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Question Number 127904 by bramlexs22 last updated on 03/Jan/21

$$\mathrm{prove}\mathrm{that}{\int }_0^{100}\frac{\mathrm{dx}}{\sqrt{\mathrm{x}(100-\mathrm{x})}}=\pi$$

Answered by liberty last updated on 03/Jan/21

$$\mathrm{let}\mathrm{x}=100{\sin }^2\mathrm{t}\Rightarrow \mathrm{dx}=200\sin \mathrm{t}\cos \mathrm{t}\mathrm{dt}$$$${\int }_0^{\frac{\pi }{2}}\frac{200\sin \mathrm{t}\cos \mathrm{t}\mathrm{dt}}{\sqrt{100\sin {}^2\mathrm{t}(100\cos {}^2\mathrm{t}})}=$$$${{\int }_0^{\frac{\pi }{2}}2\mathrm{dt}=2\mathrm{t}]}_0^{\frac{\pi }{2}}=2\times \frac{\pi }{2}=\pi$$

Answered by Dwaipayan Shikari last updated on 03/Jan/21

$${\int }_0^{100}\frac{dx}{\sqrt{x}\sqrt{100-x}}x=100u$$$$={\int }_0^1\frac{du}{\sqrt{u}\sqrt{1-u}}=\beta (\frac{1}{2},\frac{1}{2})={\mathrm{\Gamma }}^2\left(\frac{1}{2}\right)=\sqrt{\pi }.\sqrt{\pi }=\pi$$

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