Calculate $\int \frac{dx}{1-\sin^4x}$ 
Calculate $$\int \frac{dx}{1-\sin^4x}$$

My try:
\begin{align}
\int \frac{dx}{1-\sin^4x}&=\int \frac{(1-\sin^2x)+(1+\sin^2x)dx}{1-\sin^4x}
\\&=\int \frac{dx}{1-\sin^2x}+\int \frac{dx}{1+\sin^2x}
\\&=\tan x+\int \frac{dx}{1+\sin^2x}
\end{align}
How to deal with the second one?

For the 2nd integral, rewrite the integrand,
$$\frac{1}{1+\sin^2x}=\frac{\sec^2x}{2\sec^2x-1}=\frac{(\tan x)’}{2\tan^2x+1}$$
Then, integrate,
$$\int \frac{d\tan x}{2\tan^2x+1}
=\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan x) + C$$
