Exercise 7.6 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download

Ex 7.6 Question 1.

$x \sin x$

Answer.

$\int x \sin x d x$
$
=x \int \sin x d x-\int\left(\frac{d}{d x} x \int \sin x d x\right) d x
$
[Applying product rule]
$
\begin{aligned}
& =x(-\cos x)-\int 1(-\cos x) d x \\
& =-x \cos x-\int-\cos x d x \\
& =-x \cos x+\int \cos x d x \\
& =-x \cos x+\sin x+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 2.

$x \sin 3 x$

Answer.

$\int x \sin 3 x d x$
$
=x \int \sin 3 x d x-\int\left(\frac{d}{d x} x \int \sin 3 x d x\right) d x
$
[Applying product rule]
$
=x\left(\frac{-\cos 3 x}{3}\right)-\int 1\left(\frac{-\cos 3 x}{3}\right) d x+c
$

$\begin{aligned}
& =\frac{-1}{3} x \cos 3 x+\frac{1}{3} \int \cos 3 x d x+c \\
& =\frac{-1}{3} x \cos 3 x+\frac{1}{3} \frac{\sin 3 x}{3}+c \\
& =\frac{-1}{3} x \cos 3 x+\frac{1}{9} \sin 3 x+c \\
& =-\frac{x}{3} x \cos 3 x+\frac{1}{9} \sin 3 x+c \text { Ans. }
\end{aligned}$

Ex 7.6 Question 3.

$x^2 e^x$

Answer.

$\int x^2 e^x d x$
$
=x^2 \int e^x d x-\int\left[\frac{d}{d x} x^2 \int e^x d x\right] d x
$
[Applying product rule]
$
\begin{aligned}
& =x^2 e^x-\int 2 x e^x d x \\
& =x^2 e^x-2 \int x e^x d x \\
& =x^2 e^x-2\left[x \int e^x d x-\int\left\{\frac{d}{d x} x \int e^x d x\right\} d x\right]
\end{aligned}
$
[Again applying product rule]
$
\begin{aligned}
& =x^2 e^x-2\left(x e^x-\int 1 \cdot e^x d x\right) \\
& =x^2 e^x-2\left(x e^x-\int e^x d x\right) \\
& =x^2 e^x-2 x e^x+2 \int e^x d x \\
& =x^2 e^x-2 x e^x+2 e^x+c
\end{aligned}
$

$=e^x\left(x^2-2 x+2\right)+c \text { Ans. }$

Ex 7.6 Question 4.

$x \log x$

Answer.

$\int x \log x d x$
$
=\int(\log x) x d x
$
[Applying product rule]
$
\begin{aligned}
& =(\log x) \int x d x-\int\left[\frac{d}{d x} \log x \int x d x\right] d x \\
& =(\log x) \frac{x^2}{2}-\int \frac{1}{x} \cdot \frac{x^2}{2} d x \\
& =\frac{1}{2} x^2 \log x-\frac{1}{2} \int x d x \\
& =\frac{1}{2} x^2 \log x-\frac{1}{2} \frac{x^2}{2}+c \\
& =\frac{x^2}{2} \log x-\frac{x^2}{4}+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 5.

$x \log 2 x$
$
\begin{aligned}
& \text { Answer. } \int x \log 2 x d x \\
& =\int(\log 2 x) x d x \\
& =(\log 2 x) \int x d x-\int\left[\frac{d}{d x} \log 2 x \int x d x\right] d x
\end{aligned}
$
[Applying product rule]

$\begin{aligned}
& =(\log 2 x) \frac{x^2}{2}-\int \frac{1}{2 x} \cdot 2 \cdot \frac{x^2}{2} d x \\
& =\frac{1}{2} x^2 \log 2 x-\frac{1}{2} \int x d x \\
& =\frac{1}{2} x^2 \log 2 x-\frac{1}{2} \frac{x^2}{2}+c \\
& =\frac{x^2}{2} \log 2 x-\frac{x^4}{4}+c \text { Ans. }
\end{aligned}$

Ex 7.6 Question 6.

$x^2 \log x$

Answer.

$\int x^2 \log x d x$
$
\begin{aligned}
& =\int(\log x) x^2 d x \\
& =\log x \int x^2 d x-\int\left(\frac{d}{d x} \log x \int x^2 d x\right) d x
\end{aligned}
$
[Applying product rule]
$
\begin{aligned}
& =(\log x) \frac{x^3}{3}-\int \frac{1}{x} \cdot \frac{x^3}{3} d x \\
& =\frac{x^3}{3} \log x-\frac{1}{3} \int x^2 d x \\
& =\frac{x^3}{3} \log x-\frac{1}{3} \frac{x^3}{3}+c \\
& =\frac{x^3}{3} \log x-\frac{x^3}{9}+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 7.

