Show that:
$\displaystyle \int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)} dx$
is independent of $\displaystyle \alpha$.
This took me quite a while to figure out.
I think it's the following:
$\displaystyle \int_{0}^{\pi /2} \frac{1}{1+\tan^{\alpha}x} \ dx = \int^{\pi /2}_{0}\frac{\cos^{\alpha}x}{\cos^{\alpha}x + \sin^{\alpha} x} \ dx $
Let $\displaystyle x = \frac{\pi}{2}-x $
$\displaystyle = -\int^{0}_{ \pi /2}\frac{\cos^{\alpha}(\pi/ 2-x)}{\cos^{\alpha}(\pi / 2-x) + \sin^{\alpha} (\pi / 2 -x)} \ dx = \int^{\pi /2}_{0}\frac{\sin^{\alpha}x}{\sin^{\alpha}x + \cos^{\alpha} x} \ dx $
so $\displaystyle \int_{0}^{\pi /2} \frac{1}{1+\tan^{\alpha}x} \ dx = \frac{1}{2}\int^{\pi /2}_{0}\frac{\cos^{\alpha}x + \sin^{\alpha} x}{\cos^{\alpha}x + \sin^{\alpha} x} \ dx = \frac{1}{2}\int^{\pi /2}_{0} dx = \frac{\pi}{4}$
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