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Math Help - Trigonometric substitution

  1. #1
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    Trigonometric substitution

    Can someone help me with this trig subst please x^3/Sqrt(9-x^2) with respect to dx. Thanks
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  2. #2
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    Hello, Link88!

    Integrate: . \int\frac{x^3}{\sqrt{9-x^2}}\,dx
    Let: x \:=\:3\sin\theta \quad\Rightarrow\quad dx \:=\:3\cos\theta\,d\theta . . . and: . \sqrt{9-x^2} \:=\;3\cos\theta

    Substitute: . \int\frac{27\sin^3\!\theta}{3\cos\theta}\,3\cos\th  eta\,d\theta \;=\;27\int\sin^3\!\theta\,d\theta \;=\;27\int(\sin^2\!\theta)(\sin\theta\,d\theta)

    . . = \;27\int(1 - \cos^2\!\theta)(\sin\theta\,d\theta) \;=\;27\bigg[\int\sin\theta\,d\theta - \int\cos^2\!\theta(\sin\theta\,d\theta) \bigg]

    . . = \;27\bigg[-\cos\theta + \frac{1}{3}\cos^3\!\theta\bigg] + C \;=\;9\cos\theta\left(\cos^2\!\theta - 3\right) + C


    Back-substitute: . \sin\theta \,=\,\frac{x}{3}\quad\Rightarrow\quad \cos\theta \,=\,\frac{\sqrt{9-x^2}}{3}

    and we have: . 9\cdot\frac{\sqrt{9-x^2}}{3}\left(\frac{9-x^2}{9} - 3\right) + C  \;=\;3\sqrt{9-x^2}\left(\frac{9-x^2-27}{9}\right) + C

    . . = \;\;\sqrt{9-x^2}\left(\frac{-x^2-18}{3}\right) \;\;=\;\;\boxed{-\frac{1}{3}\sqrt{9-x^2}\,(x^2+18) + C}

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