Is my answer to this integral correct

**simplependulum** · May 18th 2009, 03:30 AM

$\int_0^{\pi} \cos(\tan(x)) ~dx ~= \frac{\pi}{e}$ ???

**NonCommAlg** · May 18th 2009, 04:03 AM

Originally Posted by simplependulum

$\int_0^{\pi} \cos(\tan(x)) ~dx ~= \frac{\pi}{e}$ ???

yes, it's correct. $\int_0^{\pi} \cos(\tan(x)) \ dx= 2 \int_0^{\frac{\pi}{2}} \cos(\tan x) \ dx = 2\int_0^{\infty} \frac{\cos x}{1+x^2} \ dx = \frac{\pi}{e}.$ (a very well-known result!)

**simplependulum** · May 19th 2009, 01:25 AM

Thank you so much Mr NonCommAlg

This result really surprised me !

**Kaitosan** · May 19th 2009, 12:37 PM

Originally Posted by NonCommAlg

yes, it's correct. $\int_0^{\pi} \cos(\tan(x)) \ dx= 2 \int_0^{\frac{\pi}{2}} \cos(\tan x) \ dx = 2\int_0^{\infty} \frac{\cos x}{1+x^2} \ dx = \frac{\pi}{e}.$ (a very well-known result!)

Hey! Will you please explain the last step for me? I don't understand how you converted $cos(tan(x))$ .

**Moo** · May 19th 2009, 12:57 PM

Originally Posted by Kaitosan

Hey! Will you please explain the last step for me? I don't understand how you converted $cos(tan(x))$ .

A simple substitution $x=\tan(x)$

**Kaitosan** · May 19th 2009, 01:07 PM

Huh?

$cos(tan(x)) dx$ = $cos(u) du$

$du = sec^2x dx$ which doesn't make sense.

How does $cos(tan(x)) = cos(x)/(1+x^2)$ ?

Please clarify. Maybe there's some trig information I'm missing or something.