TITLE: Integrate $\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$
QUESTION [39 upvotes]: Evaluate the integral $$\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$$

REPLY [2 votes]: Well, this problem has a nice symmetry.
The integral can be rewritten as,
$$ I=\int_{0}^{\pi/2}\frac{\cos^{\alpha}(x)}{\sin^{\alpha}(x)+\cos^{\alpha}(x)} dx \tag{1}$$
By the property of definite integrals this Integral is same as,
$$\int_{0}^{\pi/2}\frac{\cos^{\alpha}(\frac{\pi}{2}-x)}{\sin^{\alpha}(\frac{\pi}{2}-x)+\cos^{\alpha}(\frac{\pi}{2}-x)} dx$$
$$\int_{0}^{\pi/2}\frac{\sin^{\alpha}(x)}{\sin^{\alpha}(x)+\cos^{\alpha}(x)} dx \tag{2}$$
Adding $1,2$ gives
$$ 2I=\int_{0}^{\frac{\pi}{2}} dx$$
$$ 2I=\frac{\pi}{2}$$
$$\boxed{I=\frac{\pi}{4}}$$