$\text {} x \sin ^{-1} x$

Answer.

Let $\mathrm{I}=\int x \sin ^{-1} x d x$
Putting $x=\sin \theta$
$
\begin{aligned}
& \Rightarrow d x=\cos \theta d \theta \\
& \therefore \mathrm{I}=\int \sin \theta \cdot \theta \cdot \cos \theta d \theta \\
& =\frac{1}{2} \int \theta \cdot 2 \sin \theta \cos \theta d \theta \\
& =\frac{1}{2} \int \theta \cdot \sin 2 \theta d \theta \\
& =\frac{1}{2}\left[\theta\left(\frac{-\cos 2 \theta}{2}\right)-\int 1 \cdot\left(\frac{-\cos 2 \theta}{2}\right) d \theta\right]
\end{aligned}
$
[Integrating by parts]
$
\begin{aligned}
& =\frac{1}{4}\left[-\theta \cos 2 \theta+\int \cos 2 \theta d \theta\right] \\
& =\frac{1}{4}\left[-\theta \cos 2 \theta+\frac{\sin 2 \theta}{2}\right]+c \\
& =\frac{1}{4}\left[-\theta\left(1-2 \sin ^2 \theta\right)+\sin \theta \cos \theta\right]+c \\
& =\frac{1}{4}\left[-\sin ^{-1} x \cdot\left(1-2 x^2\right)+x \sqrt{1-x^2}\right]+c \\
& =\frac{1}{4}\left(2 x^2-1\right) \sin ^{-1} x+\frac{x \sqrt{1-x^2}}{4}+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 8.

$x \tan ^{-1} x$

Answer.

Let $\mathrm{I}=\int x \tan ^{-1} x d x$

$\begin{aligned}
& =\int\left(\tan ^{-1} x\right) \cdot x d x \\
& =\left(\tan ^{-1} x\right) \cdot \frac{x^2}{2}-\int \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x \\
& =\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2}{1+x^2} d x \\
& =\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1-1}{x^2+1} d x \\
& =\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int\left(1-\frac{1}{x^2+1}\right) d x \\
& =\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c \\
& =\frac{1}{2}\left[x^2 \tan ^{-1} x-x+\tan ^{-1} x\right]+c \\
& =\frac{x^2}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x+c \text { Ans. }
\end{aligned}$

Ex 7.6 Question 9.

$x \cos ^{-1} x$

Answer.

Let $\mathrm{I}=\int x \cos ^{-1} x d x$ $\qquad$
Putting $\cos ^{-1} x=\theta$
$
\begin{aligned}
& \Rightarrow x=\cos \theta \\
& \Rightarrow \frac{d x}{d \theta}=-\sin \theta \\
& \Rightarrow d x=-\sin \theta d \theta
\end{aligned}
$
$\therefore$ From eq. (i),

$
\begin{aligned}
\mathrm{I} & =\int(\cos \theta) \cdot \theta \cdot(-\sin \theta d \theta) \\
& =\frac{-1}{2} \int \theta \cdot(2 \sin \theta \cos \theta) d \theta \\
& =-\frac{1}{2} \int \theta \cdot \sin 2 \theta d \theta
\end{aligned}
$
[Applying product rule]
$
\begin{aligned}
& =\frac{-1}{2}\left[\theta\left(\frac{-\cos 2 \theta}{2}\right)-\int 1\left(\frac{-\cos 2 \theta}{2}\right) d \theta\right] \\
& =\frac{-1}{2}\left[\frac{-1}{2} \theta \cdot \cos 2 \theta+\frac{1}{2} \int \cos 2 \theta d \theta\right] \\
& =\frac{1}{4} \theta \cdot \cos 2 \theta-\frac{1}{4}\left(\frac{\sin 2 \theta}{2}\right)+c \\
& =\frac{1}{4} \theta \cdot \cos 2 \theta-\frac{1}{8}(2 \sin \theta \cos \theta)+c \\
& =\frac{1}{4} \theta \cdot\left(2 \cos ^2 \theta-1\right)-\frac{1}{4} \sqrt{1-\cos ^2 \theta} \cdot \cos \theta+c
\end{aligned}
$

Putting $\cos \theta=x$ and $\theta=\cos ^{-1} x$
$
\begin{aligned}
& =\frac{1}{4}\left(\cos ^{-1} x\right) \cdot\left(2 x^2-1\right)-\frac{1}{4} \sqrt{1-x^2} x+c \\
& =\left(2 x^2-1\right) \frac{\cos ^{-1} x}{4}-\frac{x}{4} \sqrt{1-x^2}+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 10.

