Question

Evaluate : $\int \frac{\cos x}{1-\cos x}dx$

Answer 1

We have,

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

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

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

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

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

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

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

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

Hence, this is the answer.