∫ 1/(1+3 sin ^2 x) d x is equal to

We have,

$I =\int \frac{1}{1+3 \sin ^2 x} d x=\int \frac{\sec ^2 x}{\sec ^2 x+3 \tan ^2 x} d x$

$\Rightarrow I =\frac{1}{2} \int \frac{1}{(2 \tan x)^2+1^2} d(2 \tan x)$

$\Rightarrow I =\frac{1}{2} \tan ^{-1}(2 \tan x)+C$