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.

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- Dec 6th 2009, 05:59 PMhjorturFun integral
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. - Dec 6th 2009, 06:06 PMKrizalid
and took me 1 second to tell ya that it's a quite known problem!

so leave others to try it! (i'm telling you people who already knows the answer.) - Dec 6th 2009, 07:52 PMrrm74001
Did anyone else do a facepalm when they finally figured it out? I stared blankly at my paper for 15 minutes trying to figure it out, and I feel dumb now for not seeing it much sooner.

I was never good at integrals to begin with, however (Rofl). - Dec 6th 2009, 11:13 PMsimplependulum
- Mar 26th 2010, 05:09 PMchiph588@
How is this integral done? I'm thinking complex contour integration, but I bet there's an easier way.

Also, I've noticed that this works for all $\displaystyle \alpha\in\mathbb{C} $. - Mar 26th 2010, 05:14 PMKrizalid
use $\displaystyle \tan x=\frac{\sin x}{\cos x}$ and substitute $\displaystyle x\mapsto \frac\pi2-x.$

- Mar 26th 2010, 05:54 PMchiph588@
- Mar 26th 2010, 06:54 PMRandom Variable
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 $\displaystyle \int_{0}^{\pi /2} \frac{1}{1+\tan^{\alpha}x} \ dx = \frac{1}{2}\int^{\pi /2}_{0}\frac{\cos^{\alpha}x + \sin^{\alpha} x}{\cos^{\alpha}x + \sin^{\alpha} x} \ dx = \frac{1}{2}\int^{\pi /2}_{0} dx = \frac{\pi}{4}$ - Mar 26th 2010, 09:59 PMmr fantasticPlease read the criteria for posting in the Challenge subforum
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