Question

Solve $\int \frac{1}{1-\sin 2x}dx$.

Answer 1

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

$\sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}$

Substituting this in I, we get

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

$=\int \frac{1+{\tan }^{2}x}{(1-\tan x{)}^2}dx=\int \frac{{\sec }^{2}xdx}{(1-\tan x{)}^2}$

Let $\tan x=t$ $\Rightarrow {\sec }^{2}xdx=dt$

$\Rightarrow I=\int \frac{dt}{(1-t{)}^2}=(1-t{)}^{-1}+c$

$=\frac{1}{(1-\tan x)}+c$.