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Math Help - integration of x^3*sqrt(90x^2)dx using sub x=sin(u)

  1. #1
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    Question integration of x^3*sqrt(90x^2)dx using sub x=sin(u)

    Question states using the substitution x=3 sin(x) the integral \int {x^3 \sqrt {9 - x^2 } dx} becomes?

    Answer is given as <br />
3^4 \int {\sin ^3 (u) - \sin ^5 (u)} du<br />

    I don't understand how to work this out
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  2. #2
    o_O
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    I think you're missing a 3 there: {\color{red}x = 3\sin u} \ \Rightarrow \ {\color{blue}dx = 3\cos u \ du}

    So: \int {\color{red}x}^3 \sqrt{9-{\color{red}x}^2} \ {\color{blue}dx} = \int ({\color{red}3 \sin u})^3\sqrt{9-({\color{red}3\sin u})^2} \ {\color{blue}3 \cos u \ du}
     = \int 3^3 \sin^3 u \sqrt{9\left(1 -\sin^2 u\right)} 3\cos u \ du = 3^5 \int \sin^3 u \cos^2 u \ du

    Convert \cos^2 u into terms of \sin u and you should be fine.
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