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

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- May 18th 2009, 03:30 AMsimplependulumIs my answer to this integral correct
$\displaystyle \int_0^{\pi} \cos(\tan(x)) ~dx ~= \frac{\pi}{e} $ ???

- May 18th 2009, 04:03 AMNonCommAlg
- May 19th 2009, 01:25 AMsimplependulum
Thank you so much Mr NonCommAlg

This result really surprised me ! (Surprised) - May 19th 2009, 12:37 PMKaitosan
- May 19th 2009, 12:57 PMMoo
- May 19th 2009, 01:07 PMKaitosan
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 PMMoo
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 PMSpec
Another way to look at it: $\displaystyle u=\tan x,\ x= \arctan u,\ dx=\frac{du}{1+u^2}$

- May 19th 2009, 01:30 PMKaitosan
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 PMMoo
- May 19th 2009, 01:34 PMKaitosan
Well. In that case, I feel better! For a moment I felt like crap. :p