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Integration by substitution question

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data
    I've been working on a problem from Apostol "Calculus" Volume 1 (not homework but self study). The problem is Section 5.8, Number 25 (Page 217) and states:
    If [tex]$\m$[\tex] is a positive integer, show that:

    [itex]\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = 2^{-m}\int_0^{\frac{\pi}{2}} cos^m x dx[/itex]

    2. Relevant equations


    3. The attempt at a solution
    My thinking (so far) has been that I have two consider two cases,
    1. m odd
    2. m even
    When I have tackled $$\m$$ odd, the following occurred (and this is what confused me):
    I use the substitution $$u = cos x, du = -sin x dx $
    [itex]\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx =
    - \int_0^{\frac{\pi}{2}} cos^m x sin^{m-1} x (-sin x) dx
    = -\int_0^{\frac{\pi}{2}}cos^m x (sin^2 x)^{\frac{m-1}{2}}(-sin x) dx
    = -\int_0^{\frac{\pi}{2}}cos^m x (1 - cos^2 x)^{\frac{m-1}{2}}(-sin x) dx
    = -\int_1^0 u^m(1 - u^2)^{\frac{m-1}{2}} du
    = \int_0^1 u^m(1 - u^2)^{\frac{m-1}{2}} du
    [/itex]
    Since this is clearly a binomial expansion within the integral, it seems to me that I am moving away from the answer rather than towards the answer. I know that if $$m$$ odd, I would need to use the half angle formulas for $$sin x$$ and $$ cos x$$. In my opinion, Im clearly missing the point, so if anyone can point me in the right direction, that would be appreciated.
     
  2. jcsd
  3. Mar 21, 2014 #2

    Simon Bridge

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    Hmmm ... tidying that up first:
    [tex]\int_0^{\frac{\pi}{2}} cos^m x sin^m x dx = 2^{-m}\int_0^{\frac{\pi}{2}} cos^m x dx[/tex]

    2. Relevant equations


    3. The attempt at a solution
    My thinking (so far) has been that I have two consider two cases,
    1. m odd
    2. m even
    When I have tackled ##m## odd, the following occurred (and this is what confused me):
    I use the substitution $$u = \cos x, du = -\sin x dx $$

    $$\int_0^{\frac{\pi}{2}} \cos^m x \sin^m x dx \\
    = - \int_0^{\frac{\pi}{2}} \cos^m x \sin^{m-1} x (-\sin x) dx\\
    = -\int_0^{\frac{\pi}{2}}\cos^m x (\sin^2 x)^{\frac{m-1}{2}}(-\sin x) dx\\
    = -\int_0^{\frac{\pi}{2}}\cos^m x (1 - \cos^2 x)^{\frac{m-1}{2}}(-\sin x) dx\\
    = -\int_1^0 u^m(1 - u^2)^{\frac{m-1}{2}} du\\
    = \int_0^1 u^m(1 - u^2)^{\frac{m-1}{2}} du
    $$


    Since this is clearly a binomial expansion within the integral, it seems to me that I am moving away from the answer rather than towards the answer. I know that if ##m## odd, I would need to use the half angle formulas for ##\sin x## and ##\cos x##. In my opinion, I'm clearly missing the point, so if anyone can point me in the right direction, that would be appreciated.
     
  4. Mar 21, 2014 #3

    Simon Bridge

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    I would have used the double-angle formulas myself - specifically: $$\sin 2x = 2\sin x\cos x$$
    http://en.wikipedia.org/wiki/List_o...le.2C_triple-angle.2C_and_half-angle_formulae

    There's a reduction formula for ##\int \sin^n x\; dx## and ##\int \cos^n x\; dx##

    You could also use the power reduction formulae on the integrands.
    This divides into m odd vs m even.

    Of course there is also: Note: $$\int \sin^nx\cos^nx\;\dx = -\frac{\sin^{n-1}x\cos^{n+1}x}{2n}+\frac{n-1}{2n}\int \sin^{n-2}x\cos^nx\;dx\; :\; n>0$$
    http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions

    You only need to show that the LHS = RHS after the definitite integral BTW: i.e. the areas must be the same, not the integrands.
     
    Last edited: Mar 21, 2014
  5. Mar 21, 2014 #4

    utkarshakash

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    I would suggest using Walli's Formula rather than doing the integration manually.
     
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