# 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!

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Try $x=\dfrac{1+\sin\,u}{2}$ –  Guess who it is. Jul 13 '12 at 10:33

Put $x=\sin^2\theta\implies dx= \sin2\theta d\theta$.Therefore, your integral becomes, $$\int_0^{\pi/2}\sin\theta \cos\theta \sin2\theta d\theta=(1/2)\int_0^{\pi/2}\sin^22\theta d\theta=(1/4)\int_0^{\pi/2}(1-\cos4\theta) d\theta=\pi/8$$

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Thank you very much! –  Grozav Alex Ioan Jul 16 '12 at 4:47

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

$1-2x=\sin u$

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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.

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