Question

Evaluate the following definite integral:

${\int }_1^2\frac{1}{\sqrt{(x-1)(2-x)}}dx$

Answer 1

$I={\int }_1^2\frac{1}{\sqrt{-x^2+3x-2}}dx={\int }_1^2\frac{1}{\sqrt{-\{{(x-\frac{3}{2})}^2-{\left(\frac{1}{2}\right)}^2\}}}dx={\int }_1^2\frac{1}{\sqrt{{\left(\frac{1}{2}\right)}^2-{(x-\frac{3}{2})}^2}}dx$

$\int \frac{1}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}{\int }_1^2\frac{1}{\sqrt{{\left(\frac{1}{2}\right)}^2-{(x-\frac{3}{2})}^2}}={{\sin }^{-1}\left(\frac{x-\frac{3}{2}}{\frac{1}{2}}\right)|}_1^2{{\sin }^{-1}(2x-3)|}_1^2={\sin }^{-1}\left(1\right)-{\sin }^{-1}(-1)=\frac{\pi }{2}+\frac{\pi }{2}=\pi$