Question Number 33027 by prof Abdo imad last updated on 09/Apr/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{5}} }{dx}. \\ $$
Answered by Joel578 last updated on 14/Apr/18
$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{3}} }{\mathrm{1}\:+\:{x}^{\mathrm{5}} }\:{dx} \\ $$$${u}\:=\:{x}^{\mathrm{5}} \:\:\rightarrow\:\:{du}\:=\:\mathrm{5}{x}^{\mathrm{4}} \:{dx} \\ $$$$ \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{−\mathrm{1}} }{\mathrm{1}\:+\:{u}}\:{du}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{−\frac{\mathrm{1}}{\mathrm{5}}} }{\mathrm{1}\:+\:{u}}\:{du} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{\frac{\mathrm{4}}{\mathrm{5}}\:−\:\mathrm{1}} }{\mathrm{1}\:+\:{u}}\:{du}\:\:\:\:\:\left(\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}\:−\:\mathrm{1}} }{\mathrm{1}\:+\:{t}}\:{dt}\:=\:\frac{\pi}{\mathrm{sin}\:\left(\pi{a}\right)}\right) \\ $$$$\:\:\:=\:\frac{\pi}{\mathrm{5sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)} \\ $$$$\:\:\:=\:\frac{\mathrm{4}\pi}{\mathrm{5}\sqrt{\mathrm{10}\:−\:\mathrm{2}\sqrt{\mathrm{5}}}}\:\approx\:\mathrm{1}.\mathrm{07} \\ $$
Commented by Joel578 last updated on 10/Apr/18
$$\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:=\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{5}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{10}\:−\:\mathrm{2}\sqrt{\mathrm{5}}} \\ $$