TITLE: Hard integral, low hints...
QUESTION [5 upvotes]: $$\int_{ - \pi /2}^{\pi /2} \frac1{2007^{x} + 1}\cdot \frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx .$$
This integral stuns me for a while, I just can't solve it! I tried integration by recurrence since here it seems that we have something that looks like a recurrence, but nothing comes up. I also tried all the integration techniques I'm aware of but they just don't work.
So I fear like this integral uses something like Fourier, or some Laplace transform thing or something of that sort but I don't know them.
That's why I will be very thankful if they could explain me how to solve it.
Thank you.

REPLY [8 votes]: You can use this way to do. Let
$$ I=\int_{ - \pi /2}^{\pi /2} \frac1{2007^{x} + 1}\cdot \frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx.\tag{1}$$
Now use the substitution $x=-u$ to get
\begin{eqnarray*}
I&=&-\int_{\pi /2}^{-\pi /2} \frac1{2007^{-u} + 1}\cdot \frac {\sin^{2008}(-u)}{\sin^{2008}(-u) + \cos^{2008}(-u)} \, (-du)\\
&=&\int_{-\pi /2}^{\pi /2} \frac{2007^u}{2007^{u} + 1}\cdot \frac {\sin^{2008}u}{\sin^{2008}u + \cos^{2008}u} \, dx.\tag{2}
\end{eqnarray*}
So adding (1) and (2) will give
$$ 2I=\int_{- \pi/2}^{\pi /2}\frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx$$
or
$$ I=\frac{1}{2}\int_{- \pi/2}^{\pi /2}\frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx=\int_{0}^{\pi /2}\frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx.\tag{3}$$
Now use the substitution $u=\frac{\pi}{2}-x$ to get
$$ I=\int_{0}^{\pi /2}\frac {\cos^{2008}u}{\sin^{2008}u + \cos^{2008}u} \, du.\tag{4}$$
Now adding (3) and (4) will give you $ 2I=\pi/2$ or $I=\pi/4$.