Exercise 11.7 - Chapter 11 - Integral Calculus - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 11.7
Integrate the following with respect to $\mathrm{x}$

Question 1.
(i) $9 \mathrm{xe}^{3 \mathrm{x}}$
(ii) $x \sin 3 x$
(iii) $25 \mathrm{xe}^{-5 \mathrm{x}}$
(iv) $\mathrm{x} \sec \mathrm{x} \tan \mathrm{x}$
Solution:
(i) $\int 9 x e^{3 x} d x$
$
\begin{aligned}
I & =\int 9 x e^{3 x} d x \\
& =9 \int x e^{3 x} d x
\end{aligned}
$
Applying Bernoulli's formula
$
\begin{aligned}
\int u d v & =u v-u^{\prime} v_1+\ldots \\
I & =9\left[\frac{x e^{3 x}}{3}-(1) \frac{\left(e^{3 x}\right)}{9}\right]
\end{aligned}
$
$
\begin{aligned}
=\frac{9}{3}\left(x e^{3 x}-\frac{e^{3 x}}{3}\right) & =3\left(x e^{3 x}-\frac{e^{3 x}}{3}\right) \\
& =3 x e^{3 x}-e^{3 x}=e^{3 x}(3 x-1)+c
\end{aligned}
$

(ii) $\mathrm{I}=\int x \sin 3 x d x$
Applying Bernoulli's formula

$
\begin{aligned}
\int u d v & =u v-u^{\prime} v_1+\ldots \\
I & =(x)\left(\frac{-\cos 3 x}{3}\right)-(1)\left(\frac{-\sin 3 x}{9}\right)+c \\
& =\frac{-x \cos 3 x}{3}+\frac{\sin 3 x}{9}+c .
\end{aligned}
$
(iii) $\mathrm{I}=\int 25 x e^{-5 x} d x=25 \int x e^{-5 x} d x$
Applying Bernoulli's formula
$
\int u d v=u v-u^{\prime} v_1+\ldots
$

$
\begin{aligned}
\mathrm{I} & =25\left[(x)\left(\frac{e^{-5 x}}{-5}\right)-(1)\left(\frac{e^{-5 x}}{25}\right)\right]+c \\
& =-5 x e^{-5 x}-e^{-5 x}+c \\
& =-e^{-5 x}[5 x+1]+c
\end{aligned}
$
(iv) $\mathrm{I}=\int x \sec x \tan x d x$
Applying Bernoulli's formula
$
\begin{aligned}
\int u d v & =u v-u^{\prime} v_1+\ldots \\
\mathrm{I} & =x \sec x-\log |\sec x+\tan x|+c .
\end{aligned}
$

Question 2.
(i) $\mathrm{x} \log \mathrm{x}$
(ii) $27 \mathrm{x}^2 \mathrm{e}^{3 \mathrm{x}}$
(iii) $x^2 \cos x$
(iv) $x^3 \sin x$
Solution
(iii) $\mathrm{I}=\int 25 x e^{-5 x} d x=25 \int x e^{-5 x} d x$
Applying Bernoulli's formula
$
\int u d v=u v-u^{\prime} v_1+\ldots
$

$
d u=\frac{1}{x} d x \quad \Rightarrow \quad v=\frac{x^2}{2}
$
Applying Integration by parts,
$
\begin{aligned}
\int u d v & =u v-\int v d u \\
& =(\log x)\left(\frac{x^2}{2}\right)-\int\left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) d x \\
& =\frac{x^2}{2} \log |x|-\frac{x^2}{4}+c
\end{aligned}
$

(ii) I $\mathrm{I}=\int 27 x^2 e^{3 x} d x=27 \int x^2 e^{3 x} d x$
Applying Bernoulli's formula

$
\begin{aligned}
\int u d v & =u v-u^{\prime} v_1+u^{\prime \prime} v_2-\ldots \\
& =27\left[\frac{x^2 e^{3 x}}{3}-\frac{2 x e^{3 x}}{9}+\frac{2 e^{3 x}}{27}\right]+c \\
& =27 e^{3 x}\left[\frac{x^2}{3}-\frac{2 x}{9}+\frac{2}{27}\right]+c \\
& =e^{3 x}\left[9 x^2-6 x+2\right]+c .
\end{aligned}
$
(iii) $\mathrm{I}=\int x^2 \cos x d x$
Applying Bernoulli's formula

$
\begin{aligned}
\int u d v= & u v-u^{\prime} v_1+u^{\prime \prime} v_2-\ldots \\
= & \left(x^2\right)(\sin x)-(2 x)(-\cos x) \\
& +(2)(-\sin x)+c \\
= & x^2 \sin x+2 x \cos x-2 \sin x+c
\end{aligned}
$
(iv) $\mathrm{I}=\int x^3 \sin x d x$
Applying Integration by parts,

$
\begin{aligned}
\int u d v= & u v-u^{\prime} v_1+u^{\prime \prime} v_2-\ldots \\
= & \left(x^3\right)(-\cos x)-\left(3 x^2\right)(-\sin x) \\
& +(6 x)(\cos x)-(6) \sin x+c \\
= & -x^3 \cos x+3 x^2 \sin x \\
& +6 x \cos x-6 \sin x+c
\end{aligned}
$

