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Integration by Substitution

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int sin^{5}x cosx dx[/tex]



    2. Relevant equations
    None


    3. The attempt at a solution
    I tried setting u=sin^5(x) but this ended up yielding [tex]\frac{1}{5}\int u cos^{3}x du[/tex] and I cannot think of a better substitution. Any tips?
     
  2. jcsd
  3. Dec 4, 2009 #2
    You're doing a substitution first and then trying to see if it works. That's almost never the way to do it. In almost all cases, you have to see what the solution is before doing the substitution. In simple cases like this, you then don't need to actually substitute anything.

    The derivative of f[g(x)] is f'[g(x)]g'(x). When you want to integrate f'[g(x)]g'(x)dx , you want to substititute u = g(x). But then you need to "see" the g(x), i.e. recognize the chain rule structure of the integrand. If you see this, you can directly write down the integral.
     
  4. Dec 4, 2009 #3
    Does this apply in this case? For any substitution I visualize in my head I still foresee an 'x' being in the integral after the substitution.
     
  5. Dec 4, 2009 #4
    Ahhh yes I got it now. Your clue helped a lot. I substituted u=sinx and ended up with the answer of [tex] \frac{sin^{6}x}{6} +C [/tex].

    Thank you very much!
     
  6. Dec 4, 2009 #5
    It works in this case. You have to forget about doing some substitution to "get rid of x". Because that's not the way to "see" what substitution you need to do. All you need to do is to look at the formula:

    sin^5(x) cos(x)

    and compare that to the chain rule formula:

    f[g(x)] g'(x)

    What do you think you should choose for g(x)?
     
  7. Dec 4, 2009 #6
    Well done! So, you see that you can "spot" the solution by simply looking at the integrand!
     
  8. Dec 5, 2009 #7
    Yes sir, it's a lot easier to solve once you use that technique. Thanks again!
     
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