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:
$\displaystyle \left ( \frac{4 - x^2}{2} \right ) ^{3/2} - \frac{1}{3}\left ( \frac{4 - x^2}{2} \right ) ^{3/2} = \frac{2}{3}\left ( \frac{4 - x^2}{2} \right ) ^{3/2}$
Now you want:
$\displaystyle \int_{-2}^2 \frac{2}{3}\left ( \frac{4 - x^2}{2} \right ) ^{3/2} dx = \frac{2}{3 \cdot 2^{3/2}}\int_{-2}^2 (4 - x^2)^{3/2} dx$
Let $\displaystyle x = 2 sin(\theta)$ then $\displaystyle dx = 2cos(\theta) d\theta$:
So
$\displaystyle \int_{-2}^2 (4 - x^2)^{3/2} dx = \int_{-\pi/2}^{\pi/2} (4 - 4sin^2(\theta))^{3/2} \cdot 2cos(\theta) d\theta$
$\displaystyle = 2 \cdot 4^{3/2} \int_{-\pi/2}^{\pi/2} (1 - sin^2(\theta))^{3/2} \cdot cos(\theta) d\theta$
$\displaystyle = 16 \int_{-\pi/2}^{\pi/2} cos^4(\theta) d\theta$
Can you take it from here?
-Dan