# Proving that $\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$

How do I prove that $$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi?$$ I'm just wondering if LHS even equal to the RHS in the first place? Thanks for the help!

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What if $a=b=0$? – M.B. Sep 9 '13 at 16:03
@M.B. Who cares? =) – Pedro Tamaroff Sep 9 '13 at 22:00

Hint: Substitute $x=a\cos^2\phi+b\sin^2\phi$.

(Quite an overkill solution ...)

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This is such a brilliant substitution! – L. F. Sep 9 '13 at 16:06
Amazing, may I ask how you came up with this substitution? – Spine Feast Sep 9 '13 at 16:35
I'm having two analysis class right now ... – user67258 Sep 9 '13 at 16:38
Certainly they've done you some good. (+1) – Chris Sep 9 '13 at 17:33

Hint:$$\int_a^b{(u-a)^{x-1}(b-v)^{y-1}}=(b-a)^{x+y-1}\frac{\Gamma(x).\Gamma(y)}{\Gamma(x+y)}$$

$$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=(b-a)^{\frac12+\frac12-1}\frac{\Gamma(\frac12).\Gamma(\frac12)}{\Gamma(1)}=\pi$$

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Here is another method using square completion: \begin{align} \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}&=\int_a^b \frac{dx}{\sqrt{-\left(x^2-(a+b)x+ab\right)}}\\ &=\int_a^b \frac{dx}{\sqrt{\frac{1}{4}(a-b)^2-\left(x-\frac{b}{2}-\frac{a}{2}\right)^2}} \, \text{square completion}\\ &=\int_{\frac{a-b}{2}}^{\frac{b-a}{2}}\frac{2du}{\sqrt{(a-b)^2-4u^2}}, \,\text{substituing } u=x-b/2-a/2 \end{align} Let $u=\frac{(a-b)}{2}\sin(v)$ to complete.

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