1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Calculus help
  1. Help with calculus (Replies: 1)

  2. Calculus help (Replies: 2)

  3. Calculus help (Replies: 9)

  4. Calculus help ! (Replies: 14)

  5. Help on calculus! (Replies: 0)