d x=\frac{1}{3} x^3+x+C$
View full question & answer→Question 271 Mark
Evaluate : $\int \frac{d x}{\sqrt{x^2-3 x+2}}$
Answer$\text { (b) : We have, } \int \frac{d x}{\sqrt{x^2-3 x+2}}=\int \frac{d x}{\sqrt{\left(x^2-3 x+\frac{9}{4}\right)-\frac{1}{4}}}$
$=\int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}=\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C$
View full question & answer→Question 281 Mark
Evaluate: $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$
Answer(c): Let $\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x$ Put $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$ When, $x=0 \Rightarrow \theta=0$ and $x=1 \Rightarrow \theta=\frac{\pi}{4}$ $I=\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \sec ^2 \theta d \theta$$=\int_0^{\pi / 4} \theta \sin \theta d \theta=[-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta$[Integrating by parts]
$=[-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=\frac{4-\pi}{4 \sqrt{2}}$
View full question & answer→Question 291 Mark
Which of these is equal to $\int_0^1\left\{e^x+\sin \frac{\pi x}{4}\right\} d x ?$
Answer$\text { (b) : We have, } \int_0^1\left\{e^x+\sin \frac{\pi x}{4}\right\} d x$
$=\left[e^x\right]_0^1+\frac{4}{\pi}\left[-\cos \frac{\pi}{4} x\right]_0^1=e-1-\frac{4}{\sqrt{2} \pi}+\frac{4}{\pi}$
$=e-1-\frac{2 \sqrt{2}}{\pi}+\frac{4}{\pi}$
View full question & answer→Question 301 Mark
Which of these is equal to $\int x^2(a x+b)^{-2} d x$, where $C$ is the constant of integration?
Answer14. (a) : Let $I=\int \frac{x^2}{(a x+b)^2} d x$
Put $a x+b=t \Rightarrow d x=\frac{1}{a} d t$
$
\begin{aligned}
\therefore \quad I & =\frac{1}{a^3} \int \frac{(t-b)^2}{t^2} d t=\frac{1}{a^3} \int\left(1+\frac{b^2}{t^2}-\frac{2 b}{t}\right) d t \\
& =\frac{1}{a^3}\left(t-\frac{b^2}{t}-2 b \log t\right)+C \\
& =\frac{1}{a^3}\left(a x+b-\frac{b^2}{a x+b}-2 b \log (a x+b)\right)+C
\end{aligned}
$
View full question & answer→Question 311 Mark
Evaluate: $\int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x$
Answer$\text { (b) : We have } \int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x$
$=5 \int x^3 d x+2 \int x^{-5} d x-7 \int x d x+\int x^{-1 / 2} d x+5 \int \frac{1}{x} d x$
$=5 \cdot \frac{x^4}{4}+2 \cdot \frac{x^{-4}}{(-4)}-7 \cdot \frac{x^2}{2}+\frac{x^{1 / 2}}{(1 / 2)}+5 \log |x|+C$
$=\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C$
View full question & answer→Question 321 Mark
Evaluate: $\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x$
Answer$\text { (c) : Let } I=\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x$
$=\int\left(e^{\log a^x}+e^{\log x^a}+e^{\log a^a}\right) d x=\int\left(a^x+x^a+a^a\right) d x$
$=\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^a x+c \quad\left[\because e^{\log y=y]}\right]$
View full question & answer→Question 331 Mark
Evaluate: $\int \sqrt{(x-3)(5-x)} d x$
Answer$\text { (b) : Let } I=\int \sqrt{(x-3)(5-x)} d x=\int \sqrt{-x^2+8 x-15} d x$
$\Rightarrow I=\int \sqrt{-\left\{x^2-8 x+16-16+15\right\}} d x$
$\Rightarrow I=\int \sqrt{-\left\{(x-4)^2-1^2\right\}} d x=\int \sqrt{1^2-(x-4)^2} d x$
$\Rightarrow I=\frac{1}{2}(x-4) \sqrt{(x-3)(5-x)}+\frac{1}{2} \sin ^{-1}\left(\frac{x-4}{1}\right)+C$
View full question & answer→Question 341 Mark
Which of these is equal to $\int 2^{(x+3)} d x$, where $C$ is the constant of integration?
