#### Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 65  Maths Textbook Solution.

$I=\frac{1}{\sqrt{3}} \log \left|\frac{\tan \left(\frac{x}{2}\right)-2-\sqrt{3}}{\tan \left(\frac{x}{2}\right)-2+\sqrt{3}}\right|+c$

Hint:

To solve the above equation we will write sin x as $\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}$

Given:

$\int \frac{1}{1-2 \sin x} d x$

Solution:

$I=\int \frac{1}{1-2 \sin x} d x$

$I=\int \frac{1}{1-\frac{4 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}} d x$

$I=\int \frac{1+\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x$                                        $\left[\because \sin x=\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}\right.$

$I=\int \frac{\sec ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x$

$I=\int \frac{2}{2} \frac{\sec ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)-4 \tan \left(\frac{x}{2}\right)} d x$

$I=2 \int \frac{d t}{1+t^{2}-4 t}$                                $\left[\because t=\tan \left(\frac{x}{2}\right), \frac{d t}{d x}=\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right), d t=\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x\right]$

$I=2 \int \frac{d t}{\left(t^{2}-2 \cdot 2 \cdot t+2^{2}\right)-2^{2}+1}$

$I=2 \int \frac{d t}{(t-2)^{2}-3}$

$I=2 \int \frac{d u}{u^{2}-(\sqrt{3})^{2}}$                            $[\because t-2=u, d t=d u]$

$\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right.$

$I=2 \times \frac{1}{2 \sqrt{3}} \log \left|\frac{u-\sqrt{3}}{u+\sqrt{3}}\right|+c$

$I=\frac{1}{\sqrt{3}} \log \left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right|+c$

$I=\frac{1}{\sqrt{3}} \log \left|\frac{\tan \left(\frac{x}{2}\right)-2-\sqrt{3}}{\tan \left(\frac{x}{2}\right)-2+\sqrt{3}}\right|+c$