Exercise 11.7-Additional Questions - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
$
\mathrm{x} \operatorname{cosec}^2 \mathrm{x}
$
Solution:
$
\begin{aligned}
\mathrm{I}=\int x \operatorname{cosec}^2 x d x ; \quad u & =x ; \quad \therefore d u=d x ; \quad \operatorname{cosec}^2 x d x=d v \\
\therefore \quad \int & =\int \operatorname{cosec}^2 x d x=-\cot x \\
\int u d v & =u v-\int u d v \\
\mathrm{I}=\int x d(-\cot x) & =x(-\cot x)-\int-\cot x d x=x(-\cot x)-\int-\cot x d x \\
\mathrm{I} & =-x \cot x+\int \cot x d x=-x \cot x+\log \sin x+c
\end{aligned}
$

Question 2.
$x \cos 5 x \cos 2 x$
Solution:
$
\begin{aligned}
& I=\int x \cos 5 x \cos 2 x d x \\
& \cos 5 x \cos 2 x=\frac{1}{2}\{2 \cos 5 x \cos 2 x\} \\
&=\frac{1}{2}\{\cos (5 x+2 x)+\cos (5 x-2 x)\} \\
&=\frac{1}{2}[\cos 7 x+\cos 3 x] \\
& \therefore \quad \int x \cos 5 x \cos 2 x d x=\int x \frac{1}{2}(\cos 7 x+\cos 3 x) d x \\
&=\frac{1}{2}\left\{\int x \cos 7 x d x+\int x \cos 3 x d x\right\}=\frac{1}{2}\left\{I_1+I_2\right\}
\end{aligned}
$
Now $\mathrm{I}_1=\int x \cos 7 x d x$ put $u=x ; d u=d x$
$
\cos 7 x d x=d \nu \quad \therefore \quad v=\frac{\sin 7 x}{7}
$
$
\begin{aligned}
\therefore \quad \mathrm{I}_1 & =\int x \cos 7 x d x=\int x d\left(\frac{\sin 7 x}{7}\right) \\
& =x\left(\frac{\sin 7 x}{7}\right)-\int\left(\frac{\sin 7 x}{7}\right) d x=\frac{x}{7} \sin 7 x-\frac{1}{7}\left\{\frac{-\cos 7 x}{7}\right\} \\
& =\frac{x}{7} \sin 7 x+\frac{\cos 7 x}{49} \\
\mathrm{I}_2 & =\int x \cos 3 x d x \text { put } u=x ; d u=d x
\end{aligned}
$

$
\begin{aligned}
& \cos 3 x d x=d v ; v=\frac{\sin 3 x}{3} \\
\therefore \quad \mathrm{I}_2 & =\int x \cos 3 x d x=\int x d\left(\frac{\sin 3 x}{3}\right) \\
& =x\left(\frac{\sin 3 x}{3}\right)-\int \frac{\sin 3 x}{3} d x=\frac{x}{3} \sin 3 x-\frac{1}{3}\left\{\frac{-\cos 3 x}{3}\right\} \\
& =\frac{x}{3} \sin 3 x+\frac{1}{9} \cos 3 x
\end{aligned}
$
Substituting (2) and (3) in (1) we get,
$
\begin{aligned}
\int x \cos 5 x \cos 2 x d x & =\frac{1}{2}\left[\frac{x}{7} \sin 7 x+\frac{\cos 7 x}{49}+\frac{x}{3} \sin 3 x+\frac{\cos 3 x}{9}\right]+c \\
& =\frac{1}{2}\left[\left\{\frac{x}{7} \sin 7 x+\frac{x}{3} \sin 3 x\right\}+\frac{\cos 7 x}{49}+\frac{\cos 3 x}{9}\right]+c
\end{aligned}
$

Question 3.
$\mathrm{x}^2 \mathrm{e}^{2 \mathrm{x}}$
Solution:
$
\begin{aligned}
& \int x^2 e^{2 x} d x \quad u=x^2 ; d u=2 x ; d v=e^{2 x} d x \\
& v=\frac{e^{2 x}}{2} \\
& I=\int x^2 e^{2 x} d x \int x^2 d=\left(\frac{e^{2 x}}{2}\right)=x^2 \frac{e^{2 x}}{2}-\int \frac{e^{2 x}}{2}(2 x d x) \\
& =\frac{x^2 e^{2 x}}{2}-\int x e^{2 x} d x=\frac{x^2 e^{2 x}}{2}-\mathrm{I}_1 \text { (say) } \\
& \mathrm{I}_1=\int x e^{2 x} d x \quad u=x ; d u=d x ; e^{2 x} d x=d v \\
& \therefore \quad v=\frac{e^{2 x}}{2} \\
& \mathrm{I}_1=\int x e^{2 x} d x=\int x d\left(\frac{e^{2 x}}{2}\right)=\frac{x e^{2 x}}{2}-\int \frac{e^{2 x}}{2} d x \\
& =\frac{x e^{2 x}}{2}-\frac{1}{2}\left[\frac{e^{2 x}}{2}\right] \\
&
\end{aligned}
$
Substituting (2) in (1) we get,
$
\int x^2 e^{2 x} d x=\frac{x 2 e^{2 x}}{2}-\left\{\frac{x e^{2 x}}{2}-\frac{e^{2 x}}{4}\right\}=\frac{e^{2 x}}{2}\left\{x^2-x+\frac{1}{2}\right\}+c
$

