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?

Hint:
Bioche's rules say you should  set $t=\tan x$. Indeed, with some trigonometry,
$$\frac{\sin^2x}{1+\sin^2x}=\frac{\cfrac{t^2}{1+t^2}}{1+\cfrac{t^2}{1+t^2}}=\cfrac{t^2}{1+2t^2},\qquad\mathrm dx=\frac{\mathrm dt}{1+t^2},$$
 so the indefinite integral becomes
$$\int\frac{t^2\,\mathrm dt}{(1+2t^2)(1+t^2)}$$
Can you continue?
