I think you need this:
cos2x = 1 - 2sin^2x = 2cos^2x - 1
$from this we get $ sin^2x = \frac{1}{2}(1 - cos2x)
cos^2x = \frac{1}{2}(1 + cos2x)
For the first one..
\int^1_{-1} \sqrt{4 - x^2}dx
= 2\int^1_0 \sqrt{4 - x^2}dx
$Let $ x = 2sin\theta
dx = 2cos\theta d\theta
x = 1, \theta =...