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Integral of (xsinx)^2?

  1. Oct 12, 2009 #1
    hi all,
    I've been trying to integrate this thing for ages.i tried using integration by parts using u=x^2,
    dv/dx=sin^2 x but it just doesn't seem to end.any help or pointers much appreciated.

  2. jcsd
  3. Oct 12, 2009 #2
    Switch your u=.. and dv=.. around.
    Another method would be to try an identity, such as the double angle identity for your (sinx)^2
  4. Oct 12, 2009 #3
    hmm i hadn't thought of the identity, i'll give it a go.
  5. Oct 12, 2009 #4
    I would use your original u-substitution (u=x2), and then use the half-angle formula for sin2x to integrate the dv. Then after you complete the integration by parts the first time, you'll get a sum of two functions in the integral term. One of them you'll be able to immediately integrate, while for the other one you can use integration by parts again.
  6. Oct 13, 2009 #5
    i managed to integrate it but i think i may ave made an error.
    heres the result:

    don't know how to put it in proper formulae.
    anyway thanks for the help.
  7. Oct 13, 2009 #6
    emin - I come up with a slightly different answer. Can you provide details of your solution. The starting approach I took is as follows:

    I = \int x^2 \cdot \sin^2(x)\,dx = \int x^2 \cdot \frac{(1-\cos(2x))}{2} \, dx

    then let

    2I = \int x^2 \cdot (1-\cos(2x)) \, dx = \frac{x^3}{3} - I_2

    I_2 = \int x^2 \cdot \cos(2x) \, dx = \int \left (\frac{2x}{2} \right )^2 \cdot \cos(2x) \, d\left ( \frac{2x}{2} \right ) = (1/8) \int s^2 \cdot \cos(s) \, ds

    where [itex]s=2x[/itex]
  8. Oct 13, 2009 #7
    i did get my fractions mixed didn't i?
    i was doing it on an a4 page and got all the working jumbled.
    yes that is the approach i took, i should be able to run it through mathematica when i get the chance, and see what answer it comes up with.
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