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Trig Integral

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\int[/itex]sin[itex]^{5}[/itex]x cos x dx

    2. Relevant equations

    sin[itex]^{2}[/itex]x + cos[itex]^{2}[/itex] x = 1

    3. The attempt at a solution

    I've at least written down that sin[itex]^{5}[/itex]x = (sin[itex]^{2}[/itex]x)[itex]^{2}[/itex] sin x. Then I set sin[itex]^{2}[/itex]x equal to 1 - cos[itex]^{2}[/itex]x.

    I then did a u-substitution, setting u equal to cos x to remove the sin x, and preceeded to integrate with respect to u. I ended up with [itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

    The answer I received was [itex]\frac{1}{6}[/itex]sin[itex]^{6}[/itex]x + C

    Can I get some insight on how to obtain that?

    My textbook does give a procedure for this specific case (where one of the powers is odd), but I don't get the following instructions: "Then we combine the single sin x with dx in the integral and set sinxdx equal to -d(cosx)" which is stated after using the identity to replace the (sin[itex]^{2}[/itex]x)
     
  2. jcsd
  3. Oct 17, 2012 #2

    SammyS

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    Do a u substitution directly on the problem as it was initially given to you.

    What is the derivative of sin(x) ?
     
  4. Oct 17, 2012 #3
    ...

    Right, I feel kinda "slow" now.

    Any way to use the trig identity to solve this though?
     
  5. Oct 17, 2012 #4

    Zondrina

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    .... Make the substitution u = sinx and find du just like sammy said. The rest will follow.
     
  6. Oct 17, 2012 #5
    Yeah, I did that. Just curious if I can still use the identity to solve this though. Otherwise, I suppose that's it
     
  7. Oct 17, 2012 #6

    Dick

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    Yes, you can do it that way as well. Though I think there's a sign error in the answer you gave.
     
  8. Oct 17, 2012 #7
    -[itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

    I think I was missing that negative. Is this correct?
     
  9. Oct 17, 2012 #8

    Dick

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    Yep.
     
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