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

  • Thread starter Electro
  • Start date
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. :smile:
Thank you
 
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Take the derivative of your indefinite result. If it is correct, you will get your integrand
 

dextercioby

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U need to integrate this

[tex] \int \sin^{2}x \ dx [/tex]

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

Daniel.
 
use [tex] \sin^{2}x = \frac{(1- \cos{2x})}{2} [/tex]
 
integrate by parts
Answer comes out to be [tex] \frac{\pi^2}{16} -1/2 [/tex]
 
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Isnt the integral [tex] \int_0^{\pi/2}{xsin^2(x)}{dx} [/tex] ?
 

dextercioby

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

Daniel.
 

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