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.

Applying $\displaystyle\int_c^df(x)dx=\int_c^df(c+d-x)\ \ \ \ (1)dx$

$$I=\int_{-b}^b\frac1{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$=\int_{-b}^b\frac1{a^{-b+b-x}+1}\cdot\frac{\left(\sin(-b+b-x)\right)^{2n}}{\left(\cos(-b+b-x)\right)^{2n}+\left(\sin(-b+b-x)\right)^{2n}}dx$$

$$=\int_{-b}^b\frac{a^x}{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$ as $\displaystyle\sin(-x)=-\sin x,\cos(-x)=\cos x$

$$\implies I+I=\int_{-b}^b\frac1{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx+\int_{-b}^b\frac{a^x}{a^x+1}\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$2I=\int_{-b}^b1\cdot\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

Now setting $b=\dfrac\pi2,$

$$2I=\int_{-\dfrac\pi2}^{\dfrac\pi2}\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$=\int_{-\dfrac\pi2}^0\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx +\int_0^{\dfrac\pi2}\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

$$\text{For }I_1=\int_{-\dfrac\pi2}^0\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$

if $g(x)=\sin^{2n}x,g\left(-\dfrac\pi2+0-x\right)=\cdots=\cos^{2n}x$

using $(1),$ $$\text{For }I_1=\int_{-\dfrac\pi2}^0\frac{g(x)}{g\left(-\dfrac\pi2+0-x\right)+g(x)}dx=\int_{-\dfrac\pi2}^0\frac{g\left(-\dfrac\pi2+0-x\right)}{g(x)+g\left(-\dfrac\pi2+0-x\right)}dx$$

$$\implies I_1+I_1=\int_{-\dfrac\pi2}^0 dx=\cdots$$

Similarly, for $$I_2=\int_0^{\dfrac\pi2}\frac{\sin^{2n}x}{\cos^{2n}x+\sin^{2n}x}dx$$