Question

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

Answer 1

We have,

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

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

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

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

$I=\int \cot xcosecxdx+\int {\cot }^{2}xdx$

$I=-cosecx+\int (cose{c}^{2}x-1)dx$

$I=-cosecx-\cot x-x+C$.

Hence, this is the answer.