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Integration of (xsinx)^2

  1. Oct 2, 2004 #1
    Hi ppl, I need a little pointer with the integral of (xsinx)^2. I tried by parts but it just doesn't stop and I've tried writing sin^2x in it's other forms but that yields similar results. Any help please?

  2. jcsd
  3. Oct 2, 2004 #2


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    If it doesn't stop, it is because you switch what's "u" and "v".
    It is definitely smart to use the double-angle formula!
  4. Oct 2, 2004 #3


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    [tex] sin^{2}x = \frac{1-cos2x}{2} [/tex]
  5. Oct 2, 2004 #4
    I used the identity suggested, but I don't get anywhere, could u please just take the problem a little further?
  6. Oct 2, 2004 #5


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    Show in some detail why you don't get anywhere!
  7. Oct 3, 2004 #6
    By the way, may I "join"? In the first place, do you already know the answer in the first hand? (I mean, you got the answer somewhere else? =) )

    Since the members here are not giving you a direct answer to the question... I'll give you some hints... I only did some integration by parts....

    1.) let u = sin^2 x and dv = x^2dx... so, du = sin2x dx and v = x^3/3...
    2.) You get the 2nd integral, right? If you've done the first step (of mine) correctly, then your on the track..
    let u = x^3 and dv = sin[2x] dx... so, du = 3x^2 dx and v = -cos[2x] / 2.
    3.) keep doing integration by parts a little more... later, you'll notice something - integration by simple substitution... and then if there's anything to simplify, do so.

    Show your work... so that others can guide you...

    Cyclovenom's hint for you, unfortunately, does not apply in my strategy that I've written for you here... but still, both different approach, if correctly done, can lead you to the same answer. =) Good luck!
  8. Oct 3, 2004 #7
    Joe, I hope that helps... I'll be watching here.. time to time. =)
  9. Oct 7, 2004 #8
    Thanks for the help irony, I managed it now, cheers,
  10. Oct 8, 2004 #9
    Welcome... :D
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