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Math Help - (x^2)cosx

  1. #1
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    (x^2)cosx

    Calculate the integral from 0 to pi of (x^2)cosxdx
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  2. #2
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    Quote Originally Posted by TexasGirl
    Calculate the integral from 0 to pi of (x^2)cosxdx
    I will find the anti-derivative then you can complete the problem.
    ---
    \int x^2\cos x dx.
    Let u=x^2 and v'=\cos x
    Then,
    u'=2x and v=\sin x
    Thus, by parts,
    x^2\sin x-2\int x \sin xdx
    Now, you do it again,
    Let u=x and v'=\sin x
    Then,
    u'=1 and v=-\cos x
    Thus, by parts,
    x^2\sin x-2\left( -x\cos x-\int -\cos xdx \right)
    Simplify,
    x^2\sin x+2x\cos x-2\int \cos xdx
    Thus,
    x^2\sin x+2x\cos x-2\sin x+C
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  3. #3
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    Hello, TexasGirl!

    You need to integrate by parts . . . twice!


    I \;= \;\int_0^{\pi} x^2\cos x\,dx

    Let: u = x^2\qquad dv = \cos x\,dx

    Then: du = 2x\,dx\quad v = \sin x

    And we have: . I\;=\;x^2\sin x - 2\int x\sin x\,dx


    Let: u = x\qquad dv = \sin x\,dx

    Then: du = dx\quad v = -\cos x

    And we have: . I \;= \;x^2\sin x - 2\left[-x\cos x - \int(-\cos x)\,dx\right]
    . . . . . . . . . . I \;= \;x^2\sin x + 2x\cos x - 2\int\cos x\,dx

    Therefore:. . . \boxed{I \;= \;x^2\sin x + 2x\cos x - 2\sin x + C}


    Edit: Too fast for me, TPHacker!
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