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Question Number 132554 by mohammad17 last updated on 15/Feb/21

Commented by MJS_new last updated on 15/Feb/21

$$\mathrm{we}\mathrm{had}\mathrm{this}\mathrm{many}\mathrm{times}\mathrm{before}.\mathrm{simply}$$$$\mathrm{substitute}t=\sqrt{\tan \theta }\mathrm{and}\mathrm{there}\mathrm{you}\mathrm{go}$$

Commented by liberty last updated on 15/Feb/21

$$\mathrm{I}={\int }_0^{\pi /2}\sqrt{\tan \mathrm{x}}\mathrm{dx}={\int }_0^{\pi /2}\sqrt{\cot \mathrm{x}}\mathrm{dx}$$$$2\mathrm{I}={\int }_0^{\pi /2}\frac{\sin \mathrm{x}+\cos \mathrm{x}}{\sqrt{\sin \mathrm{xcos}\mathrm{x}}}\mathrm{dx}$$$$\mathrm{let}\mathrm{u}=\sin \mathrm{x}-\cos \mathrm{x}\Rightarrow \mathrm{du}=(\sin \mathrm{x}+\cos \mathrm{x})\mathrm{dx}$$$$\{\begin{array}{l}\mathrm{u}=1\\ \mathrm{u}=-1\end{array}\Rightarrow {\mathrm{u}}^2=1-2\sin \mathrm{xcos}\mathrm{x}$$$$\sin \mathrm{xcos}\mathrm{x}=\frac{1-{\mathrm{u}}^2}{2}$$$$2\mathrm{I}={\int }_{-1}^1\frac{\sqrt{2}\mathrm{du}}{\sqrt{1-{\mathrm{u}}^2}}$$$$\mathrm{I}=\frac{1}{\sqrt{2}}\int \frac{\mathrm{du}}{\sqrt{1-{\mathrm{u}}^2}}$$$$\mathrm{I}=\frac{1}{\sqrt{2}}{\left[\arcsin \left(\mathrm{u}\right)\right]}_{-1}^1=\frac{1}{\sqrt{2}}[\frac{\pi }{2}+\frac{\pi }{2}]$$$$=\frac{\pi }{\sqrt{2}}$$

Answered by Ñï= last updated on 15/Feb/21

$${\int }_0^{\frac{\pi }{2}}\sqrt{tanx}dx$$$$={\int }_0^{\frac{\pi }{2}}{sin}^{\frac{1}{2}}{xcos}^{-\frac{1}{2}}xdx$$$$=\frac{1}{2}B(\frac{1+\frac{1}{2}}{2},\frac{1-\frac{1}{2}}{2})$$$$=\frac{1}{2}\mathrm{\Gamma }\left(\frac{3}{4}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)$$$$=\frac{\pi }{2sin\frac{\pi }{4}}$$$$=\frac{\pi }{\sqrt{2}}$$

Answered by Dwaipayan Shikari last updated on 15/Feb/21

$${\int }_0^{\frac{\pi }{2}}{sin}^{\frac{1}{2}}x{cos}^{-\frac{1}{2}}xdx=\frac{\mathrm{\Gamma }\left(\frac{3}{4}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)}{2\mathrm{\Gamma }\left(1\right)}=\frac{\pi }{\sqrt{2}}$$

Answered by mathmax by abdo last updated on 15/Feb/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\sqrt{\tan \theta }\mathrm{d}\theta \mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{\tan \theta }=\mathrm{t}\Rightarrow \tan \theta ={\mathrm{t}}^2\Rightarrow$$$$\theta =\arctan \left({\mathrm{t}}^2\right)\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\infty }}\mathrm{t}.\frac{2\mathrm{t}}{1+{\mathrm{t}}^4}\mathrm{dt}=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2}{1+{\mathrm{t}}^4}\mathrm{dt}$$$${=}_{{\mathrm{t}}^4=\mathrm{z}}2.\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{{\left({\mathrm{z}}^{\frac{1}{4}}\right)}^2}{1+\mathrm{z}}{\mathrm{z}}^{\frac{1}{4}-1}\mathrm{dz}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{2}+\frac{1}{4}-1}}{1+\mathrm{z}}\mathrm{dz}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{3}{4}-1}}{1+\mathrm{z}}=\frac{1}{2}\times \frac{\pi }{\sin \left(\frac{3\pi }{4}\right)}=\frac{\pi }{2.\frac{\sqrt{2}}{2}}=\frac{\pi }{\sqrt{2}}$$

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