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Question Number 133016 by metamorfose last updated on 18/Feb/21

$${\int }_0^{\frac{\pi }{2}}{\left(tan\left(x\right)\right)}^{\frac{1}{n}}dx...$$

Answered by Ar Brandon last updated on 18/Feb/21

$$\mathcal{I}={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{tanx}\right)}^{\frac{1}{\mathrm{n}}}\mathrm{dx}={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{sinx}\right)}^{\frac{1}{\mathrm{n}}}{\left(\mathrm{cosx}\right)}^{-\frac{1}{\mathrm{n}}}\mathrm{dx}$$$$=\frac{1}{2}\mathrm{\Gamma }\left(\frac{\mathrm{n}+1}{2\mathrm{n}}\right)\mathrm{\Gamma }\left(\frac{\mathrm{n}-1}{2\mathrm{n}}\right)=\frac{1}{2}\cdot \frac{\pi }{\sin \left(\frac{\mathrm{n}+1}{2\mathrm{n}}\pi \right)},\mathrm{n}>1$$

Answered by Dwaipayan Shikari last updated on 18/Feb/21

$${\int }_0^{\frac{\pi }{2}}\sqrt[n]{tanx}dx=\frac{\mathrm{\Gamma }(\frac{1}{2}-\frac{1}{2n})\mathrm{\Gamma }(\frac{1}{2}+\frac{1}{2n})}{2}=\frac{\pi }{2sin(\frac{\pi }{2}+\frac{\pi }{2n})}$$

Answered by mathmax by abdo last updated on 20/Feb/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{tanx}\right)}^{\frac{1}{\mathrm{n}}}\mathrm{dx}\mathrm{changement}\mathrm{tanx}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}}}{1+{\mathrm{t}}^2}\mathrm{dt}{=}_{\mathrm{t}={\mathrm{z}}^{\frac{1}{2}}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{2\mathrm{n}}}}{1+\mathrm{z}}\frac{1}{2}{\mathrm{z}}^{-\frac{1}{2}}\mathrm{dz}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{2\mathrm{n}}+\frac{1}{2}-1}}{1+\mathrm{z}}\mathrm{dz}$$$$=\frac{1}{2}\frac{\pi }{\sin \left(\pi (\frac{1}{2\mathrm{n}}+\frac{1}{2})\right)}=\frac{\pi }{2\sin (\frac{\pi }{2\mathrm{n}}+\frac{\pi }{2})}=\frac{\pi }{2\cos \left(\frac{\pi }{2\mathrm{n}}\right)}(\mathrm{with}\mathrm{n}>1)$$

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