How to calculate $\pi\int^1_0\sqrt{x(1-x)} \mathrm \, dx$ =? I need to find the value of the following integral
$$\pi\int^1_0\sqrt{x(1-x)} \mathrm \, dx = ? $$
I tried parts integration, $a\sin x$ substitution and none of them seems work.
Can someone get me on the right path? 
Thank you very much!
 A: Put $x=\sin^2\theta\implies dx= \sin2\theta d\theta$.Therefore, your integral becomes, $$ \int_0^{\pi/2}\sin\theta \cos\theta \sin2\theta d\theta=(1/2)\int_0^{\pi/2}\sin^22\theta d\theta=(1/4)\int_0^{\pi/2}(1-\cos4\theta) d\theta=\pi/8$$
A: HINT:
$\sqrt{x(1-x)} = \sqrt{\frac{1}{4} -(\frac{1}{2}-x)^2}= \frac{1}{2}\sqrt{1 -(1-2x)^2} $
Then change variable
$1-2x=\sin u$
A: Let $y=\sqrt{x(1-x)},0\leq x\leq 1,$ and this is the upper semicircle of radius $\frac12$ and centered $(1/2,0)$.
From the geometric meaning of definite integral, this integral is equal to
$$\int^1_0\sqrt{x(1-x)} \mathrm \, dx=\frac12\pi\left(\frac12\right)^2=\frac{\pi}{8}.$$
A: $$y=\sqrt{x(1-x)}$$ describes a half circle of radius $\frac12$. Hence the integral must be
$$\pi\frac12\pi\left(\frac12\right)^2.$$
A: Hint: Try to bring this integral to form: $\int \sqrt{a^{2}-t^{2}}dt$. Than use $\int \frac{t^{2}}{\sqrt{a^{2}-t^{2}}}dt$ to get result. You should get system of two linear equations with two unknown (those two integrals) - which is easy to solve by elimination. 
