Definite integration using even and odd functions How to integrate this:
$$\int_{-\pi /2}^{\pi /2} \frac{\sin^{2002}x }{(2003^x +1)(\sin^{2002}x + \cos^{2002}x)}dx$$
I tried to replace $x$ with $-x$ but that was a dead end. I know that this should be separated into one even and one odd function but I can't figure it out.. 

Calculate the integral from $-\pi/2$ to $0$, making your suggested substitution $u=-x$. After some algebra, we end up with
$$\int_0^{\pi/2} \frac{2003^u\sin^{2002}u}{(2003^u+1)(\sin^{2002} u+\cos^{2002} u)}\,du.$$
Change the dummy variable back to $x$, and add the integral from $0$ to $\pi/2$. There is some cancellation, and we end up with
$$\int_0^{\pi/2}\frac{\sin^{2002}x}{\sin^{2002} x+\cos^{2002} x}\,dx.$$
Call this integral $I$. One nasty part of the original integrand is gone!
Split $I$ into two parts, from $0$ to $\pi/4$ and $\pi/4$ to $\pi/2$. Call the first integral $J$, and the second $K$. For the second integral, make the change of variable $u=\pi/2-x$. We end up with
$$K=\int_0^{\pi/4}\frac{\cos^{2002}u}{\sin^{2002} u+\cos^{2002} u}\,du.$$
Change the dummy variable back to $x$, and add $J$. We get wonderful cancellation, and find that 
$$I=\int_0^{\pi/4} 1\,dx.$$ 
