Question

Integrate:$\int \frac{\cos x}{1+\cos x}dx$

Answer 1

Consider the given integral.

$I=\int \frac{\cos x}{1+\cos x}dx$

Put $t=\tan \frac{x}{2}$. So,

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

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

$dx=\frac{2}{1+t^2}dt$

Therefore,

$I=\int \frac{\cos x}{1+\cos x}dx$

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

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

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

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

$I=-t+2{\tan }^{-1}t+C$

$I=-\tan \frac{x}{2}+x+C$

Hence, this is the required value of the integral.