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

Solution 1:

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

(Quite an overkill solution ...)

Solution 2:

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

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

Solution 3:

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.