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Thread: substitution

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    substitution

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

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


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


    You can rewrite your integral as

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

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

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

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

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