\(\int \frac{dx}{1+3\sin^{2}x}=\)

\(\frac{1}{2}\tan^{-1}(2 \tan x)\)

$I=\int \frac{dx}{1+3 \sin^{2}x}=\int \frac{\sec^{2}xdx}{\sec^{2}x+3\tan^{2}x}=\int\frac{\sec^2xdx}{1+4\tan^2x}$

let $\tan x= y$

$⇒dy=\sec^2xdx$

$⇒I=\int\frac{dy}{1+(2y)^2}=\frac{1}{4}\int\frac{1}{(\frac{1}{2})^2+y^2}dy$

$=\frac{1}{4×\frac{1}{2}}\tan^{-1}\frac{y}{\frac{1}{2}}+C=\frac{1}{2}\tan^{-1}2y+C$

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