How to show that $\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$ I was reading Advanced Integration Techniques, and found that$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$$
The book provides one method using residue theorem and Laurent expansion. However, I wonder if there are other techniques that I can use to evaluate this integral. 
The most direct method would be solving the integral and plug in the limits. 
$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx = \left[\frac{\arcsin(2x-1)+\sqrt{(1-x)x}(4x-2)}{8}\right]_{0}^{1}=\frac{\pi}{8}$$

J.G. has the elementary method down. If you see a $1-x^2$ term inside your integrand, it might be wise to give a trig substitution a try. In this case, letting $x=\sin^2\theta$ works out beautifully.
There’s another less elementary way by utilizing the beta function and the Gamma function. Let me know if you need a proof (I proudly have an elementary proof of it).

Beta Function:$$\operatorname{B}\left(m,n\right)=\int\limits_0^1\mathrm dx\,x^{m-1}(1-x)^{n-1}=\frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$

The integral under question is then simply$$\begin{align*}\mathfrak{I} & =\operatorname{B}\left(\frac 32,\frac 32\right)\\ & =\frac {1}{2}\left(\frac {\sqrt{\pi}}2\right)^2\end{align*}$$
Thus$$\int\limits_0^1\mathrm dx\, \sqrt{x(1-x)}\color{blue}{=\frac {\pi}8}$$
