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Integration with Trig (Calc 2)

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex][tex]\frac{(sin x)^{3}}{(cos x)}[/tex]dx


    2. Relevant equations
    Trigonometric identities


    3. The attempt at a solution
    [tex]\int[/tex][tex]\frac{(sin x)^{2}}{(cos x)}(sin x)[/tex]dx

    [tex]\int[/tex][tex]\frac{1-(cos x)^{2}}{(cos x)}(sin x)[/tex]dx

    u = cos x
    du = -sin x dx

    - [tex]\int[/tex][tex]\frac{1-(u)^{2}}{(u)}[/tex]du

    Where do I go from here? I am kind of stuck. Can I just simply split the two up, as in 1/u - u^2/2? I think I tried that and end up with the wrong answer. Thanks in advance for the advice!
     
  2. jcsd
  3. Sep 22, 2008 #2

    Dick

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    Sure you can split them up. But what happened to the '-' in -sin(x)dx?
     
  4. Sep 22, 2008 #3
    Oops! I had the - on my paper, I forgot it in transferring it :smile: I will edit that now!

    But the answer I am getting is not equal to the answers those online integral calculators are showing. Here is my work:

    = - [tex]\int[/tex]du/u + [tex]\int\frac{u^{2}}{u}[/tex]du
    = - ln u + (1/2)u[tex]^{2}[/tex] + c
    = - ln (cos x) + ((cos x)^2)/2 + c

    Am I making some sort of elementary mistake?
     
  5. Sep 22, 2008 #4

    Dick

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    That's perfectly fine. Differentiate it to make sure. Now take one on the online integrator solutions and do the same. There's lots of different ways to write the result that only differ by a constant. For example instead of cos(x)^2/2 you could put -sin(x)^2/2. They only differ by a constant.
     
  6. Sep 22, 2008 #5
    Thanks, Dick! It did work. I feel very silly for not thinking of doing that myself!
     
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