I have the integral $ \space \int_{-1}^1 \sqrt{1-x^2}\,dx $ and I found that it equals $ {1\over2} \arcsin x + x\sqrt{1-x^2}$
and when I placed in $ {1\over2} \arcsin{x} \space + \space x\sqrt{1-x^2} \space|_{-1}^{1} \space $ I got $ - \pi\over2$ and wolfram alpha says it's $ \pi \over 2$, where did I go wrong?
(Way I solved it):
$$ ={1\over2} \cdot {\pi\over 2} + 0 - ({1\over2} \cdot {3\pi\over2} + 0)$$
$$ = {\pi\over 4} \space - \space {3 \pi \over 4} = -{\pi\over 2}$$
Suppose you evaluate the integral $\displaystyle \int_{-1}^1 \sqrt{1-x^2\,} \,dx $ by starting with the substitution $x=\sin\theta.$ As $x$ goes from $-1$ to $1,$ what does $\theta$ do? It could go from $-\pi/2$ to $\pi/2,$ or from $3\pi/2$ to $\pi/2,$ but not from $\pi/2$ to $3\pi/2,$ since that would give you $\displaystyle \int^{-1}_1$ rather than $\displaystyle \int_{-1}^1.$
So you would need $\displaystyle \int_{3\pi/2}^{\pi/2},$ not $\displaystyle \int^{3\pi/2}_{\pi/2}$