Integration Questions

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

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({tan}\left({x}\right)\right)^{\frac{\mathrm{1}}{{n}}} {dx}\:... \\$$

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

$$\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sinx}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \left(\mathrm{cosx}\right)^{−\frac{\mathrm{1}}{\mathrm{n}}} \mathrm{dx} \\$$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}}\right)\Gamma\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2n}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\pi}{\mathrm{sin}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}}\pi\right)}\:,\:\mathrm{n}>\mathrm{1} \\$$

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

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt[{{n}}]{{tanx}}\:{dx}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}{\mathrm{2}}=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}{n}}\right)} \\$$

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

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{tanx}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\mathrm{dx}\:\mathrm{changement}\:\:\mathrm{tanx}\:=\mathrm{t}\:\mathrm{give} \\$$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:=_{\mathrm{t}=\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2}}} } \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2n}}} }{\mathrm{1}+\mathrm{z}}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{z}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+\mathrm{z}}\mathrm{dz} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi}{\mathrm{sin}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}\:=\frac{\pi}{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2n}}+\frac{\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{2cos}\left(\frac{\pi}{\mathrm{2n}}\right)}\left(\mathrm{with}\:\mathrm{n}>\mathrm{1}\right) \\$$