# Is my answer to this integral correct

• May 18th 2009, 03:30 AM
simplependulum
Is my answer to this integral correct
$\displaystyle \int_0^{\pi} \cos(\tan(x)) ~dx ~= \frac{\pi}{e}$ ???
• May 18th 2009, 04:03 AM
NonCommAlg
Quote:

Originally Posted by simplependulum

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

yes, it's correct. $\displaystyle \int_0^{\pi} \cos(\tan(x)) \ dx= 2 \int_0^{\frac{\pi}{2}} \cos(\tan x) \ dx = 2\int_0^{\infty} \frac{\cos x}{1+x^2} \ dx = \frac{\pi}{e}.$ (a very well-known result!)
• May 19th 2009, 01:25 AM
simplependulum
Thank you so much Mr NonCommAlg

This result really surprised me ! (Surprised)
• May 19th 2009, 12:37 PM
Kaitosan
Quote:

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

Hey! Will you please explain the last step for me? I don't understand how you converted $\displaystyle cos(tan(x))$.
• May 19th 2009, 12:57 PM
Moo
Quote:

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

A simple substitution $\displaystyle x=\tan(x)$ (Surprised)
• May 19th 2009, 01:07 PM
Kaitosan
Huh?

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

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

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

Please clarify. Maybe there's some trig information I'm missing or something.
• May 19th 2009, 01:11 PM
Moo
Maybe because $\displaystyle \sec^2(x)=\frac{1}{\cos^2(x)}=\frac{\cos^2(x)+\sin ^2(x)}{\cos^2(x)}=1+\tan^2(x)=1+u^2$

(Tongueout)

In fact, it can be known that the derivative of tangent is also $\displaystyle 1+\tan^2$
It depends on the way you learnt it ^^'
• May 19th 2009, 01:13 PM
Spec
Another way to look at it: $\displaystyle u=\tan x,\ x= \arctan u,\ dx=\frac{du}{1+u^2}$
• May 19th 2009, 01:30 PM
Kaitosan
ooh I see. Thanks!

One last question -

How do you integrate $\displaystyle cos(x)/(1+x^2)$? Trig sub and inverse trig function integration seem not to work.
• May 19th 2009, 01:31 PM
Moo
Quote:

Originally Posted by Kaitosan
ooh I see. Thanks!

One last question -

How do you integrate $\displaystyle cos(x)/(1+x^2)$? Trig sub and inverse trig function integration seem not to work.

Residue theorem :D
That's complex analysis.
• May 19th 2009, 01:34 PM
Kaitosan
Well. In that case, I feel better! For a moment I felt like crap. :p