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Evaluating Trigonometric Integral

  1. Jun 5, 2014 #1
    1. The problem statement, all variables and given/known data
    [itex]\int[/itex][itex]\frac{\pi}{2}[/itex]0 sin7y dy
    The bounds are from [itex]\frac{\pi}{2}[/itex] to 0.

    2. Relevant equations



    3. The attempt at a solution
    I think I did the integration correctly, but I don't really know how to evaluate this.

    [itex]\int[/itex]siny(sin2y)3 dy
    [itex]\int[/itex]siny(1-cos2y)3 dy
    u=cosy
    du=-siny dy

    -[itex]\int[/itex](1-u2)3 du
    -[itex]\int[/itex](-u6+3u4-3u3+1) du

    [itex]\frac{u7}{7}[/itex]-[itex]\frac{3u5}{5}[/itex]+[itex]\frac{3u4}{4}[/itex]-u

    [itex]\frac{cos7y}{7}[/itex]-[itex]\frac{3cos5y}{5}[/itex]+[itex]\frac{3cos4y}{4}[/itex]-cosy

    I'm a little rusty on my trig! Please help :)
     
  2. jcsd
  3. Jun 5, 2014 #2

    Ray Vickson

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    Homework Helper

    No the bounds are not from ##\pi/2## to 0, they are from 0 to ##\pi/2##. Anyway, you have the right idea; you just need to figure out the bounds after the u-substitution; that is, what are ##a## and ##b## in
    [tex] \int_0^{\pi/2} \sin^7 y \: dy = -\int_a^b (1-u^2)^3 \, du?[/tex]
     
  4. Jun 5, 2014 #3
    Oops! That's what I meant. Haha thanks, I guess it would be a lot easier to just change the bounds.
     
  5. Jun 5, 2014 #4
    Hey! I ran into another problem, I changed the bounds to 1 to 0:

    -∫ (-u6+3u4-3u3+1) du

    Then I flipped the bounds to 0 to 1 and changed the sign on the integral:

    ∫(-u6+3u4-3u3+1) du

    Then I plugged in my bounds and got 99/140, but the answer in the back of the book is 16/35. I'm having trouble figuring out what I did wrong.
     
  6. Jun 5, 2014 #5
    How do you get ##u^3##? It should be ##u^2## instead.
     
  7. Jun 6, 2014 #6
    Ooops! It was just a silly mistake. Thanks so much!
     
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