Answer(c): $\int 2^{(x+3)} d x=\int 2^x \cdot 2^3 d x=8 \int 2^x d x$
$=8 \cdot \frac{2^x}{\log 2}+C=\frac{2^{(x+3)}}{\log 2}+C$
View full question & answer→Question 351 Mark
Using fundamental theorem of calculus, which of following integrals can be solved.
(i) $\int_0^1 x^3 d x$
(ii) $\int x e^x d x$
(iii) $\int_2^3 \frac{1}{x} d x$
Answer(c) : (i) $\int_0^1 x^3 d x=\left[\frac{x^4}{4}\right]_0^1=\frac{1}{4}$
(ii) $\int x e^x d x=x e^x-e^x+C$
(iii) $\int_2^3 \frac{1}{x} d x=[\log x]_2^3=\log \frac{3}{2}$
View full question & answer→Question 361 Mark
Evaluate: $\int_0^2(x-[x]) d x$
Answer$\text { (c) : Let } I=\int_0^2(x-[x]) d x=\int_0^2 x d x-\int_0^2[x] d x$
$=\left[\frac{x^2}{2}\right]_0^2-\int_0^1[x] d x-\int_1^2[x] d x=\frac{4}{2}-\int_0^1 0 d x-\int_1^2 1 d x$
$=2-0-[x]_1^2=2-[2-1]=2-1=1$
View full question & answer→Question 371 Mark
Evaluate: $\int \tan \ x \tan 2 x \tan \ 3 x\ d x$
Answer$(d) $ : Let $I=\int \tan\ x\ \tan\ 2 x \tan \ 3 x\ d x$
Since $, \tan 3 x=\tan (2 x+x)=\frac{\tan \ 2 x+\tan x}{1-\tan x \tan \ 2 x}$
$\Rightarrow \tan \ x \tan 2 x \tan \ 3 x=\tan \ 3 x-\tan \ 2 x-\tan x ...(i)$
$\therefore I=\int(\tan 3 x-\tan 2 x-\tan x) d x \ ($From $(i))$
$=\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C$
View full question & answer→Question 381 Mark
Which of these is equal to $\int x e^{x^2} d x$, where $C$ is the constant of integration?
Answer(b) : Let $I=\int x e^{x^2} d x$
Put $x^2=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$
$\therefore \quad I=\frac{1}{2} \int e^t d t=\frac{e^t}{2}+C=\frac{e^{t^2}}{2}+C$
View full question & answer→Question 391 Mark
Evaluate : $\int_2^4 \frac{x}{x^2+1} d x$
Answer$(a) :$ Let $I=\int_2^4 \frac{x}{x^2+1} d x$
Put $x^2+1=t \Rightarrow 2 x d x=d t$
$\Rightarrow x d x=\frac{1}{2} d t$
Also, $x=2$
$\Rightarrow t=5$ and $x=4$
$\Rightarrow t=17$
$\therefore I=\frac{1}{2} \int_5^{17} \frac{d t}{t}=\frac{1}{2}[\log t]_5^{17}$
$=\frac{1}{2}[\log 17-\log 5]=\frac{1}{2} \log \left(\frac{17}{5}\right)$
View full question & answer→Question 401 Mark
Find the value of $\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$.
