Question

Solve$\int \frac{\cos x}{1+\cos x}dx$

Answer 1

Consider the given integral.

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

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

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

We know that

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

Therefore,

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

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

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

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

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

Hence, this is the answer.