Integral $ \int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx $ I am trying to solve this integral
$$
\int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx
$$
A closed form does exist despite the looks of the integrand.  This problem is from some old high school IMO training courses.  I am not sure how to solve it. The only information that may be of help is
$$
\sin x=\sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x=\sum_{n=1}^\infty \frac{ x^{2n}}{(2n)!} \quad \forall \ x
$$
Thanks, I am looking for a complete solution, not a description as of what to do.

Separate into two parts, $-\pi/2$ to $0$ and $0$ to $\pi/2$. 
For the integral from $-\pi/2$ to $0$, make the change of variable $t=-x$. We get
$$\int_0^{\pi/2} \frac{2007^t}{2007^t+1} \frac{\sin^{2008}t}{\sin^{2008} t+\cos^{2008} t}\,dt.$$
Change the variable back to $x$, and add to the integral from $0$ to $\pi/2$.We get 
$$\int_0^{\pi/2}\frac{\sin^{2008}x}{\sin^{2008} x+\cos^{2008} x}\,dx.\tag{1}$$
Progress! Now split the integral (1) into the integral from $0$ to $\pi/4$, and the integral from $\pi/4$ to $\pi/2$.
For the integral from $\pi/4$ to $\pi/2$, let $u=\pi/2-x$, and use much the same trick we already used. 
The integral from $\pi/4$ to $\pi/2$ turns out to be
$$\int_0^{\pi/4}\frac{\cos^{2008}u}{\cos^{2008} u+\sin^{2008} u}\,du.$$
Change the variable to $x$, and add to the integral from $0$ to $\pi/4$. We end up with
$$\int_0^{\pi/4} 1\,dx,$$
which is not difficult.
