Integral from 0 to pi/2 of (x*[sin x]^2) dx

[Electro]

Hello everyone,
I was solving an integral, but I am not quite sure for the final answer. If someone has the time, just take a look.

Integral from 0 to pi/2 of (x*[sin x]^2)dx

I used by parts integration; using u=(sinx)^2 du=2 sinx cosx
dv = x v = x^2/2
I used once more by parts integration and I got as a final answer pi/24.
I need some advice.
Thank you

 

[whozum]

Take the derivative of your indefinite result. If it is correct, you will get your integrand

 

[dextercioby]

U need to integrate this

\int \sin^{2}x \ dx

and the result wrt "x"...The integrations are not difficult,if u know a bit of circular trigonometry.

Daniel.

 

[tutor69]

use \sin^{2}x = \frac{(1- \cos{2x})}{2}

 

[tutor69]

integrate by parts
Answer comes out to be \frac{\pi^2}{16} -1/2

 

[whozum]

Isnt the integral \int_0^{\pi/2}{xsin^2(x)}{dx} ?

 

[dextercioby]

Yes,it is,but part integrating once,makes u integrate sine squared,just as I've written above.

Daniel.