#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 72 Maths Textbook Solution.

$\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|+\frac{1}{4} \tan ^{2} \frac{x}{2}+\tan \frac{x}{2}+C$

Hint:

To solve the given statement multiply and divide the equation by sin x.

Given:

$\int \frac{1+\sin x}{\sin x(1+\cos x)} d x$

Solution:

$I=\int \frac{(1+\sin x)}{\sin x(1+\cos x)} d x$

$\text { putting }$

$\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

$\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

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

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

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

$\text { putting } \tan \frac{x}{2}=t$

$\Rightarrow \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x=d t$

$\Rightarrow \sec ^{2}\left(\frac{x}{2}\right) d x=2 d t$

$I=\frac{1}{4} \int \frac{\left(1+t^{2}+2 t\right) \cdot(2 d t)}{t}$

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

$=\frac{1}{2}\left[\ln |t|+\frac{t^{2}}{2}+2 t\right]+C$

$=\frac{1}{2}\left[\ln \left|\tan \frac{x}{2}\right|+\frac{\tan ^{2}\left(\frac{x}{2}\right)}{2}+2 \tan \left(\frac{x}{2}\right)\right]+C\left[\because t=\tan \frac{x}{2}\right]$

$=\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|+\frac{1}{4} \tan ^{2} \frac{x}{2}+\tan \frac{x}{2}+C$