Answer
$$\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan (x))+c$$
Work Step by Step
Given $$\int \frac{d x}{1+\sin ^{2} x}$$ Let $u=\tan x\ \ \ \ \ \ du=\sec^2 xdx $, then
\begin{align*}
dx&=\frac{du}{\sec^2 x}\\
&= \frac{du}{1+\tan^2 x}\\
&=\frac{du}{1+u^2}
\end{align*}
Since
\begin{align*}
\sin ^{2}(x)&=\frac{\tan ^{2} x}{1+\tan ^{2} x}\\
&=\frac{u^{2}}{1+2u^{2}}
\end{align*}
Then
\begin{align*}
\int \frac{d x}{1+\sin ^{2} x}&= \int \frac{1+u^{2}}{1+2 u^{2}} \cdot \frac{d u}{1+u^{2}}\\
&=\int \frac{d u}{1+2 u^{2}}\\
&=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} u)+c\\
&=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan (x))+c
\end{align*}
