<s>[INST] 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!
 [/INST] 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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