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Definite integrals with trig issues

  1. Sep 27, 2014 #1
    from 0 to π/2
    ∫sin5θ cos5θ dθ

    I have been trying to solve the above for quite some time now yet can't see what Im doing wrong. I break it down using double angle formulas into:
    ∫ 1/25 sin5(2θ) dθ
    1/32 ∫sin4(2θ) * sin(2θ) dθ
    1/32 ∫(1-cos2(2θ))2 * sin(2θ) dθ

    With this I can make u = cos(2θ) and du = -sin(2θ) dθ, so insert -du
    -1/32 ∫(1-u2)2 du

    Here is where I get lost leading me into various answers, none of which are right.

    What seems correct is to expand that into 1-2u2+u4 for 3 separate integrals all multiplied by -1/32 but it gets me to π/64 - 1/24 + 1/80

    Can anyone help?
     
  2. jcsd
  3. Sep 27, 2014 #2

    ShayanJ

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    Try integration by parts. You should only find the right functions to differentiate and integrate. Then rules for sine or cosine squared will take care of the rest. Its not that hard!
     
  4. Sep 27, 2014 #3
    So v=sinθ and du=cos dθ makes dv=cosθ (or is it cosθ dθ?) and u=sinθ

    sin5θ * sin5θ - ∫sin5θ * cos5θ dθ ??

    because I'm back at the same problem

    I even tried the X=Y-X form and changed in 2X=Y/2 and still didn't get the right answer
     
    Last edited: Sep 27, 2014
  5. Sep 27, 2014 #4

    mathman

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    I did it exactly as you did and got 1/32(1- 2/3+1/5) = 1/60.

    The indefinite integral is (1/64)(u - (2/3)u^3 + (1/5)u^5) and the limits are (-1,1).

    It seems your error is having π where you should have 2.
     
  6. Sep 27, 2014 #5
    Huh? What did you do after -1/32 ∫(1-u2)2 du?

    Not sure what you mean about my error, pi/2 is the limit why would it be 2?
     
    Last edited: Sep 27, 2014
  7. Sep 27, 2014 #6

    ShayanJ

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    Try [itex] dv=\sin\theta \cos^5 \theta [/itex] or [itex] dv=\cos\theta \sin^5 \theta [/itex].
     
  8. Sep 28, 2014 #7

    mathman

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    The limits on θ are (-π/2,π/2). This becomes (-1,1) for u.
     
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