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Math Help - integration

  1. #1
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    Question integration

    8 times the integral from -2 to 2 of [(4-x^2)/2]^3/2 - 1/3[(4-x^2)/2]^3/2 dx

    Thank you very much
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  2. #2
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    Quote Originally Posted by kittycat View Post
    8 times the integral from -2 to 2 of [(4-x^2)/2]^3/2 - 1/3[(4-x^2)/2]^3/2 dx

    Thank you very much
    I'll drop the factor of 8 in the front for simplicity.

    First simplify the integrand:
    \left ( \frac{4 - x^2}{2} \right ) ^{3/2} - \frac{1}{3}\left ( \frac{4 - x^2}{2} \right ) ^{3/2} = \frac{2}{3}\left ( \frac{4 - x^2}{2} \right ) ^{3/2}

    Now you want:
    \int_{-2}^2 \frac{2}{3}\left ( \frac{4 - x^2}{2} \right ) ^{3/2} dx = \frac{2}{3 \cdot 2^{3/2}}\int_{-2}^2 (4 - x^2)^{3/2} dx

    Let x = 2 sin(\theta) then dx = 2cos(\theta) d\theta:

    So
    \int_{-2}^2 (4 - x^2)^{3/2} dx = \int_{-\pi/2}^{\pi/2} (4 - 4sin^2(\theta))^{3/2} \cdot 2cos(\theta) d\theta

    = 2 \cdot 4^{3/2} \int_{-\pi/2}^{\pi/2} (1 - sin^2(\theta))^{3/2} \cdot cos(\theta) d\theta

    = 16 \int_{-\pi/2}^{\pi/2} cos^4(\theta) d\theta

    Can you take it from here?

    -Dan
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