Question 4.
$
\left(\sin ^{-1} x\right) \frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}
$
Solution:
$
\begin{aligned}
& \int \sin ^{-1} x \frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}} d x \\
& u=t ; \quad d u=d t \\
& e^t d t=d v ; \quad v=e^t \\
& \mathrm{I}=\int t e^t d t=\int t d\left(e^t\right)=t e^t-\int e^t d t=t e^t-e^t=e^t(t-1) \\
& =e^{\sin ^{-1} x}\left\{\sin ^{-1} x-1\right\}+c \\
&
\end{aligned}
$

Question 5.
$
\operatorname{cosec}^3 \mathrm{x}
$
Solution:
$
\begin{aligned}
\mathrm{I} & =\int \operatorname{cosec}^3 x d x=\int \operatorname{cosec}^x\left(\operatorname{cosec}^2 x\right) d x \\
u & =\operatorname{cosec} x ; \quad d v=\operatorname{cosec}^2 x d x \\
d u & =-\operatorname{cosec} x \cot x d x, \quad v=-\cot x \\
\mathrm{I} & =\int \operatorname{cosec} 3 d x \\
& =\int \operatorname{cosec} x d(-\cot x) \\
& =-\operatorname{cosec} x \cot x-\int \cot x[-\operatorname{cosec} x \cot x d x] \\
& =-\operatorname{cosec} x \cot x-\int \operatorname{cosec} x \cot { }^2 x d x \\
& =-\operatorname{cosec} x \cot x-\int \operatorname{cosec} x(\operatorname{cosec} x-1) d x \\
& =-\operatorname{cosec} x \cot x-\int \operatorname{cosec}{ }^2 x+\int \operatorname{cosec} x d x \\
\mathrm{I} & =-\operatorname{cosec} x \cot x-\mathrm{I}+\int \operatorname{cosec} x d x \\
2 \mathrm{I} & =-\operatorname{cosec} x \cot x+\log \tan \frac{x}{2} \\
\mathrm{I} & =\frac{-1}{2} \operatorname{cosec} x \cot x+\frac{1}{2} \log \tan \frac{x}{2}+c
\end{aligned}
$

Question. 6
$\sec ³2 x$
Solution:

$\begin{aligned}
& I=\int \sec ^3 2 x d x=\int \sec 2 x \sec ^2 2 x d x \\
& \text { Let } u=\sec ^2 2 x ; d u=2 \sec 2 x \tan 2 x d x \\
& \sec ^2 2 x d x=d v \\
& v=\int \sec ^2 2 x d x=\left(\frac{\tan 2 x}{2}\right) \\
& I=\int \sec 2 x d\left(\frac{\tan 2 x}{2}\right) \\
& =(\sec 2 x)\left(\frac{\tan 2 x}{2}\right)-\int\left(\frac{\tan 2 x}{2}\right)(2 \sec 2 x \tan 2 x) d x \\
& \therefore \quad \mathrm{I}=\frac{1}{2} \sec 2 x \tan 2 x-\int \tan ^2 2 x \sec 2 x d x \\
& =\frac{1}{2} \sec 2 x \tan 2 x-\int\left(\sec ^2 2 x-1\right) \sec 2 x d x \\
& =\frac{1}{2} \sec 2 x \tan 2 x-\int \sec ^3 2 x d x+\int \sec 2 x d x \\
& I=\frac{1}{2} \sec 2 x \tan 2 x-I+\frac{1}{2} \log (\sec 2 x+\tan 2 x) \\
& 2 I=\frac{1}{2}\{\sec 2 x \tan 2 x+\log (\sec 2 x+\tan 2 x)\} \\
& I=\frac{1}{4}[\sec 2 x \tan 2 x+\log (\sec 2 x+\tan 2 x)]+c \\
&
\end{aligned}$