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Thread: (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.
    ---
    $\displaystyle \int x^2\cos x dx$.
    Let $\displaystyle u=x^2$ and $\displaystyle v'=\cos x$
    Then,
    $\displaystyle u'=2x$ and $\displaystyle v=\sin x$
    Thus, by parts,
    $\displaystyle x^2\sin x-2\int x \sin xdx$
    Now, you do it again,
    Let $\displaystyle u=x$ and $\displaystyle v'=\sin x$
    Then,
    $\displaystyle u'=1$ and $\displaystyle v=-\cos x$
    Thus, by parts,
    $\displaystyle x^2\sin x-2\left( -x\cos x-\int -\cos xdx \right)$
    Simplify,
    $\displaystyle x^2\sin x+2x\cos x-2\int \cos xdx$
    Thus,
    $\displaystyle x^2\sin x+2x\cos x-2\sin x+C$
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  3. #3
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    Hello, TexasGirl!

    You need to integrate by parts . . . twice!


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

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

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

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


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

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

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

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


    Edit: Too fast for me, TPHacker!
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