# Thread: Fun integral

1. ## Fun integral

Show that:

$\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)} dx$

is independent of $\alpha$.

This took me quite a while to figure out.

2. 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.)

3. 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 .

4. Originally Posted by hjortur
Show that:

$\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)} dx$

is independent of $\alpha$.

This took me quite a while to figure out.
It is also a constant when $\alpha = 0$

5. 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 $\alpha\in\mathbb{C}$.

6. use $\tan x=\frac{\sin x}{\cos x}$ and substitute $x\mapsto \frac\pi2-x.$

7. Originally Posted by Krizalid
use $\tan x=\frac{\sin x}{\cos x}$ and substitute $x\mapsto \frac\pi2-x.$
After doing that, I get $\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\alpha}(x) } = \int_0^{\frac{\pi}{2}}\frac{dx}{1+\cot^{\alpha}(x) } = \int_0^{\infty}\frac{du}{u^2+u^{2-\alpha}+u^{-\alpha}+1}$.

I'm definitely missing something here...

8. I think it's the following:

$\int_{0}^{\pi /2} \frac{1}{1+\tan^{\alpha}x} \ dx = \int^{\pi /2}_{0}\frac{\cos^{\alpha}x}{\cos^{\alpha}x + \sin^{\alpha} x} \ dx$

Let $x = \frac{\pi}{2}-x$

$= -\int^{0}_{ \pi /2}\frac{\cos^{\alpha}(\pi/ 2-x)}{\cos^{\alpha}(\pi / 2-x) + \sin^{\alpha} (\pi / 2 -x)} \ dx = \int^{\pi /2}_{0}\frac{\sin^{\alpha}x}{\sin^{\alpha}x + \cos^{\alpha} x} \ dx$

so $\int_{0}^{\pi /2} \frac{1}{1+\tan^{\alpha}x} \ dx = \frac{1}{2}\int^{\pi /2}_{0}\frac{\cos^{\alpha}x + \sin^{\alpha} x}{\cos^{\alpha}x + \sin^{\alpha} x} \ dx = \frac{1}{2}\int^{\pi /2}_{0} dx = \frac{\pi}{4}$

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