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Question Number 78549 by aliesam last updated on 18/Jan/20

Answered by ~blr237~ last updated on 18/Jan/20

$$\mathrm{let}\mathrm{named}\mathrm{it}\mathrm{A}$$$$\mathrm{state}\mathrm{u}=\frac{\mathrm{x}}{2},\mathrm{A}=2{\int }_0^{\frac{\pi }{4}}\frac{\sqrt{\tan 2\mathrm{u}}}{1+\mathrm{sinu}}\mathrm{du}$$$$\frac{\mathrm{A}}{2}={\int }_0^{\frac{\pi }{4}}\frac{\sqrt{\tan 2\mathrm{u}}}{{\cos }^2\mathrm{u}}(1-\mathrm{sinu})\mathrm{du}$$$$={\int }_0^{\frac{\pi }{4}}\sqrt{\tan 2\mathrm{u}}\left(\frac{\mathrm{du}}{{\cos }^2\mathrm{u}}\right)-{\int }_0^{\frac{\pi }{4}}\mathrm{sinu}\sqrt{\tan 2\mathrm{u}}\left(\frac{\mathrm{du}}{{\cos }^2\mathrm{u}}\right)$$$$={\int }_0^1\sqrt{\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}}\mathrm{dt}-{\int }_0^1\sqrt{\frac{{2\mathrm{t}}^3}{1-{\mathrm{t}}^4}}\mathrm{dt}\mathrm{where}\mathrm{t}=\mathrm{tanu}$$$$\frac{\mathrm{A}}{2\sqrt{2}}={\mathrm{I}}_2-{\mathrm{I}}_4\mathrm{with}{\mathrm{I}}_{\mathrm{n}}={\int }_0^1\sqrt{\frac{{\mathrm{t}}^{\mathrm{n}-1}}{1-{\mathrm{t}}^{\mathrm{n}}}}\mathrm{dt}\mathrm{for}\mathrm{n}⩾1$$$$\mathrm{let}\mathrm{state}\mathrm{y}={\mathrm{t}}^{\mathrm{n}}\Rightarrow \mathrm{dt}=\frac{1}{\mathrm{n}}{\mathrm{y}}^{\frac{1}{\mathrm{n}}-1}\mathrm{dy}$$$${\mathrm{I}}_{\mathrm{n}}={\int }_0^1{\mathrm{y}}^{\frac{\mathrm{n}-1}{2\mathrm{n}}}{(1-\mathrm{y})}^{\frac{-1}{2}}\left(\frac{1}{\mathrm{n}}{\mathrm{y}}^{\frac{1}{\mathrm{n}}-1}\mathrm{dy}\right)$$$${\mathrm{nI}}_{\mathrm{n}}={\int }_0^1{\mathrm{y}}^{\frac{1}{2\mathrm{n}}+\frac{1}{2}-1}{(1-\mathrm{y})}^{\frac{1}{2}-1}\mathrm{dy}$$$$=\mathrm{B}(\frac{1}{2\mathrm{n}}+\frac{1}{2},\frac{1}{2})=\frac{\mathrm{\Gamma }(\frac{1}{2\mathrm{n}}+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\frac{1}{2\mathrm{n}}+1)}$$$$\mathrm{so}{\mathrm{I}}_{\mathrm{n}}=\frac{2\sqrt{\pi }\mathrm{\Gamma }(\frac{1}{2\mathrm{n}}+\frac{1}{2})}{\mathrm{\Gamma }\left(\frac{1}{2\mathrm{n}}\right)}$$$$$$

Commented by mind is power last updated on 19/Jan/20

$$\mathrm{nice}\mathrm{Sir}$$

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