-1/2 times the integral (from 0 to pi/2) of cos^4 theta - 2cos^2 theta d(theta)
Thank you very much.
$\displaystyle -\frac{1}{2} \int_{0}^{\pi/2}( (cos^2( \theta ) )^2 - 2 cos^2( \theta )) d \theta$
$\displaystyle = -\frac{1}{2} \int_{0}^{\pi/2}\left ( \left ( \frac{1 + cos(2 \theta )}{2} \right )^2 - 2 \left ( \frac{ 1 + cos(2 \theta )}{2} \right ) \right ) d \theta$
$\displaystyle = -\frac{1}{8} \int_{0}^{\pi/2} (1 + cos(2 \theta ))^2 d \theta + \frac{1}{2} \int_0^{\pi/2} (1 + cos(2 \theta ) ) d \theta$
Expand the integrand in the first integral:
$\displaystyle = -\frac{1}{8} \int_{0}^{\pi/2}(1 + 2cos(2 \theta ) + cos^2(2 \theta) ) d \theta + \frac{1}{2} \int_0^{\pi/2} (1 + cos(2 \theta ) ) d \theta $
$\displaystyle = \frac{3}{8} \int_{0}^{\pi/2} d \theta + \frac{1}{4} \int_0^{\pi/2} cos(2 \theta ) d \theta - \frac{1}{8} \int_0^{\pi/2} cos^2(2 \theta) d \theta$
Now use power reduction in the last integral again:
$\displaystyle = \frac{3}{8} \int_{0}^{\pi/2} d \theta + \frac{1}{4} \int_0^{\pi/2} cos(2 \theta ) d \theta - \frac{1}{8} \int_0^{\pi/2} \frac{1 + cos(4 \theta)}{2} d \theta$
$\displaystyle = \frac{5}{16} \int_{0}^{\pi/2} d \theta + \frac{1}{4} \int_0^{\pi/2} cos(2 \theta ) d \theta - \frac{1}{16} \int_0^{\pi/2} cos(4 \theta) d \theta$
Now try it.
-Dan