$\left(\sin ^{-1} x\right)^2$

Answer.

Putting $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$
$
\therefore \int\left(\sin ^{-1} x\right)^2 d x
$

$
=\int \theta^2 \cos \theta d \theta
$
[Applying product rule]
$
\begin{aligned}
& =\theta^2 \sin \theta-\int 2 \theta \sin \theta d \theta \\
& =\theta^2 \sin \theta-2 \int \theta \sin \theta d \theta
\end{aligned}
$
[Again applying product rule]
$
\begin{aligned}
& =\theta^2 \sin \theta-2\left[\theta(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta\right] \\
& =\theta^2 \sin \theta+2 \theta \cos \theta-2 \int \cos \theta d \theta \\
& =\theta^2 \sin \theta+2 \theta \cos \theta-2 \sin \theta+c \\
& =x\left(\sin ^{-1} x\right)^2+2 \sqrt{1-x^2} \sin ^{-1} x-2 x+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 11.

$\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}$

Answer.

Let $\mathrm{I}=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^2}} d x$ $\qquad$
Putting $\cos ^{-1} x=\theta$
$
\begin{aligned}
& \Rightarrow x=\cos \theta \\
& \Rightarrow \frac{d x}{d \theta}=-\sin \theta \\
& \Rightarrow d x=-\sin \theta d \theta \\
& \therefore \text { From eq. (i), }
\end{aligned}
$
$\therefore$ From eq. (i),

$
\begin{aligned}
\mathrm{I} & =\int \frac{(\cos \theta) \theta}{\sqrt{1-\cos ^2 \theta}}(-\sin \theta d \theta) \\
& =-\int \frac{\theta \cdot \cos \theta \cdot \sin \theta}{\sin \theta} d \theta \\
& =-\int \theta \cdot \cos \theta d \theta
\end{aligned}
$
[Applying product rule]
$
\begin{aligned}
& =-\left[\theta \cdot \sin \theta-\int 1 \cdot \sin \theta d \theta\right] \\
& =-\theta \sin \theta+\int \sin \theta d \theta \\
& =-\theta \sin \theta-\cos \theta+c \\
& =-\theta \sqrt{1-\cos ^2 \theta}-\cos \theta+c \\
& =-\left(\cos ^{-1} x\right) \sqrt{1-x^2}-x+c \\
& =-\left[\sqrt{1-x^2} \cos ^{-1} x+x\right]+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 12.

$x \sec ^2 x$

Answer.
$\int x \sec ^2 x d x$
[Applying product rule]
$
\begin{aligned}
& =x \int \sec ^2 x d x-\int\left[\frac{d}{d x} x \int \sec ^2 x d x\right] d x \\
& =x \tan x-\int 1 \cdot \tan x d x \\
& =x \tan x-\int \tan x d x \\
& =x \tan x-(-\log |\cos x|)+c
\end{aligned}
$

$=x \tan x+\log |\cos x|+c$ Ans.
Ex 7.6 Question 13.

$\tan ^{-1} x$
$
\begin{aligned}
& \text { Answer. Let } \mathrm{I}=\tan ^{-1} x d x \\
& =\int\left(\tan ^{-1} x\right) \cdot 1 d x \\
& =\tan ^{-1} x x-\int \frac{1}{1+x^2} x d x \\
& =x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^2} d x \\
& =x \tan ^{-1} x-\frac{1}{2} \log \left|\left(1+x^2\right)\right|+c \\
& {\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|\right]} \\
& =x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^2\right)+c
\end{aligned}
$

Ex 7.6 Question 14.

$x(\log x)^2$

Answer.

$\int x(\log x)^2 d x$
$
\begin{aligned}
& =\int(\log x)^2 x d x \\
& =(\log x)^2 \int x d x-\int\left[\frac{d}{d x}(\log x)^2 \int x d x\right] d x
\end{aligned}
$

$=(\log x)^2 \frac{x^2}{2}-\int 2(\log x) \frac{d}{d x}(\log x) \cdot \frac{x^2}{2} d x$

$\begin{aligned}
& =(\log x)^2 \frac{x^2}{2}-\int \frac{2(\log x)}{x} \cdot \frac{x^2}{2} d x \\
& =\frac{x^2}{2}(\log x)^2-\int(\log x) x d x \\
& =\frac{x^2}{2}(\log x)^2-\left[(\log x) \frac{x^2}{2}-\int\left(\frac{1}{x} \frac{x^2}{2}\right) d x\right]+c \\
& =\frac{x^2}{2}(\log x)^2-\frac{x^2}{2} \log x+\frac{1}{2} \int x d x+c \\
& =\frac{x^2}{2}(\log x)^2-\frac{x^2}{2} \log x+\frac{x^2}{4}+c \text { Ans. }
\end{aligned}$

Ex 7.6 Question 15.

