Question

Solve: ${\int }_a^b\frac{1}{\sqrt{\left(x-a\right)\left(b-x\right)}}dx,b>a$

Answer 1

$\begin{array}{c}{\int }_a^b\frac{1}{\sqrt{\left(x-a\right)\left(b-x\right)}}dx,b>a\\ let,x=a{\sin }^2\theta +b{\cos }^2\theta \\ dx=\left(2a\sin \theta \cos \theta +2b\cos \theta \sin \theta \right)d\theta \\ =2\sin \theta \cos \theta \left(a+b\right)d\theta \\ when,x=a,then\\ a=a{\sin }^2\theta +b{\cos }^2\theta \\ \Rightarrow a{\cos }^2\theta =b{\cos }^2\theta \\ \Rightarrow \cos \theta =0^0\Rightarrow \theta =\frac{\pi }{2}\\ when,x=b,\\ b=a{\sin }^2\theta +b{\cos }^2\theta \\ \Rightarrow \theta =0^0\\ {\int }_{\frac{\pi }{2}}^0\frac{2\sin \theta \cos \theta }{\sqrt{\left(b-a\right){\cos }^2\theta }\sqrt{\left(b-a\right){\sin }^2\theta }}d\theta \\ ={\int }_{\frac{\pi }{2}}^0\frac{2\sin \theta \cos \theta \left(a-b\right)}{\left(b-a\right)\cos \theta \sin \theta }d\theta \\ =\left(a-b\right)\times 2\times \frac{-\pi }{2}=\pi (b-a)\end{array}$