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. 
$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx = \left[\frac{\arcsin(2x-1)+\sqrt{(1-x)x}(4x-2)}{8}\right]_{0}^{1}=\frac{\pi}{8}$$

Write $x=\sin^2 t$ so $dx=2\sin t\cos t dt$. Your integral becomes $$\int_0^{\pi/2}2\sin^2 t\cos^2 t dt=\int_0^{\pi/2}\frac12 \sin^2 2t dt=\int_0^{\pi/2}\frac{1-\cos 4t}{4}dt=\left[\frac{t}{4}-\frac{1}{16}\sin 4t\right]_0^{\pi/2}=\frac{\pi}{8}.$$
