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Question Number 97928 by  M±th+et+s last updated on 10/Jun/20

$$findthegeneralformula$$$${\int }_0^{\frac{\pi }{2}}{tan}^{\alpha }\left(x\right)dx$$

Answered by abdomathmax last updated on 10/Jun/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}{\tan }^{\alpha }\mathrm{xdx}\mathrm{changement}\mathrm{tanx}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\alpha }}{1+{\mathrm{t}}^2}\mathrm{dt}\mathrm{also}\mathrm{changement}\mathrm{t}={\mathrm{u}}^{\frac{1}{2}}$$$$\mathrm{give}\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{\alpha }{2}}}{1+\mathrm{u}}\frac{1}{2}{\mathrm{u}}^{-\frac{1}{2}}\mathrm{du}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{\alpha -1}{2}}}{1+\mathrm{u}}\mathrm{du}\mathrm{we}\mathrm{hsve}\mathrm{proved}\mathrm{that}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}$$$$=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\mathrm{if}0<\mathrm{a}<1\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{\frac{\alpha +1}{2}-1}}{1+\mathrm{u}}\mathrm{du}=\frac{\pi }{\sin \left(\frac{\pi }{2}(\alpha +1)\right)}=\frac{\pi }{\cos \left(\frac{\pi \alpha }{2}\right)}$$$$\mathrm{so}{\int }_0^{\frac{\pi }{2}}{\tan }^{\alpha }\mathrm{xdx}=\frac{\pi }{\cos \left(\frac{\pi \alpha }{2}\right)}$$$$\mathrm{conditions}!0<\frac{\alpha +1}{2}<1\Rightarrow$$$$0<\alpha +1<2\Rightarrow -1<\alpha <1\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{convergence}$$

Commented by  M±th+et+s last updated on 11/Jun/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 11/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}.$$

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