2 times the integral ( from 0 to pi/2) of [ sin theta (1+cos theta)^2 -sin theta ] d(theta).

Thank you very much.

2. Originally Posted by kittycat
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
$2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 - sin(\theta)] d\theta$
right?

$= 2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 ] d \theta - 2 \int_0^{\pi/2} sin(\theta) d\theta$

The second integral is trivial. For the first integral, let
$u = 1 + cos(\theta) \implies dx = -sin(\theta) d\theta$

So the first integral becomes:
$= 2 \int_0^{\pi/2} [sin(\theta)(1 + cos(\theta))^2 ] d \theta = 2 \int_2^1 u^2 \cdot (-1) du$

$= -2 \int_2^1 u^2 du = 2 \int_1^2 u^2 du$
which is another trivial integral.

-Dan

3. One approach may be to rewrite as:

$sin{\theta}(1+cos{\theta})^{2}-sin{\theta}=$

$\frac{sin(3\theta)}{4}+sin(2\theta)+\frac{sin(\the ta)}{4}$

Makes for an easy term by term integration.