$\left(x^2+1\right) \log x$
$
\begin{aligned}
& \text { Answer. } \int\left(x^2+1\right) \log x d x \\
& =\int(\log x)\left(x^2+1\right) d x
\end{aligned}
$
[Applying product rule]
$
\begin{aligned}
& =\log x\left(\frac{x^3}{3}+x\right)-\int \frac{1}{x}\left(\frac{x^3}{3}+x\right) d x \\
& =\left(\frac{x^3}{3}+x\right) \log x-\int\left(\frac{x^2}{3}+1\right) d x \\
& =\left(\frac{x^3}{3}+x\right) \log x-\frac{1}{3} \int x^2 d x-\int 1 d x \\
& =\left(\frac{x^3}{3}+x\right) \log x-\frac{1}{3} \frac{x^3}{3}-x+c
\end{aligned}
$

$=\left(\frac{x^3}{3}+x\right) \log x-\frac{x^3}{9}-x+c \text { Ans. }$

Ex 7.6 Question 16.

$e^x(\sin x+\cos x)$

Answer.

Let $\mathrm{I}=\int e^x(\sin x+\cos x) d x$
$
\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\right]
$

It is in the form of $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$ since here $f(x)=\sin x$ and $f^{\prime}(x)=\cos x$
$
\begin{aligned}
& \therefore \mathrm{I}=e^x \sin x+c \\
& {\left[\because \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+c\right]}
\end{aligned}
$

Ex 7.6 Question 17.

$\frac{x e^x}{(1+x)^2}$

Answer.

Let $\mathrm{I}=\int \frac{x e^x}{(x+1)^2} d x$
$
\begin{aligned}
& =\int \frac{(x+1)-1}{(x+1)^2} e^x d x \\
& =\int e^x\left[\frac{x+1}{(x+1)^2}-\frac{1}{(x+1)^2}\right] d x \\
& \mathrm{I}=\int e^x\left[\frac{1}{x+1}+\frac{-1}{(x+1)^2}\right] d x
\end{aligned}
$

$
\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\right]
$

It is in the form of $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$ since here $f(x)=\frac{1}{x+1}$ and
$
\begin{aligned}
& f^{\prime}(x)=\frac{-1}{(x+1)^2} \\
& \mathrm{I}=\frac{e^x}{x+1}+c \\
& {\left[\because \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+c\right]}
\end{aligned}
$

Ex 7.6 Question 18.

$e^x\left(\frac{1+\sin x}{1+\cos x}\right)$

Answer.

Let $\mathrm{I}=\int e^x \cdot \frac{1+\sin x}{1+\cos x} d x$
$
\begin{aligned}
& =\int e^x \cdot \frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x \\
& =\int e^x \cdot\left[\frac{1}{2 \cos ^2 \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right] d x \\
& =\int e^x\left(\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right) d x \\
& =\int e^x\left(\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x \\
& {\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\right]}
\end{aligned}
$

It is in the form of $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$ since here $f(x)=\tan \frac{x}{2}$ and
$
\begin{aligned}
& f^{\prime}(x)=\frac{1}{2} \sec ^2 \frac{x}{2} \\
& =e^x \tan \frac{x}{2}+c \\
& {\left[\because \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+c\right]}
\end{aligned}
$

Ex 7.6 Question 19.

$e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)$

Answer.

Let $\mathrm{I}=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$
$
\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\right]
$

It is in the form of $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$ since here $f(x)=\frac{1}{x}=x^{-1}$ and $f^{\prime}(x)=\frac{-1}{x^2}$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=e^x \frac{1}{x}+c \\
& =\frac{e^x}{x}+c \\
& {\left[\because \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+c\right]}
\end{aligned}
$

Ex 7.6 Question 20.

$\frac{(x-3) e^x}{(x-1)^3}$

Answer.

