<s>[INST] 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?
 [/INST] $$\dfrac{dx}{a\sin x\cos x+b\sin^2x+c\cos^2x}$$
Divide numerator and denominator 
by $\cos^2x$ and set $\tan x=u$
Or by $\sin^2x$ and set $\cot x=v$
Here $(1)+\sin^2x=(\cos^2x+\sin^2x)+\sin^2x=?$
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