#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 22 Maths Textbook Solution.

$\sqrt{2}\left[\tan ^{-1}\left(\frac{(\sqrt{2 \tan x}-1)}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 \tan x}+1}{\sqrt{2}}\right)\right]$

Given:

$\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$

Hint:

To solve this equation we use partial function method.

Solution:

$\int \frac{\sin x+\cos x}{\sqrt{2 \sin x \cos x}} d x \quad \because\left[\sin x=\frac{\tan x}{\sec x}, \cos x=\frac{1}{\sec x}\right]$

$\int \frac{\frac{\tan x}{\sec x}+\frac{1}{\sec x}}{\sqrt{\frac{\operatorname{stan} x}{\sec x} \times \frac{1}{\sec x}}} d x$

$\int \frac{\tan x+1}{\sqrt{2(\tan x)}} d x$

$\frac{1}{\sqrt{2}} \int \frac{\tan x+1}{\sqrt{\tan x}} \frac{\tan ^{2} x+1}{\tan ^{2} x+1} d x$

$\frac{1}{\sqrt{2}} \int \frac{\sec ^{2} x(\tan x+1)}{\sqrt{\tan x}\left(\tan ^{2} x+1\right)} d x \quad\left[\because \tan x=u, \sec ^{2} x d x=d u\right]$

$=\frac{1}{\sqrt{2}} \int \frac{u+1}{\sqrt{u}\left(u^{2}+1\right)} d u$

$\left.=\frac{1}{\sqrt{2}} \int \frac{v^{2}+1}{\sqrt{u}\left(v^{4}+1\right)} 2 \sqrt{u} d v_{\cdots . .[\text { put } u}=v^{2}, d u=2 v d v\right]$

$=\sqrt{2} \int \frac{v^{2}+1}{v^{4}+1} d v$

$=\sqrt{2} \int \frac{v^{2}+1}{\left(v^{2}-\sqrt{2} v+1\right)\left(v^{2}+\sqrt{2} v+1\right)} d v$

$=\frac{v^{2}+1}{\left(v^{2}-\sqrt{2} v+1\right)\left(v^{2}+\sqrt{2} v+1\right)}=\frac{A}{v^{2}-\sqrt{2} v+1}+\frac{B}{v^{2}+\sqrt{2} v+1}$

$=\sqrt{2}\left(\frac{1}{2}\right)\left(\int \frac{1}{v^{2}-\sqrt{2} v+1} d v+\int \frac{1}{v^{2}+\sqrt{2} v+1} d v\right)$

$=\frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}} \int \frac{1}{v-\left(\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}} d v+\int \frac{1}{v+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)} d v$

$\left[\frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]$

$=\frac{1}{\sqrt{2}} \sqrt{2} \frac{\tan ^{-1}\left(v-\left(\frac{1}{\sqrt{2}}\right)\right)}{\frac{1}{\sqrt{2}}}+\sqrt{2} \frac{\tan ^{-1}\left(v+\left(\frac{1}{\sqrt{2}}\right)\right)}{\frac{1}{\sqrt{2}}}$

$=\sqrt{2}\left[\tan ^{-1}\left(\frac{\sqrt{2 v}-1}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 v}+1}{\sqrt{2}}\right)\right]$

$=\sqrt{2}\left[\tan ^{-1}\left(\frac{\sqrt{2 v}-1}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 v}+1}{\sqrt{2}}\right)\right] \quad[\because v=\sqrt{u}, u=\tan x]$

$=\sqrt{2}\left[\tan ^{-1}\left(\frac{(\sqrt{2 \tan x}-1)}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 \tan x}+1}{\sqrt{2}}\right)\right]$