Results 1 to 2 of 2

Math Help - Don't get these 2 trig integration problems?

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    126

    Don't get these 2 trig integration problems?

    2. Integral of x^2sqrt[a^2-x^2] from the interval [-a,0]

    i SOLVED THE FIRST ONE.. bUT i DO NOT UNDERSTAND THE SECOND ONE AT ALL!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by elpermic View Post
    1. Integral of x^3 X square root(16-4x^2)

    2. Integral of x^2sqrt[a^2-x^2] from the interval [-a,0]
    The first integral doesn't require trig substitution...but if you need to have to, I would recommend using the trig sub x=2\sin\theta.

    Spoiler:
    In case you're wondering how to do the first one without trig sub, let u=16-4x^2\implies x^2=\tfrac{1}{4}(16-u). Therefore, du=-8x\,dx \implies -\frac{1}{8}\,du=dx.

    Therefore,

    \displaystyle\int x^3\sqrt{16-4x^2}\,dx \xrightarrow{u=16-x^2}{}-\tfrac{1}{8}\int x^2\sqrt{u}\,du=-\tfrac{1}{32}\int (16-u)\sqrt{u}\,du

    which can be easily integrated.


    For the second integral, I would consider using the trig sub x=a\sin\theta.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Problems with integration word problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 25th 2010, 05:39 PM
  3. Replies: 7
    Last Post: April 15th 2010, 08:12 PM
  4. Replies: 14
    Last Post: June 8th 2009, 07:48 PM
  5. More trig integration problems :(.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 22nd 2008, 09:52 AM

Search Tags


/mathhelpforum @mathhelpforum