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. 
$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx = \left[\frac{\arcsin(2x-1)+\sqrt{(1-x)x}(4x-2)}{8}\right]_{0}^{1}=\frac{\pi}{8}$$
 A: Another technique just for fun (and in the meanwhile, happy new year!). We have
$$ \frac{1}{\sqrt{1-x}}\stackrel{L^2(0,1)}{=}2\sum_{n\geq 1}P_n(2x-1) $$
hence by Bonnet's recursion formulas and symmetry
$$ \sqrt{1-x}=2\sum_{n\geq 0}\frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$
$$ \sqrt{x}=2\sum_{n\geq 0}\frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$
and these FL expansions lead to
$$ \int_{0}^{1}\sqrt{x(1-x)}\,dx = \color{blue}{4\sum_{n\geq 0}\frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$
Partial fraction decomposition and telescopic series convert the RHS into
$$ \frac{1}{2}\sum_{n\geq 0}\frac{(-1)^n}{2n+1} = \frac{1}{2}\int_{0}^{1}\sum_{n\geq 0}(-1)^n x^{2n}\,dx = \frac{1}{2}\int_{0}^{1}\frac{dx}{1+x^2}=\frac{\pi}{8}.$$
As a by-product, we got a nice, rapidly convergent representation for $\pi$ (eight times the blue one).

Don't like the Legendre base of $L^2(0,1)$? Fine, let us go with the Chebyshev one (with respect to a different inner product). Our integral is 
$$ \int_{0}^{1}x(1-x)\frac{dx}{\sqrt{x(1-x)}} $$
and the orthogonality relation for the Chebyshev base is
$$ \int_{0}^{1}T_m(2x-1)T_n(2x-1)\frac{dx}{\sqrt{x(1-x)}}=\frac{\pi}{2}\delta(m,n)(1+\delta(m)). $$
Since $x(1-x)$ decomposes as $\color{blue}{\frac{1}{8}}T_0(2x-1)-\frac{1}{8}T_2(2x-1)$, the value of our integral is $\frac{\pi}{8}$.
A: 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$$\int\limits_0^1\mathrm dx\, \sqrt{x(1-x)}\color{blue}{=\frac {\pi}8}$$
A: Write $x=\sin^2 t$ so $dx=2\sin t\cos t dt$. Your integral becomes $$\int_0^{\pi/2}2\sin^2 t\cos^2 t dt=\int_0^{\pi/2}\frac12 \sin^2 2t dt=\int_0^{\pi/2}\frac{1-\cos 4t}{4}dt=\left[\frac{t}{4}-\frac{1}{16}\sin 4t\right]_0^{\pi/2}=\frac{\pi}{8}.$$
A: Let $\sqrt{x-x^2}=y$.
Thus, $y\geq0$ and $$x^2-x+y^2=0$$ or
$$\left(x-\frac{1}{2}\right)^2+y^2=\left(\frac{1}{2}\right)^2,$$ which is a semicircle with radius $\frac{1}{2}.$
Thus, our integral is $$\frac{1}{2}\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{8}.$$
A: Substitute $x=\frac12(1+t) $
$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\ dx 
=\frac14\int_{-1}^1\sqrt{1-t^2}\ dt=\frac{\pi}{8}$$
