Question Number 42810 by maxmathsup by imad last updated on 02/Sep/18
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{5}} }{\mathrm{1}+{x}^{\mathrm{7}} }{dx}\:\:. \\ $$
Commented by prof Abdo imad last updated on 03/Sep/18
$${changement}\:\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{7}}} \:\:\:{give}\: \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{5}}{\mathrm{7}}} }{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{7}}\:{t}^{\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\frac{\mathrm{6}}{\mathrm{7}}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{7}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)}\:=\:\frac{\pi}{\mathrm{7}{sin}\left(\frac{\pi}{\mathrm{7}}\right)}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
$${x}^{\mathrm{7}} ={tan}^{\mathrm{2}} \alpha\:\:\:{x}=\left({tan}\alpha\right)^{\frac{\mathrm{2}}{\mathrm{7}}} \\ $$$${dx}=\frac{\mathrm{2}}{\mathrm{7}}{tan}^{\frac{−\mathrm{5}}{\mathrm{7}}} \alpha\:{sec}^{\mathrm{2}} \alpha\:{d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{\left({tan}\alpha\right)^{\frac{\mathrm{10}}{\mathrm{7}}} }{{sec}^{\mathrm{2}} \alpha}×\frac{\mathrm{2}}{\mathrm{7}}\left({tan}\alpha\right)^{\frac{−\mathrm{5}}{\mathrm{7}}} ×{sec}^{\mathrm{2}} \alpha{d}\alpha \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \left({tan}\alpha\right)^{\frac{\mathrm{5}}{\mathrm{7}}} \alpha{d}\alpha \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \left({sin}\alpha\right)^{\frac{\mathrm{5}}{\mathrm{7}}} \left({cos}\alpha\right)^{\frac{−\mathrm{5}}{\mathrm{7}}} {d}\alpha \\ $$$${using}\:{formula} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \alpha{cos}^{\mathrm{2}{q}−\mathrm{1}} \alpha\:{d}\alpha\:=\frac{\lceil{p}\:×\lceil{q}}{\mathrm{2}\lceil\left({p}+{q}\right)} \\ $$$${here}\:\mathrm{2}{p}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{7}}\:\:\mathrm{2}{p}=\frac{\mathrm{12}}{\mathrm{7}}\:\:\:{p}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\mathrm{2}{q}−\mathrm{1}=−\frac{\mathrm{5}}{\mathrm{7}}\:\:\:\mathrm{2}{q}=\frac{\mathrm{2}}{\mathrm{7}}\:\:\:{q}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \left({sin}\alpha^{} \right)^{\mathrm{2}×\frac{\mathrm{6}}{\mathrm{7}}−\mathrm{1}} \left({cos}\alpha\right)^{\frac{\mathrm{2}×\mathrm{1}}{\mathrm{7}}−\mathrm{1}} {d}\alpha \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}×\frac{\lceil\left(\frac{\mathrm{6}}{\mathrm{7}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}\right)}{\mathrm{2}\lceil\left(\frac{\mathrm{6}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}}\right)}=\frac{\mathrm{1}}{\mathrm{7}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}\right)×\lceil\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{7}}×\frac{\Pi}{{sin}\left(\frac{\Pi}{\mathrm{7}}\right)} \\ $$$$ \\ $$$$\lceil\left({p}\right)×\lceil\left(\mathrm{1}−{p}\right)=\frac{\Pi}{{sin}\left({p}\Pi\right)} \\ $$