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

  1. Mar 29, 2005 #1
    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
     
  2. jcsd
  3. Mar 29, 2005 #2
    Take the derivative of your indefinite result. If it is correct, you will get your integrand
     
  4. Mar 29, 2005 #3

    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.
     
  5. Mar 29, 2005 #4
    use [tex] \sin^{2}x = \frac{(1- \cos{2x})}{2} [/tex]
     
  6. Mar 29, 2005 #5
    integrate by parts
    Answer comes out to be [tex] \frac{\pi^2}{16} -1/2 [/tex]
     
  7. Mar 29, 2005 #6
    Isnt the integral [tex] \int_0^{\pi/2}{xsin^2(x)}{dx} [/tex] ?
     
  8. Mar 29, 2005 #7

    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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