Results 1 to 2 of 2

Math Help - Integral with complex numbers

  1. #1
    Newbie Capric's Avatar
    Joined
    Mar 2014
    From
    ohio
    Posts
    1

    Integral with complex numbers

    I know how to do this problem the easy way, where I have trouble though is when I apply Euler's formula.

    integral Cos^3(x)dx= Sin(x) - (1/3) Sin^3(x)+c

    however when I try to do this with complex numbers I get stuck at this point.
    I can do the integral I am just not sure how to go back to trig functions.

    Cos^3(x) = (1/4) ((e^3ix - e^-3ix)/2 - 3(e^ix - e^-ix)/2)

    I think this part -3(e^ix - e^-ix)/2) becomes -3Cos(x)
    but what about the rest?

    ps- I'm on spring break or else I would ask my professor (calc 2)




    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1149

    Re: Integral with complex numbers

    $\cos^3(x) = \left(\dfrac{e^{ \imath x}+e^{ -\imath x}}{2}\right)^3=$

    $\dfrac{3}{4}\dfrac{e^{\imath x}+e^{-\imath x}}{2}+\dfrac{1}{4}\dfrac{e^{\imath 3x}+e^{-\imath 3x}}{2}=$

    $\dfrac{3}{4}\cos(x)+\dfrac{1}{4}\cos(3x)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: March 25th 2011, 11:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 04:14 PM
  3. Replies: 2
    Last Post: February 7th 2009, 07:12 PM
  4. Replies: 1
    Last Post: May 24th 2007, 04:49 AM
  5. Complex Numbers- Imaginary numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 24th 2007, 01:34 AM

Search Tags


/mathhelpforum @mathhelpforum