Answer$\text { (c) : We have, } \int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$
$=\int\left(\sec ^2 x-\operatorname{cosec}^2 x\right) d x=\tan x+\cot x+C$
View full question & answer→Question 411 Mark
Evaluate: $\int\left(2^x+2^{-x}\right)^2 d x$
Answer$\text { (b) : We have, } \int\left(2^x+2^{-x}\right)^2 d x=\int\left(2^{2 x}+2^{-2 x}+2\right) d x$
$=\frac{2^{2 x}}{(\log 2) \times 2}+\frac{2^{-2 x}}{(\log 2)(-2)}+2 \cdot x+C$
$=\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$
View full question & answer→Question 421 Mark
Evaluate:$\int\left(3 \sin x-2 \cos x+4 \sec ^2 x-5 \operatorname{cosec}^2 x\right) d x$
Answer$\text { (a) : Let } I=\int\left(3 \sin x-2 \cos x+4 \sec ^2 x-5 \operatorname{cosec}^2 x\right) d x$
$\Rightarrow I=3 \int \sin x d x-2 \int \cos x d x+4 \int \sec ^2 x d x-5 \int \operatorname{cosec}^2 x d x$
$\Rightarrow I=-3 \cos x-2 \sin x+4 \tan x+5 \cot x+C$
View full question & answer→Question 431 Mark
Evaluate: $\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$
Answer(a) : Let $I=\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$
$
\Rightarrow \quad I=\int \frac{x\left(1-\frac{1}{x^3}\right)^{\frac{1}{4}}}{x^5} d x=\int \frac{\left(1-\frac{1}{x^3}\right)^{\frac{1}{4}}}{x^4} d x
$
Put $1-\frac{1}{x^3}=t \Rightarrow \frac{3}{x^4} d x=d t$
View full question & answer→Question 441 Mark
Evaluate: $\int \frac{\cot x}{\sqrt[3]{\sin x}} d x$
Answer$\text { (a) : Let } I=\int \frac{\cot x}{\sqrt[3]{\sin x}} d x=\int \frac{\cos x}{\sin ^{1 / 3} x \cdot \sin x} d x$
$=\int \frac{\cos x}{\sin ^{4 / 3} x} d x=\int \sin ^{-4 / 3} x \cdot \cos x d x$
Put $\sin x=t \Rightarrow \cos x d x=d t$
$\Rightarrow I=\int t^{-4 / 3} d t=\frac{t^{-1 / 3}}{-1 / 3}+C=\frac{-3}{\sqrt[3]{\sin x}}+C$
View full question & answer→Question 451 Mark
Evaluate : $\int_0^2 e^{3-4 x} d x$
Answer$\text { (d) : We have, } \int_0^2 e^{3-4 x} d x=\left[\frac{e^{3-4 x}}{-4}\right]_0^2$
$=-\frac{1}{4}\left[e^{3-8}-e^{3-0}\right]=\frac{-1}{4}\left[e^{-5}-e^3\right]$
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Find the value of $\int_{-\pi / 2}^{\pi / 2}|\sin x| d x$.
Answer(c) : $\because|\sin x|$ is an even function.$\therefore \quad \int_{-\pi / 2}^{\pi / 2}|\sin x| d x=2 \int_0^{\pi / 2}|\sin x| d x=2 \int_0^{\pi / 2} \sin x d x$ $=-2[\cos x]_0^{\pi / 2}=-2(0-1)=2$
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Evaluate: $\int \frac{d x}{5-8 x-x^2}$
Answer$\text { (b) : Let } I=\int \frac{d x}{5-8 x-x^2}=\int \frac{d x}{21-(x+4)^2}$
$=\int \frac{d x}{(\sqrt{21})^2-(x+4)^2}$
$=\frac{1}{2 \sqrt{21}} \log \left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C$
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Evaluate: $\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x$
Answer$(c):$ We have,
$ \int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int \frac{\cos ^2(x / 2)-\sin ^2(x / 2)}{\{\cos (x / 2)+\sin (x / 2)\}^3} d x$
$\text { Put } t=\cos \frac{x}{2}+\sin \frac{x}{2}$
$\Rightarrow 2 d t=\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x$
$\Rightarrow \int \frac{\cos (x / 2)-\sin (x / 2)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x=2 \int \frac{1}{t^2} d t$
$\quad=\frac{-2}{t}+C=\frac{-2}{\cos (x / 2)+\sin (x / 2)}+C$
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Evaluate: $\int_{-\pi}^\pi x^{10} \sin ^7 x d x$
Answer(d) : Let $I=\int_{-\pi}^\pi x^{10} \sin ^7 x d x$
Let $f(x)=x^{10} \sin ^7 x$
and $f(-x)=(-x)^{10}[\sin (-x)]^7=-x^{10} \sin ^7 x=-f(x)$
$\therefore f(x)$ is an odd function.
$\therefore \quad I=\int_{-\pi}^\pi x^{10} \sin ^7 x d x=0$
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Evaluate: $\int \frac{\sin x}{1+\sin x} d x$
Answer(d) : Let $I=\int \frac{\sin x}{1+\sin x} d x=\int \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)} d x$
$=\int \frac{\sin x-\sin ^2 x}{\cos ^2 x} d x=\int \sec x \tan x d x-\int \tan ^2 x d x$
$=\int \sec x \tan x d x-\int\left(\sec ^2 x-1\right) d x=\sec x-\tan x+x+C$
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