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Thread: Evaluating Integrals

  1. June 1st 2009, 01:18 PM #1
    Myers3490
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    Evaluating Integrals

    square root x sin^2(x^(3/2))dx, u=1-r^3
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  2. June 1st 2009, 02:01 PM #2
    Spec
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    \int \sqrt{x}\sin ^2 x^{3/2}\ dx=\left\{ u=x^{3/2},\ du=\frac{3}{2}\sqrt{x}\ dx \implies dx=\frac{2du}{3\sqrt{x}} \right\}=\int \frac{2}{3}\sin^2 u \ du

    Use the identity \sin^2 u = \frac{1-\cos 2u}{2}
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