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Question Number 142763 by lapache last updated on 05/Jun/21

Commented by Ar Brandon last updated on 05/Jun/21

$$\mathcal{I}=\int \sqrt{\mathrm{tanx}}\mathrm{dx},{\mathrm{t}}^2=\mathrm{tanx}\Rightarrow 2\mathrm{tdt}=(1+{\mathrm{t}}^4)\mathrm{dx}$$$$=\int \frac{{2\mathrm{t}}^2\mathrm{dt}}{1+{\mathrm{t}}^4}=\int \frac{({\mathrm{t}}^2+1)+({\mathrm{t}}^2-1)}{{\mathrm{t}}^4+1}\mathrm{dt}$$$$=\int \frac{1+\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}+\int \frac{1-\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}=\int \frac{1+\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}-\frac{1}{\mathrm{t}})}^2+2}\mathrm{dt}+\int \frac{1-\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}+\frac{1}{\mathrm{t}})}^2-2}\mathrm{dt}$$$$=\int \frac{\mathrm{du}}{{\mathrm{u}}^2+2}+\int \frac{\mathrm{dv}}{{\mathrm{v}}^2-2}=\frac{\arctan \left(\mathrm{u}/\sqrt{2}\right)}{\sqrt{2}}-\frac{\mathrm{arctanh}\left(\mathrm{v}/\sqrt{2}\right)}{\sqrt{2}}+\mathrm{C}$$$$=\frac{1}{\sqrt{2}}\arctan \left(\frac{{\mathrm{t}}^2-1}{\sqrt{2}\mathrm{t}}\right)-\frac{1}{2\sqrt{2}}\ln \mid \frac{{\mathrm{t}}^2+\sqrt{2}\mathrm{t}+1}{{\mathrm{t}}^2-\sqrt{2}\mathrm{t}+1}\mid +\mathrm{C}$$$$=\frac{1}{\sqrt{2}}\arctan \left(\frac{\mathrm{tanx}-1}{\sqrt{2\mathrm{tanx}}}\right)-\frac{1}{2\sqrt{2}}\ln \mid \frac{\mathrm{tanx}+\sqrt{2\mathrm{tanx}}+1}{\mathrm{tanx}-\sqrt{2\mathrm{tanx}}+1}\mid +\mathrm{C}$$

Commented by Tinku Tara last updated on 05/Jun/21

$$\mathrm{Please}\mathrm{updste}\mathrm{to}\mathrm{latest}\mathrm{version}.$$$$\mathrm{Watermark}\mathrm{has}\mathrm{been}\mathrm{removed}\mathrm{in}$$$$\mathrm{latest}\mathrm{version}\left(\mathrm{free}\right)$$

Answered by Ar Brandon last updated on 05/Jun/21

$$\mathcal{I}={\int }_0^{\frac{\pi }{2}}\sqrt{\mathrm{tanx}}\mathrm{dx},{\mathrm{t}}^2=\mathrm{tanx}\Rightarrow 2\mathrm{tdt}=(1+{\mathrm{t}}^4)\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{2\mathrm{t}}^2\mathrm{dt}}{1+{\mathrm{t}}^4}={\int }_0^{\mathrm{\infty }}\frac{({\mathrm{t}}^2+1)+({\mathrm{t}}^2-1)}{{\mathrm{t}}^4+1}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}+{\int }_0^{\mathrm{\infty }}\frac{1-\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}-\frac{1}{\mathrm{t}})}^2+2}\mathrm{dt}+{\int }_0^{\mathrm{\infty }}\frac{1-\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}+\frac{1}{\mathrm{t}})}^2-2}\mathrm{dt}$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+2}+{\int }_{+\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dv}}{{\mathrm{v}}^2-2}={\left[\frac{\arctan \left(\mathrm{u}/\sqrt{2}\right)}{\sqrt{2}}\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{1}{\sqrt{2}}(\frac{\pi }{2}+\frac{\pi }{2})=\frac{\pi }{\sqrt{2}}$$

Commented by mathmax by abdo last updated on 06/Jun/21

$$\mathrm{\Phi }={\int }_0^{\frac{\pi }{2}}\sqrt{\mathrm{tanx}}\mathrm{dx}\mathrm{changement}\sqrt{\mathrm{tanx}}=\mathrm{t}\mathrm{give}\mathrm{tanx}={\mathrm{t}}^2\Rightarrow \mathrm{x}=\arctan \left({\mathrm{t}}^2\right)\Rightarrow$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}.\left(2\mathrm{t}\right)}{1+{\mathrm{t}}^4}\mathrm{dt}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^2}{1+{\mathrm{t}}^4}\mathrm{dt}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{{\mathrm{z}}^4+1}$$$$\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{({\mathrm{z}}^2-\mathrm{i})({\mathrm{z}}^2+\mathrm{i})}=\frac{{\mathrm{z}}^2}{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}$$$$\mathrm{residus}\mathrm{theorem}\mathrm{give}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})\}$$$$\mathrm{we}\mathrm{have}\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=\frac{\mathrm{i}}{{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\left(2\mathrm{i}\right)}=\frac{1}{4}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}$$$$\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})=\frac{-\mathrm{i}}{(-{2\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(-2\mathrm{i})}=-\frac{1}{4}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\Rightarrow$$$${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=-2\mathrm{i}\pi \{\frac{1}{4}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}-\frac{1}{4}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}\}$$$$=-\frac{\pi \mathrm{i}}{2}\left(2\mathrm{isin}\left(\frac{\pi }{4}\right)\right)=\pi .\frac{1}{\sqrt{2}}=\frac{\pi }{\sqrt{2}}\Rightarrow \mathrm{\Phi }=\frac{\pi }{\sqrt{2}}$$

Answered by Ar Brandon last updated on 05/Jun/21

$${\int }_0^{\frac{\pi }{2}}\sqrt{\tan \theta }\mathrm{d}\theta =\frac{1}{2}\beta (\frac{3}{4},\frac{1}{4})=\frac{\pi }{2\sin \left(\frac{\pi }{4}\right)}=\frac{\pi }{\sqrt{2}}$$

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