Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By Prove It

Math Help - Integration trig help

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    Trinidad
    Posts
    49

    Integration trig help

    How do i integrate SinxCosX by parts?

    I am stuck
    Last edited by Benja303; April 18th 2013 at 02:11 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,610
    Thanks
    1576
    Awards
    1

    Re: Integration trig help

    Quote Originally Posted by Benja303 View Post
    How do i integrate SinxCosX by parts?

    Forget parts. Not needed.

    What is the derivative of \tfrac{1}{2}\sin^2(x)~?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2012
    From
    Trinidad
    Posts
    49

    Re: Integration trig help

    -1/2 cos^2(x)?

    BTW how do i know when to use parts or not?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,610
    Thanks
    1576
    Awards
    1

    Re: Integration trig help

    Quote Originally Posted by Benja303 View Post
    -1/2 cos^2(x)?
    BTW how do i know when to use parts or not?
    That is also a possible answer. It is not the derivative as I had in mind.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2012
    From
    Trinidad
    Posts
    49

    Re: Integration trig help

    ? what other way can u do it?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1393

    Re: Integration trig help

    Quote Originally Posted by Benja303 View Post
    How do i integrate SinxCosX by parts?

    I am stuck
    Plato is right that using Integration by Parts here is overkill, but if you MUST use it...

    \displaystyle \begin{align*} I &= \int{\sin{(x)}\cos{(x)}\,dx} \\ I &= \sin^2{(x)} - \int{\cos{(x)}\sin{(x)}\,dx} \\ I &= \sin^2{(x)} - I \\ 2I &= \sin^2{(x)} \\ I &= \frac{1}{2}\sin^2{(x)} + C  \end{align*}

    I personally prefer to avoid integration by parts where possible. This is easier integrated if you change the integrand using a double angle formula.

    \displaystyle \begin{align*} \int{\sin{(x)}\cos{(x)}\,dx} &= \int{\frac{1}{2}\sin{(2x)}\,dx} \\ &= -\frac{1}{4}\cos{(2x)} + C \\ &= -\frac{1}{4} \left[ 1 - 2\sin^2{(x)} \right] + C \\ &= \frac{1}{2}\sin^2{(x)} - \frac{1}{4} + C \end{align*}

    As you can see, these answers differ only by a constant, and since the constants are arbitrary anyway, they can be considered to be equivalent
    Thanks from Benja303
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2012
    From
    Trinidad
    Posts
    49

    Re: Integration trig help

    I was not taught these double angle things. Can u give me some insight?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration with Trig
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 16th 2010, 05:07 PM
  2. Trig Integration help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 2nd 2010, 06:09 PM
  3. trig integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 1st 2009, 10:36 PM
  4. Integration without Trig Sub
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 11th 2009, 10:31 AM
  5. trig integration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 5th 2009, 05:49 PM

Search Tags


/mathhelpforum @mathhelpforum