# integrate x/(1+cos(x)^2) from 0 to pi

• Oct 24th 2012, 07:28 AM
phigirl
integrate x/(1+cos(x)^2) from 0 to pi
integrate x/(1+cos(x)^2) from 0 to pi
I thought integration by parts udv = uv - vdu, so x is u and 1/(1+cos(x)^2) is v. But I have trouble finding the trick to integrate v. Thanks for any help or insight
• Oct 24th 2012, 07:33 AM
Siron
Re: integrate x/(1+cos(x)^2) from 0 to pi
You'll need a numerical integration method to solve this integral.
• Oct 24th 2012, 09:15 AM
JJacquelin
Re: integrate x/(1+cos(x)^2) from 0 to pi
Is it (cos(x))^2 or cos(x^2) ?
Is it a school work ? If YES, what exactly is the wording ?
• Oct 24th 2012, 09:18 AM
phigirl
Re: integrate x/(1+cos(x)^2) from 0 to pi
(cos(x))^2

show the integral x/(1+(cos(x))^2) = (pi^2)/2sqrt(2)
• Oct 24th 2012, 12:53 PM
JJacquelin
Re: integrate x/(1+cos(x)^2) from 0 to pi
Quote:

Originally Posted by phigirl
(cos(x))^2
show the integral x/(1+(cos(x))^2) = (pi^2)/2sqrt(2)

OK. That is clear.
You can solve it thanks to integration by parts with u=x and v(x)=primitive of 1/(1+(cos(x))^2)
A primitive of 1/(1+(cos(x)^2)) is v(x)=(1/sqrt(2))*arctan(tan(x)/sqrt(2))
u'=1. So you should have to integrate u'*v=v, wich is very difficult (involving special functions).
But, you don't need to do it explicitly. Since the function v(x) is periodic, one can see that its definite integal from x=0 to x=pi is equal to 0.
So, only x*v(x) is remaining. But there is a discontinuity at x=pi/2. Since v' is even then x*v is also even, the definite integral is :
2*(x*v(x) at x=pi/2) = 2*(pi/2)*(1/sqrt(2))*arctan(tan(pi/2)/sqrt(2)) = 2*(pi/2)*(1/sqrt(2)*(pi/2) = piČ/(2*sqrt(2))