<s>[INST] Calculate $\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx$ 
Calculate $$\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx$$

I tried to do this task in several ways, but none of them proved to be effective. For example:

$$\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx=\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2(1-(\sin x)^2)}{1-(\sin x)^4}dx=\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2}{1+(\sin x)^2}dx=\int ^{4\pi} _{-4\pi} \frac{1}{1+\frac{1}{(\sin x)^2}}dx$$
However I don't know what I can do the next to finish this task. When I use $u=(\sin x)^2 $ I have $du=\cos x dx$ so I can't use it.Have you got some intelligent way to do this task?
 [/INST] Hint:
$$ \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}= 1-\frac{1-\sin^2 x}{1-(\sin x)^4}=1-\frac{1}{1+\sin^2x}=1-\frac{\sec^2x}{1+2\tan^2x}$$
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