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Math Help - Having trouble evaluating an integral

  1. #1
    s3a
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    Having trouble evaluating an integral

    How do I go from the step indicated to the other step (Red arrow)? The red arrow points to the first part that's incorrect according to wolfram alpha.

    My work is attached.

    Any help in catching my mistake would be GREATLY appreciated!
    Thanks in advance!

    P.S.
    The answer I am supposed to get is 4*pi.
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  2. #2
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    Re: Having trouble evaluating an integral

    Quote Originally Posted by s3a View Post
    How do I go from the step indicated to the other step (Red arrow)? The red arrow points to the first part that's incorrect according to wolfram alpha.

    My work is attached.

    Any help in catching my mistake would be GREATLY appreciated!
    Thanks in advance!

    P.S.
    The answer I am supposed to get is 4*pi.
    \displaystyle \begin{align*} 4\int_0^{\sqrt{2}}{4\sqrt{2 - x^2} - 2x^2\sqrt{2 - x^2} - \frac{2}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} &= 4\int_0^{\sqrt{2}}{2\sqrt{2 - x^2}\left(2 - x^2\right) - \frac{2}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} \\ &= 4\int_0^{\sqrt{2}}{2\left(2 - x^2\right)^{\frac{3}{2}} - \frac{2}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} \\ &= 4\int{\frac{4}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} \\ &= \frac{16}{3}\int_0^{\sqrt{2}}{\left(2 - x^2\right)^{\frac{3}{2}}\,dx}\end{align*}

    Now let \displaystyle \begin{align*} x = \sqrt{2}\sin{\theta} \implies dx = \sqrt{2}\cos{\theta}\,d\theta \end{align*} and note that when \displaystyle \begin{align*} x = 0, \theta = 0 \end{align*} and when \displaystyle \begin{align*} x = \sqrt{2}, \theta = \frac{\pi}{2} \end{align*} and the integral becomes

    \displaystyle \begin{align*} \frac{16}{3}\int_0^{\sqrt{2}}{\left(2 - x^2\right)^{\frac{3}{2}}\,dx} &= \frac{16}{3}\int_0^{\frac{\pi}{2}}{\left[2 - \left(\sqrt{2}\sin{\theta}\right)^2\right]^{\frac{3}{2}}\,\sqrt{2}\cos{\theta}\,d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{\left(2 - 2\sin^2{\theta}\right)^{\frac{3}{2}}\,\cos{\theta}  \,d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{\left[2\left(1 - \sin^2{\theta}\right)\right]^{\frac{3}{2}}\,\cos{\theta}\,d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{\left(2  \cos^2{\theta}\right)^{\frac{3}{2}}\,\cos{\theta} \, d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{2\sqrt{  2}\cos^3{\theta}\cos{\theta}\,d\theta} \\ &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\cos^4{\theta} \, d\theta} \\ &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\left(\frac{1}{  2} + \frac{1}{2}\cos{2\theta}\right)^2\,d\theta} \end{align*}

    \displaystyle \begin{align*} &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\frac{1}{4} + \frac{1}{2}\cos{2\theta} + \frac{1}{4}\cos^2{2\theta}\,d\theta} \\ &= \frac{64}{3} \int_0^{\frac{\pi}{2}}{ \frac{1}{4} + \frac{1}{2} \cos{ 2 \theta } + \frac{1}{4}\left( \frac{1}{2} + \frac{1}{2} \cos{ 4\theta } \right) \,d\theta} \\ &= \frac{64}{3} \int_0^{\frac{\pi}{2}}{\frac{1}{4} + \frac{1}{2}\cos{2\theta} + \frac{1}{8} + \frac{1}{8}\cos{4\theta}\,d\theta} \\ &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\frac{3}{8} + \frac{1}{2}\cos{2\theta} + \frac{1}{8}\cos{4\theta}\,d\theta} \\ &= \frac{8}{3}\int_0^{\frac{\pi}{2}}{3 + 4\cos{2\theta} + \cos{4\theta}\,d\theta} \\ &= \frac{8}{3}\left[3\theta + 2\sin{2\theta} + \frac{1}{4}\sin{4\theta}\right]_0^{\frac{\pi}{2}} \\ &= \frac{8}{3}\left[\left(\frac{3\pi}{2} + 2\sin{\pi} + \frac{1}{4}\sin{2\pi} \right) - \left(3 \cdot 0 + 2\sin{0} + \frac{1}{4}\sin{0} \right)\right] \\ &= 4\pi \end{align*}
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  3. #3
    s3a
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    Re: Having trouble evaluating an integral

    Thank you!
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