The antiderivative of $\frac{1}{{\sin }^{2}x+{\tan }^{2}x}$ is

The integrand can be written as $\frac{{\sec }^{2}x}{{\tan }^{2}x\left(2+{\tan }^{2}x\right)}$

putting $\tan x=t$, the integral reduces to 

$\begin{array}{l}\int \frac{dt}{t^2\left(2+t^2\right)}=\frac{1}{2}\int \left(\frac{1}{t^2}-\frac{1}{2+t^2}\right)dt\\ =\frac{1}{2}\left[-\frac{1}{t}-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}\right]+C\\ =\left(-\frac{1}{2}\right)\left[\cot x+\left(\frac{1}{\sqrt{2}}\right){\tan }^{-1}\left(\left(\frac{1}{\sqrt{2}}\right)\tan x\right)\right]+C\end{array}$