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Stuck on this integral (cos and sin)

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Integrate (sin5x)(cos4x)dx


    2. Relevant equations



    3. The attempt at a solution
    I tried substituting u as (sin5x), but wound up with something way more complicated than what I started with. Same thing happens if I substitute cos^4 for u.

    I have no idea what to try next... is there some way that I can break up the integral?
     
  2. jcsd
  3. Feb 2, 2009 #2

    Dick

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    Yes. Pick u=cos(x). du=(-sin(x))*dx. That leaves you with sin(x)^4. Use cos(x)^2=1-sin(x)^2.
     
  4. Feb 2, 2009 #3
    Then I end up with

    -(1-u^2)(1-u^2)(u^4)du

    That doesn't seem any easier to integrate.
     
  5. Feb 2, 2009 #4

    Dick

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    I consider polynomials pretty easy to integrate, but maybe that's just me. Multiply it out.
     
  6. Feb 2, 2009 #5
    Ooh I see, when I multiply it out I get just 1 there. So it's the integral of

    1 + (u^4)

    Which gives

    u + (u^5)/5 or cos x + (cos^5(x))/5. Multiplied by a negative, of course.

    But the integral is actually definite and when I plug in the bounds it's giving me the wrong answer.
     
  7. Feb 2, 2009 #6

    Dick

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    You don't get 1+u^4. You get a more complicated polynomial. But not all THAT much more complicated. It only has three terms.
     
  8. Feb 2, 2009 #7
    Ok I think I have it.

    (cos^5(x))/5 - (2cos^7(x))/7 + (cos^9(x))/(9)

    just wondering: did you know to use u = cos x just by doing lots of practice and recognizing a pattern, or is there a rule you used?
     
  9. Feb 2, 2009 #8

    Dick

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    Yeah, basically. Watch your overall sign there.
     
  10. Feb 2, 2009 #9

    Dick

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    It's a standard thing. If you've got an odd power of cos or sin times an even power of the other one, you peel off one of the odd power to save for the du and then just convert it to a polynomial. As you did. And check the sign again.
     
  11. Feb 2, 2009 #10
    Oh yeah, forgot to multiply by the negative. The sign of each term should just be flipped, then.

    Thanks for your help, I really appreciate it.
     
  12. Feb 2, 2009 #11
    What if sin and cos are both even powers? Or both odd, for that matter?
     
  13. Feb 2, 2009 #12

    Dick

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