Integrate without expansion? I want to evaluate 
$$ \int_0^1 ( 1 - x^2)^{10} dx $$
One way I can do this is by expanding out $(1 - x^2)^{10}$ term by term, but is there a better way to do this? 

Similar to dxdydz's answer, let $x=\sin(t)$
$$\int_0^1 ( 1 - x^2)^{10}\, dx=\int_0^{\frac \pi 2}\cos^{21}(t)\,dt$$ and remember that
$$\int_0^{\frac \pi 2}\cos^{n}(t)\,dt=\frac{\sqrt{\pi }}2 \frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
   \left(\frac{n+2}{2}\right)}$$
