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?
 A: Doing $x=\arctan t$ and $\mathrm dx=\frac1{1+t^2}\,\mathrm dt$, you get$$\int\frac{\sin^2x}{1+\sin^2x}\,\mathrm dx=\int\frac{t^2}{2 t^4+3 t^2+1}\,\mathrm dt=\int\frac{1}{t^2+1}-\frac{1}{2 t^2+1}\,\mathrm dt.$$Can you take it from here?
A: Hint:
$$ \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}= 1-\frac{1-\sin^2 x}{1-(\sin x)^4}=1-\frac{1}{1+\sin^2x}=1-\frac{\sec^2x}{1+2\tan^2x}$$
A: You're integrating a polynomial in $\sin$ and $\cos$; the $\tan(t/2)$ substitution (http://www-math.mit.edu/~djk/18_01/chapter24/section03.html) will convert this to a rational function of a variable $t$. 
(You can also slightly simplify by noting that your integrand is an even function, so you can just double the integral from $0$ to $4\pi$; then you can note that the integrand is periodic with period $\pi$, so you get $8$ times the integral from $0$ to $\pi$. But probably neither of these saves much effort once you do the $\tan(t/2)$ substitution.)
A: 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?
