Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 26 Maths Textbook Solution.

$-2 \sqrt{1-\sin x}+\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$

Given:

$\int \frac{\sin x}{\sqrt{1+\sin x}} d x$

Hint:

To solve the statement we have to use formula like sin x and 1+cos x

Solution:

$I=\int \sqrt{\sin x+1}-\frac{1}{\sqrt{\sin x+1}} d x$

$=\int \sqrt{\sin x+1} d x-\int \frac{d x}{\sqrt{\sin x+1}}$

$I_{1}=\int \sqrt{\sin x+1} d x$

$=\int \sqrt{1+\sin x} \cdot \frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}} d x$

$=\frac{\sqrt{(1+\sin x)(1-\sin x)}}{\sqrt{1-\sin x}} d x$

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

$=\frac{\cos x}{\sqrt{1-\sin x}} d x$

$=\sqrt{1-\sin x}=u$

$\frac{1}{2 \sqrt{1-\sin x}} \cos x d x=d u$

$\frac{\cos x}{\sqrt{1-\sin x}}=-2 d u$

$I_{1}=\int-2 d u=-2 u=-2 \sqrt{1-\sin x}$

$I_{2}=\int \frac{d x}{\sqrt{1+\sin x}}=\int \frac{d x}{\sqrt{1+\cos \left(\frac{\pi}{2}-x\right)}}$                            $\left[\because \sin x=\cos \left(\frac{\pi}{2}-x\right)\right],\left[1+\cos x=2 \cos ^{2} x \frac{x}{2}\right]$

$=\int \frac{d x}{\sqrt{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}}$

$=\int \frac{d x}{\sqrt{2}\left(\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)\right.}$                                            $\left[\because \int \sec x d x=\ln |\sec x+\tan x|+c\right]$

$=\int \frac{1}{\sqrt{2}} \sec \left(\frac{\pi}{4}-\frac{x}{2}\right) d x$

$=\frac{1}{\sqrt{2}} \cdot \frac{1}{-\frac{1}{2}} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$

$=-\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$

$\mathrm{I}_{1}-\mathrm{I}_{2}=\mathrm{l}=-2 \sqrt{1-\sin x}+\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c$