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Math Help - substitution

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    substitution

    Thanks!
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    Quote Originally Posted by gralla55 View Post
    Thanks!
    Is this

    \int{2\pi x \sqrt{4 - x^2}\,dx}?
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    Yep!
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    \int{2\pi x \sqrt{4 - x^2}\,dx} = \int{2\pi x (4 - x^2)^{\frac{1}{2}}\,dx}.


    Let u = 4 - x^2 so that \frac{du}{dx} = -2x.


    You can rewrite your integral as

    -\pi \int{ (4 - x^2)^{\frac{1}{2}}(-2x)\,dx}

     = -\pi\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx}

     = -\pi \int{u^{\frac{1}{2}}\,du}

     = -\frac{2\pi}{3}u^{\frac{3}{2}} + C

     = C -\frac{2\pi \sqrt{(4 - x^2)^3}}{3}.
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