Question

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

Answer 1

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

$=\int \frac{{\sec }^{2}x}{\tan x}dx$

substitute $u=\tan x\to du={\sec }^{2}xdx$

$=\int \frac{1}{u}du$

$=\ln u+C$

$=\ln \left(\tan x\right)+C$