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Homework Help: Calculus help

  1. Mar 18, 2004 #1
    Hey all--
    I had an Integration of Parts quiz today and got stuck on a few problems-- was wondering if you could explain the steps involved in solving these integrals:

    int( (sin(3x))^3 * (cos(3x))^3 dx)


    int( (tan(4x))^4) dx)

  2. jcsd
  3. Mar 18, 2004 #2
    [tex]\int \sin^3{3x} \cos^3{3x}\,dx[/tex]

    First things first, get rid of the 3x's with a u substitution. It's easy to see that all it does is change the solution by a factor of 1 over 3. So we have,

    [tex]\frac{1}{3}\int \sin^3{u} \cos^3{u}\,du[/tex]

    If we had only 1 sine term or only 1 cosine term, we'd be gold. Problem solved. But we got 2 too many. So let's get rid of them!

    [tex]\sin^2{x} + \cos^2{x} = 1[/tex]

    Use this to turn the integral into

    [tex]\frac{1}{3}\int \sin^3{u} (1 - \sin^2{u}) \cos{u}\,du[/tex]

    which is easily separated and solved by substitution.

    Have another shot at the second one, keeping in mind that

    [tex]\frac{d}{dx}\tan{x} = \sec^2{x}[/tex]

  4. Mar 18, 2004 #3


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    Where did you get [tex]\cos^3=1-\sin^2[/tex]
    I would probably use the half angle formulas:

    Now you've got:


    Oh, by parts...
  5. Mar 18, 2004 #4
    I didn't. I only took two of the cosines and I left the third for the u substitution.

    u = sinx
    du = cosxdx

    It's used in the du.

    Edit: By parts?
