Results 1 to 3 of 3

Math Help - integration by parts

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    5

    integration by parts

    can someone show me step by step in solving these problems?

    \int{x^2 sin4x^3} {dx}

    and

    \int{sin^3 2x cos^2 2x} {dx}

    thx in advance of ur help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by kokolily View Post
    can someone show me step by step in solving these problems?

    \int{x^2 sin4x^3} {dx}

    and

    \int{sin^3 2x cos^2 2x} {dx}

    thx in advance of ur help
    Neither of these are integration by part, but are u substitions.

    For the first one use the u sub u=4x^3 \implies du=12x^2dx to get

    \int{x^2 sin4x^3} {dx} =\frac{1}{12}\int \sin(u)du

    For the 2nd one you need a trig identity...

    \int{\sin^3 2x \cos^2 2x} {dx} =\int \sin(2x)[1-\cos^2(2x)]\cos^2(2x)dx=

    \int \sin(2x)\cos^2(2x)dx-\int \sin(2x)\cos^4(2x)dx

    Now just let u=\cos(2x) and you are off to the races.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    5
    thanks for ur help

    how do u know that it is not int by parts but exactly int by substitution?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 03:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 03:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum