Exercise 9.4 - Chapter 8 Differentials and Partial Derivatives 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $9.5$
Evaluate the following:
Question 1.
$
\int_0^1 x^3 e^{-2 x} d x
$
Solution:
$
\mathrm{u}=\mathrm{x}^3 ; \mathrm{dv}=\mathrm{e}^{-2}
$
Applying Bernoulli's formula, we get,
$
\begin{aligned}
\int_0^1 x^3 e^{-2 x} d x & =\left[u v-u^{\prime} v_1+u^{\prime \prime} v_2-u^{\prime \prime} v_3\right]_0^1 \\
& =\left[x^3\left(\frac{e^{-2 x}}{-2}\right)-3 x^2\left(\frac{e^{-2 x}}{4}\right)+6 x\left(\frac{e^{-2 x}}{-8}\right)-6\left(\frac{e^{-2 x}}{16}\right)\right]_0^1 \\
& =\frac{e^{-2}}{-2}-\frac{3 e^{-2}}{4}-\frac{3 e^{-2}}{4}-\frac{3 e^{-2}}{8}+\frac{3}{8}=\frac{3-19 e^{-2}}{8}
\end{aligned}
$
Question 2.
$
\int_0^1 \frac{\sin \left(3 \tan ^{-1} x\right) \tan ^{-1} x}{1+x^2} d x
$

Solution:
Let $t=\tan ^{-1} x$
Differentiate with respect to $x$
$
\begin{gathered}
d t=\frac{1}{1+x^2} d x=\frac{d x}{1+x^2} \\
\int_0^1 \frac{\sin \left(3 \tan ^{-1} x\right) \tan ^{-1} x}{1+x^2} d x=\int_0^{\pi / 4} \sin (3 t) \cdot t d t
\end{gathered}
$

Take $u=t ; d v=\sin 3 t$
Applying Bernoulli's formula, we get,
$
\begin{array}{rlr}
\int_0^{\pi / 4} t \sin (3 t) d t & =\left[u v-u^1 v_1\right]_0^{\frac{\pi}{4}} & \cos \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}} \\
& =\left[t\left(\frac{-\cos 3 t}{3}\right)-\left(\frac{-\sin 3 t}{9}\right)\right]_0^{\frac{\pi}{4}} & \sin \frac{3 \pi}{4}=\frac{1}{\sqrt{2}} \\
& =\frac{1}{9}[\sin 3 t-3 t \cos 3 t]_0^{\frac{\pi}{4}} \\
& =\frac{1}{9}\left[\sin \frac{3 \pi}{4}-\frac{3 \pi}{4} \cos \frac{3 \pi}{4}\right] \\
& =\frac{1}{9}\left[\frac{1}{\sqrt{2}}-\frac{3 \pi}{4}\left(\frac{-1}{\sqrt{2}}\right)\right]=\frac{1}{\sqrt{2}}\left[\frac{1}{9}+\frac{3 \pi}{36}\right] \\
& =\frac{1}{\sqrt{2}}\left[\frac{1}{9}+\frac{\pi}{12}\right] .
\end{array}
$

Question 3.
$
\int_0^{\frac{1}{\sqrt{2}}} \frac{e^{\sin ^{-1} x} \sin ^{-1} x}{\sqrt{1-x^2}} d x
$
Solution:

$
\begin{aligned}
\int_0^{\frac{1}{\sqrt{2}}} \frac{e^{\sin ^{-1} x} \sin ^{-1} x}{\sqrt{1-x^2}} d x \\
t=\sin ^{-1} x
\end{aligned}
$

Differentiate with respect to $x$
$
\begin{aligned}
\frac{d t}{d x} & =\frac{1}{\sqrt{1-x^2}} \\
d t & =\frac{d x}{\sqrt{1-x^2}}
\end{aligned}
$
Substitute (2) and (3) in (1),
(1) $\Rightarrow \int_0^{\frac{\pi}{4}} e^t t d t$
Applying Bernoulli's formula, we get,
$
\begin{aligned}
& =\left(e^t[t-1]\right)_0^{\frac{\pi}{4}} \\
& =e^{\pi / 4}\left[\frac{\pi}{4}-1\right]-e^o[0-1]=e^{\pi / 4}\left[\frac{\pi}{4}-1\right]+1
\end{aligned}
$

Question 4.
$
\int_0^{\frac{\pi}{2}} x^2 \cos 2 x d x
$

Solution:
Take $u=x^2 ; d v=\cos 2 x$
Applying Bernoulli's formula, we get,
$
\begin{array}{rlr}
\int_0^{\frac{\pi}{2}} x^2 \cos 2 x d x & =\left[u v-u^{\prime} v_1+u^{\prime \prime} v_2\right]_0^{\frac{\pi}{2}} & \sin \pi=0 \\
& =\left[x^2\left(\frac{\sin 2 x}{2}\right)-2 x\left(\frac{-\cos 2 x}{4}\right)+2\left(\frac{-\sin 2 x}{8}\right)\right]_0^{\frac{\pi}{2}} \\
& =0-2\left(\frac{\pi}{2}\right)\left(\frac{1}{4}\right)+0=-\frac{\pi}{4}
\end{array}
$