Question

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

Answer 1

$I={\int }_1^2\frac{dx}{\sqrt{(x-1)(2-x)}}={\int }_1^2\frac{dx}{\sqrt{2x-x^2-2+x}}={\int }_1^2\frac{dx}{\sqrt{x^2-3x+2}}={\int }_1^2\frac{dx}{\sqrt{-[x^2-2.\frac{3}{2}x{\left(\frac{3}{2}\right)}^2÷2-\frac{9}{4}]}}={\int }_1^2\frac{dx}{\sqrt{-\{{(x-\frac{3}{2})}^2-{\left(\frac{1}{2}\right)}^2\}}}={\int }_1^2\frac{dx}{\sqrt{{\left(\frac{1}{2}\right)}^2-{(x-\frac{3}{2})}^2}}={\left[si{n}^{-1}(x-\frac{x-\frac{3}{2}}{\frac{1}{2}})\right]}_1^2=\left[si{n}^{-1}\right(2x-3){]}_1^2=si{n}^{-1}1-si{n}^{-1}(-1)=\frac{\pi }{2}+\frac{\pi }{2}[\therefore sin\frac{\pi }{2}=1andsin(-\theta )=-sin\theta ]=\pi$