Let $\mathrm{I}=\int \frac{(x-3) e^x}{(x-1)^3} d x$

$
\begin{aligned}
& =\int \frac{(x-1)-2}{(x-1)^3} e^x d x \\
& =\int e^x\left[\frac{x-1}{(x-1)^3}-\frac{2}{(x-1)^3}\right] d x \\
& \Rightarrow \mathrm{I}=\int e^x\left[\frac{1}{(x-1)^2}+\frac{-2}{(x-1)^3}\right] d x \\
& {\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\right]}
\end{aligned}
$

It is in the form of $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$ since here $f(x)=\frac{1}{(x-1)^2}$ and
$
\begin{aligned}
& f^{\prime}(x)=\frac{d}{d x}\left\{(x-1)^{-2}\right\} \\
& =-2(x-1)^{-3} \\
& =\frac{-2}{(x-1)^3} . \\
& \Rightarrow \mathrm{I}=\frac{e^x}{(x-1)^2}+c \text { Ans. } \\
& {\left[\because\left\{e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+c\right]\right.}
\end{aligned}
$

Ex 7.6 Question 21.

$e^{2 x} \sin x$

Answer.

Let $\mathrm{I}=\int e^{2 x} \sin x d x$
[Applying product rule]
$
=e^{2 x}(-\cos x)-\int e^{2 x} \cdot 2 \cdot(-\cos x) d x
$

$
\Rightarrow \mathrm{I}=-e^{2 x} \cos x+2 \int e^{2 x} \cos x d x
$
[Again applying product rule]
$
\begin{aligned}
& \Rightarrow \mathrm{I}=-e^{2 x} \cos x+2\left[e^{2 x} \sin x-\int 2 e^{2 x} \sin x d x\right] \\
& \Rightarrow \mathrm{I}=-e^{2 x} \cos x+2 e^{2 x} \sin x-4 \int e^{2 x} \sin x d x \\
& =e^{2 x}(-\cos x+2 \sin x)-4 \mathrm{I} \\
& \Rightarrow 5 \mathrm{I}=e^{2 x}(-\cos x+2 \sin x) \\
& \Rightarrow \mathrm{I}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+c \text { Ans. }
\end{aligned}
$

Ex 7.6 Question 22.

$\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$

Answer.

Putting $x=\tan \theta$
$
\begin{aligned}
& \Rightarrow d x=\sec ^2 \theta d \theta \\
& \therefore \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x \\
& =\int \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \cdot \sec ^2 \theta d \theta \\
& =\int \sin ^{-1}(\sin 2 \theta) \cdot \sec ^2 \theta d \theta \\
& =\int 2 \theta \sec ^2 \theta d \theta \\
& =2 \int \theta \sec ^2 \theta d \theta
\end{aligned}
$
[Applying product rule]

$\begin{aligned}
& =2\left[\theta \cdot \tan \theta-\int 1 \cdot \tan \theta d \theta\right] \\
& =2\left[\theta \tan \theta-\int \tan \theta d \theta\right] \\
& =2[\theta \tan \theta-\log \sec \theta]+c \\
& =2\left[\tan ^{-1} x x-\log \sqrt{1+x^2}\right]+c \\
& =2\left[x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^2\right)\right]+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c \text { Ans. }
\end{aligned}$

Ex 7.6 Question 23.

$\int x^2 e^{x^4} d x$ equals to
(A) $\frac{1}{3} e^{x^3}+\mathrm{C}$
(B) $\frac{1}{3} e^{x^2}+\mathrm{C}$
(C) $\frac{1}{2} e^{x^3}+C$
(D) $\frac{1}{2} e^{x^2}+\mathrm{C}$

Answer.

Let $\mathrm{I}=\int x^2 e^{x^3} d x$
$
=\frac{1}{3} \int e^{\left(x^3\right)}\left(3 x^2\right) d x\left[\because \frac{d}{d x} x^3=3 x^2\right]
$

Putting $x^3=t$

$
\begin{aligned}
& \Rightarrow 3 x^2=\frac{d t}{d x} \\
& \Rightarrow 3 x^2 d x=d t \\
& \therefore \text { From eq. (i), I }=\frac{1}{3} \int e^t d t \\
& =\frac{1}{3} e^t+\mathrm{C} \\
& =\frac{1}{3} e^{x^3}+\mathrm{C}
\end{aligned}
$

Therefore, option (A) is correct.

Ex 7.6 Question 24.

$\int e^x \sec x(1+\tan x) d x$ equals:
(A) $e^x \cos x+\mathrm{C}$
(B) $e^x \sec x+C$
(C) $e^x \sin x+C$
(D) $e^x \tan x+C$

Answer.

Let $\mathrm{I}=\int e^x \sec x(1+\tan x) d x$ $=\int e^x(\sec x+\sec x \tan x) d x$
It is in the form of $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$ since here $f(x)=\sec x$ and
$
\begin{aligned}
& f^{\prime}(x)=\sec x \tan x \\
& \therefore \mathrm{I}=e^x \sec x+c \\
& {\left[\because \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+c\right]}
\end{aligned}
$

Therefore, option (B) is correct.