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Math Help - Trig Substitution again

  1. #1
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    Trig Substitution again

    The integral from 0 to a of (x^2)(((a^2)-(x^2))^.5) dx
    It must be done using trig substitution. I have it down to
    (((a^4)(2sqrt(2))-(a^4))/3 + ((a^4)-((a^4)(4sqrt(2))))/5

    Is there any way to simply further?
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Hellacious D View Post
    The integral from 0 to a of (x^2)(((a^2)-(x^2))^.5) dx
    It must be done using trig substitution. I have it down to
    (((a^4)(2sqrt(2))-(a^4))/3 + ((a^4)-((a^4)(4sqrt(2))))/5

    Is there any way to simply further?

    \int_{0}^{a} x^2\sqrt{a^2-x^2} dx

    let a\sin u = x \Rightarrow a\cos u\cdot du = dx

    \cos u = \frac{\sqrt{a^2-x^2} }{a}

    \int a^2\sin ^2u (a\cos u )(a\cos u) du

    \int a^4 \sin ^2u \cos ^2u\cdot du

    \int \frac{a^4 \sin ^2(2u)}{4} du

    \int \frac{a^4 (1-\cos 4u)}{8} du

    \frac{a^4 (u)}{8} - \frac{a^4\sin 4u}{32}

    but u=\sin ^{-1} \frac{x}{a}

    \frac{a^4 \sin ^{-1} \frac{x}{a}}{8} - \frac{a^4\sin 4(\sin ^{-1} \frac{x}{a})}{32} \mid ^{a}_{0}

    \frac{a^4 \sin ^{-1} (1)}{8} - \frac{a^4 \sin 4(\sin ^{-1} (1))}{32} - \frac{a^4 \sin ^{-1} (0)}{8} + \frac{a^4 \sin 4(\sin ^{-1} (0))}{32}

    \sin ^{-1} (1) = \frac{\pi}{2}

    \sin ^{-1} (0) = 0
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