Integrate $\sqrt{1-x^2}$ on $x=-1$ to $1$ I have been given the task to compute $\int_{-1}^1 \sqrt{1-x^2} dx$ by means of calculus. We got the hint to substitute $x=\sin u$, but that only seems to make things more complicated:
$$\int_{-1}^1 \sqrt{1-x^2}dx = \int_{\arcsin-1}^{\arcsin1} \sqrt{1-\sin^2u} \frac{d\arcsin u}{du} du = \int_{\arcsin-1}^{\arcsin1} \sqrt{\frac{(1-\sin u)(1+\sin u)}{(1-u)(1+u)}} du.$$
I know that the answer should be $\frac\pi2$, because this is the area under one half of the unit circle, but how to arrive there by means of calculus is completely unclear to me. Could someone point me in the right direction?
 A: Hint
$$\int_{-1}^1\sqrt{1-x^2}dx=\int_{\arcsin(-1)}^{\arcsin(1)}\sqrt{1-\sin^2(x)}\cos(x)dx=...$$
A: $$\begin{align*} \int_{-1}^{1}\sqrt{1-x^2}\,dx &= 2\int_{0}^{1}\sqrt{1-x^2}\,dx \\&= 2\int_{0}^{\pi/2}\cos^2\theta\,d\theta\\&=\int_{0}^{\pi/2}(\cos(2\theta)+1)\,d\theta\\&=\int_{0}^{\pi/2}1\,d\theta=\color{red}{\frac{\pi}{2}}\end{align*} $$
as expected. We exploited the parity of the function $\sqrt{1-x^2}$, the substitution $x=\sin\theta$, the duplication formula for the cosine function and the periodicity of the cosine function.
A: This is how it works:
$$\left\{ \matrix{
  x = \sin (u) \hfill \cr 
  x = 1\, \to \,\,\,\,\,\,\,u = {\pi  \over 2} \hfill \cr 
  x =  - 1 \to u =  - {\pi  \over 2} \hfill \cr}  \right.$$
and also
$$\sqrt {1 - {x^2}}  = \sqrt {1 - {{\sin }^2}(u)}  = \sqrt {{{\cos }^2}(u)}  = \left| {\cos (u)} \right| = \cos (u)$$
The last equality holds since $ - {\pi  \over 2} \le u \le {\pi  \over 2}$. One more thing
$$dx = \cos (u)du$$
Finally, putting all this into the integral you find
$$\eqalign{
  & \int_{ - {\pi  \over 2}}^{{\pi  \over 2}} {{{\cos }^2}(u)du}  = 2\int_0^{{\pi  \over 2}} {{{\cos }^2}(u)du}  = 2\int_0^{{\pi  \over 2}} {\left( {{{1 - \cos (2u)} \over 2}} \right)du}   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{{\pi  \over 2}} {\left( {1 - \cos (2u)} \right)du}  = \left. {\left( {u - {1 \over 2}\sin (2u)} \right)} \right|_0^{{\pi  \over 2}}  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\pi  \over 2} \cr} $$
