Evaluate the following integrals:

first divide nominator by denominator -

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

$=\int \frac{1}{\sin x(1+\cos x)} d x+\int \frac{1}{1+2 \cos ^{2} x-1} d x$

: To solve this type of solution, we are going to substitute the value of $\sin x$ and $\cos x$ in terms of tan( $x / 2)$

$\sin x=\frac{2\left[\tan \frac{x}{2}\right]}{1+\tan ^{2} \frac{x}{2}}$

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

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

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

In this type of equations we apply substitution method so that equation may be solve in simple way

Let $\tan (x / 2)=t$

$1 / 2 \cdot \sec ^{2}(x / 2) d x=d t$

Put these terms in above equation, we get $I=\int \frac{d t}{2 t}$

Substitute these terms in above equation gives-

$I=\frac{1}{2} \int \frac{d t}{t}$

$I=\frac{-1}{2 t^{2}}$

Now put the value of $t, t=\tan (x / 2)$ in above equation gives us the finally solution

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