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Question Number 47976 by peter frank last updated on 17/Nov/18

Commented by maxmathsup by imad last updated on 17/Nov/18

$$letA={\int }_{\frac{1}{4}}^{\frac{3}{4}}\frac{{\left(\sqrt{x-x^2}\right)}^{-1}}{{\left(dx\right)}^{-2}}{\left(dx\right)}^{-1}\Rightarrow A={\int }_{\frac{1}{4}}^{\frac{3}{4}}\frac{dx}{\sqrt{x-x^2}}$$$$={\int }_{\frac{1}{4}}^{\frac{3}{4}}\frac{dx}{\sqrt{-(x^2-2x\frac{1}{2}+\frac{1}{4}-\frac{1}{4})}}={\int }_{\frac{1}{4}}^{\frac{3}{4}}\frac{dx}{\sqrt{\frac{1}{4}-{(x-\frac{1}{2})}^2}}$$$${=}_{x-\frac{1}{2}=\frac{t}{2}}\frac{1}{2}{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{2dt}{\sqrt{1-t^2}}={\left[arcsin\left(t\right)\right]}_{-\frac{1}{2}}^{\frac{1}{2}}=\{\frac{\pi }{6}+\frac{\pi }{6}\}=\frac{2\pi }{6}=\frac{\pi }{3}.$$

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