Question

Solve:$\int \frac{1}{\sin x.\cos x}$

Answer 1

$\int \frac{1}{\sin x.\cos x}dx$

$=\int \frac{2}{2\sin x.\cos x}dx$

$=2\int \frac{1}{\sin 2x}dx$

$=2\int cosec2xdx$

$=2\left[\frac{1}{2}\log |cosec2x-\cot 2x|\right]+C$

$=\log |cosec2x-\cot 2x|+C$