Question

Find the integral of the function
$\frac{\cos x}{1+\cos x}$

Answer 1

We have,

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

We know that

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

Therefore,

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

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

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

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

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

$I=\frac{x}{2}-\frac{1}{2}[\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x]+C$

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

$I=\frac{3x}{2}-\tan \frac{x}{2}+C$

Hence, this is the answer.