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Proof of equation with integrals

  1. Dec 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that
    \int cos^mxdx = \frac{cos^{m-1}x sinx}{m} + \frac{m-1}{m}\int cos^{m-2}xdx

    2. Relevant equations
    \int f(x)g'(x) = f(x)g(x) - \int g(x)f'(x) dx

    3. The attempt at a solution
    Going through any of the integrals provides constants that seem to be a problem in proving the above.
    Starting from the left side and applying the integral product rule (see 2. -- not sure how the rule is called in English!) we get:

    \int cos^mxdx = \int cos^{m-1}x \cdot cosx dx
    With [tex]f(x) = cos^{m-1}x[/tex], [tex]g'(x) = cosx[/tex], [tex]g(x) = sinx[/tex], [tex]f'(x) = (m-1)cos^{m-2}x[/tex] we get:

    cos^{m-1}x \cdot sinx - \int sinx(m-1)cos^{m-2}x dx = cos^{m-1}x \cdot sinx - (m-1)\int cos^{m-2}x \cdot sinx dx

    With [tex]f(x) = cos^{m-2}x[/tex], [tex]g'(x) = sinx[/tex], [tex]g(x) = -cosx[/tex], [tex]f'(x) = (m-2)cos^{m-3}x[/tex] we get:

    cos^{m-1}x \cdot sinx - (m-1)\left( cos^{m-2}x (-cosx) - \int (-cosx)(m-2)cos^{m-3}x dx \right) =

    cos^{m-1}x \cdot sinx - (m-1)\left( -cos^{m-1}x + (m-2)\int cos^{m-2}x dx \right) =

    Any ideas on how to continue?
  2. jcsd
  3. Dec 27, 2008 #2


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    Homework Helper

    In your first parts integration you should have f'(x)=(m-1)*cos(x)^(m-2)*(-sin(x)). You forgot to use the chain rule. That gives you a sin(x)^2 in the parts integral. Change it to 1-cos(x)^2 and rearrange.
  4. Dec 27, 2008 #3
    Ah..so true :) Thank you!!
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