Question

Find the integral of the function
$\frac{\cos x}{1+\cos x}$ w.r.t x

Answer 1

We have,

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

We know that

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

Therefore,

$I=\int \frac{2{\cos }^2\frac{x}{2}-1}{1+1-2{\sin }^2\frac{x}{2}}dx$

$I=\int \frac{2{\cos }^2\frac{x}{2}-1}{2-2{\sin }^2\frac{x}{2}}dx$

$I=\int \frac{{\cos }^2\frac{x}{2}-\frac{1}{2}}{1-{\sin }^2\frac{x}{2}}dx$

$I=\int \frac{{\cos }^2\frac{x}{2}-\frac{1}{2}}{{\cos }^2\frac{x}{2}}dx$

$I=\int 1dx-\frac{1}{2}\int \frac{1}{{\cos }^2\frac{x}{2}}dx$

$I=\int 1dx-\frac{1}{2}\int {\sec }^2\frac{x}{2}dx$

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

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

Hence, this is the answer.