2 times the integral ( from 0 to pi/2) of [ sin theta (1+cos theta)^2 -sin theta ] d(theta).
Thank you very much.
This is
$\displaystyle 2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 - sin(\theta)] d\theta$
right?
$\displaystyle = 2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 ] d \theta - 2 \int_0^{\pi/2} sin(\theta) d\theta$
The second integral is trivial. For the first integral, let
$\displaystyle u = 1 + cos(\theta) \implies dx = -sin(\theta) d\theta$
So the first integral becomes:
$\displaystyle = 2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 ] d \theta = 2 \int_2^1 u^2 \cdot (-1) du$
$\displaystyle = -2 \int_2^1 u^2 du = 2 \int_1^2 u^2 du$
which is another trivial integral.
-Dan