Question 3.
(i) $\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}$
(ii) $\mathrm{x}^2 \mathrm{e}^{x^2}$
(iii) $\tan ^{-1}\left(\frac{8 x}{1-16 x^2}\right)$
(iv) $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$
Solution:

(i) $\mathrm{I}=\int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x$
$
\begin{aligned}
& t=\sin ^{-1} x \\
& \Rightarrow \quad \sin t=x \\
& \frac{d t}{d x}=\frac{1}{\sqrt{1-x^2}} \\
& \sqrt{1-x^2} d t=d x \\
&
\end{aligned}
$

Sub (2) and (3) in (1)
$
\begin{aligned}
& \Rightarrow \int \frac{x t}{\sqrt{1-x^2}} \sqrt{1-x^2} d t \\
&=\int t \sin t d t
\end{aligned}
$
Applying Bernoulli's formula
$
\begin{aligned}
& =(t)(-\cos t)-(1)(-\sin t)+c \\
& =-t \cos t+\sin t+c
\end{aligned}
$

$
\text { since, } \begin{aligned}
\sin t & =\sin \left(\sin ^{-1} x\right)=x \\
\cos t & =\sqrt{1-\sin ^2 t}=\sqrt{1-x^2}
\end{aligned}
$
(ii) $\mathrm{I}=\int x^5 e^{x^2} d x$
D.w.r. to ' $x$ '
$
\begin{aligned}
& \frac{d t}{d x}=2 x \\
& \frac{d t}{2 x}=d x
\end{aligned}
$
Sub (2) and (3) in (1)
$
\begin{aligned}
\Rightarrow \quad \int x^5 e^t \frac{d t}{2 x} & =\frac{1}{2} \int x^4 e^t d t \\
& =\frac{1}{2} \int t^2 e^t d t
\end{aligned}
$
Applying Bernoulli's formula
$
\begin{array}{rc} 
& d v=e^t d t \\
u=t^2, & v=e^t \\
u^{\prime}=2 t, & v_1=e^t \\
u^{\prime \prime}=2, & v_2=e^t
\end{array}
$
$
=\frac{1}{2}\left[t^2 e^t-2 t e^t+2 e^t\right]+c
$

$
\begin{aligned}
& =\frac{e^t}{2}\left[t^2-2 t+2\right]+c \\
& =\frac{e^{x^2}}{2}\left[x^4-2 x^2+2\right]+c
\end{aligned}
$
$
\begin{aligned}
& I=\int \tan ^{-1}\left(\frac{8 x}{1-16 x^2}\right) d x \\
& \text { put } 4 x=\tan \theta \\
& 4 d x=\sec ^2 \theta d \theta \\
& I=\int \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\left(\frac{\sec ^2 \theta d \theta}{4}\right) \\
& =\frac{1}{4} \int \tan ^{-1}(\tan 2 \theta) \sec ^2 \theta d \theta \\
& =\frac{1}{4} \int 2 \theta \sec ^2 \theta d \theta \\
& =\frac{1}{2} \int \theta \sec ^2 \theta d \theta \\
&
\end{aligned}
$
(iii)
Applying by integration parts

$
\begin{aligned}
& =\frac{1}{2}\left[\theta \tan \theta-\int \tan \theta d \theta\right] \\
& =\frac{1}{2}[\theta \tan \theta-\log |\sec \theta|]+c \\
\int \tan ^{-1}( & \left.\frac{8 x}{1-16 x^2}\right) d x \\
& =\frac{1}{2}\left[4 x \tan ^{-1}(4 x)-\log \left|\sqrt{1+\tan ^2 \theta}\right|\right]+c \\
& =\frac{1}{2}\left[4 x \tan ^{-1}(4 x)-\log \left|\sqrt{1+16 x^2}\right|\right]+c
\end{aligned}
$
(iv) $\mathrm{I}=\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$
$
\text { put } \begin{aligned}
x & =\tan \theta \\
d x & =\sec ^2 \theta d \theta
\end{aligned}
$

$\begin{aligned}
\mathrm{I} & =\int \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \sec ^2 \theta d \theta \\
& =\int \sin ^{-1}(\sin 2 \theta) \sec ^2 \theta d \theta
\end{aligned}$

$
\begin{aligned}
& =\int 2 \theta \sec ^2 \theta d \theta \\
& =2 \int(\theta)\left(\sec ^2 \theta d \theta\right) \\
& I=2\left[\theta \tan \theta-\int \tan \theta d \theta\right] \\
& =2 \theta \tan \theta-2 \log |\sec \theta|+c \quad\left(\sec \theta=\sqrt{1+\tan ^2 \theta}\right) \\
& \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=2 x \tan ^{-1} x-2 \log \left|\sqrt{1+x^2}\right|+c \\
& =2\left[x \tan ^{-1} x-\log \left|\sqrt{1+x^2}\right|\right]+c \\
&
\end{aligned}
$