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Trigonometric Integrals

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The integral of cos[x]*(sin[x])^5dx


    2. Relevant equations



    3. The attempt at a solution

    I split it into cos[x]*(sin[x])^2 * (sin[x])^2 * sin[x]

    then each sin^2[x] term becomes (1-cos[x]^2)

    so you end up with the integral of cos[x]*(1-cos[x]^2)^2*sin[x]dx

    let u=cos[x]
    -du=sin[x]dx

    so it comes to the integral of -u*(1-u^2)^2 which foils out to be

    -u+2u^3-u^5 then integrating this gives

    -(1/2)u^2 +(2/4)*u^4 - (1/6)*u^6+c

    replacing all u's with cos[x] of course at the end I just want to make sure this is correct because I have to hand it in tomarrow and this is the first of these problems I have tried and I don't want to do them all wrong.
     
  2. jcsd
  3. Feb 13, 2008 #2
    [tex]\int\cos x \sin^{5}xdx[/tex]

    Yes?

    Let [tex]u=\sin xdx[/tex] and it's solved.
     
  4. Feb 13, 2008 #3
    I'm sorry I wasn't clear enough in the explanation you must use "the method for odd powers to evaluate the integral" since we are just learning this technique.
     
  5. Feb 13, 2008 #4
    Well then your answer is correct!
     
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