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Math Help - recursive integration formula

  1. #1
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    need help with this integration by parts

    I need help with this question

    Use integration by parts to express:

    I(n) = (int) sin^n(x) dx

    in terms of I(n - 2). Hence show that:

    (int) between pi/4 & pi/2 1/ sin^4(x) dx =4 / 3


    help would be much appreciated

    cheers.
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  2. #2
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    Ok, I think I've worked out your explanation.

    \int^{\frac {\pi}{2}}_{\frac{\pi}{4}} \sin^{n}x  dx

    Integral in terms of x using n is

    [\frac {-\cos^{n+1}x}{n+1}]^{\frac {\pi}{2}}_{\frac{\pi}{4}}

    And on you go, hopefully....if I can work LaTeX
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  3. #3
    Member Glaysher's Avatar
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    Quote Originally Posted by sterps
    I need help with this question

    Use integration by parts to express:

    I(n) = (int) sin^n(x) dx

    in terms of I(n - 2). Hence show that:

    (int) between pi/4 & pi/2 1/ sin^4(x) dx =4 / 3


    help would be much appreciated

    cheers.
    \int sin^n x dx = \int sin^{n-1}xsin x dx

    u = sin^{n-1}x and v' = sin x

    u' = (n - 1)sin^{n-2}xcosx and v = -cosx

    Integral = -sin^{n-1}xcosx + (n-1)\int sin^{n-2}xcos^2xdx

    Integral = -sin^{n-1}xcosx + (n-1)\int sin^{n-2}x(1-sin^2x)dx

    Integral = -sin^{n-1}xcosx + (n-1)\int sin^{n-2}xdx  - (n-1)\int sin^nxdx

    I_n = -sin^{n-1}xcosx + (n-1)I_{n-2} - (n-1)I_n

    nI_n = -sin^{n-1}xcosx + (n-1)I_{n-2}

    I_n = \frac{-sin^{n-1}xcosx + (n-1)I_{n-2}}{n}

    See if you can do the rest
    Last edited by Glaysher; August 30th 2006 at 08:14 AM.
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  4. #4
    Member Glaysher's Avatar
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    Quote Originally Posted by Random333
    Ok, I think I've worked out your explanation.

    \int^{\frac {\pi}{2}}_{\frac{\pi}{4}} \sin^{n}x  dx

    Integral in terms of x using n is

    [\frac {-\cos^{n+1}x}{n+1}]^{\frac {\pi}{2}}_{\frac{\pi}{4}}

    And on you go, hopefully....if I can work LaTeX
    This I'm afraid is not correct
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  5. #5
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    Hello, sterps!

    This is a Reduction Formula derivation.
    It's quite long and elaborate . . . better sit down.


    Use integration by parts to express:

    I(n) = \int \sin^n\!x\\, dx in terms of I(n - 2).

    Given: . I(n)\;=\;\int\sin^n\!x\,dx

    Let: u = \sin^{n-1}\!x\qquad\qquad\qquad\qquad dv = \cos x\,dx

    Then: du = (n-1)\sin^{n-2}\!x\cos x\,dx\qquad v = -\sin x

    We have: . I(n)\;=\;-\sin^{n-1}\!x\cos x - \int[-\cos x][(n-1)\sin^{n-1}\!x\cdot\cos x\,dx

    . . . . . . . . I(n)\;= \;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x\cos^2\!x\,dx

    . . . . . . . . I(n) \;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x(1 - \sin^2\!x)\,dx

    . . . . . . . . I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\left[\sin^{n-2}\!x - \sin^n\!x\right]\,dx

    . . . . . . . . I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\int\sin^{n-2}\!x\,dx -  (n-1)\!\underbrace{\int\sin^n\!x\,dx}
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is I(n)

    We have: . I(n) \;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x\,dx - (n-1)\!\cdot\!I(n)

    I(n) + (n-1)\!\cdot\!I(n) \;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\underbrace{\int\sin^{n-2}\!x\,dx}

    . . . . . . . . n\!\cdot\!I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\cdot\!I(n-2)


    Therefore: . \boxed{I(n) \;=\;-\frac{1}{n}\sin^{n-1}\!x\cos x + \frac{n-1}{n}\!\cdot\!I(n-2)}

    Whew! . . . I need a nap . . .

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  6. #6
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    thanks for the help guys.

    ive never heard of Reduction Formula derivation.
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  7. #7
    Member Glaysher's Avatar
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    Well it is the name of what you are trying to do.

    Kind of turning an integration problem into an iterative problem
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