Question Number 151954 by mnjuly1970 last updated on 24/Aug/21
$$ \\ $$$$\:\:\:\:{nice}…{calculus} \\ $$$$\: \\ $$$$\:\:\:\:\:\boldsymbol{\phi}\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}^{\:\mathrm{3}} .\:{cot}\:\left({x}\:\right){dx}\:=\frac{{a}}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\::=? \\ $$$${m}.{n}… \\ $$
Answered by Kamel last updated on 24/Aug/21
$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{3}} {cot}\left({x}\right){dx}\overset{{IBP}} {=}β\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} {Ln}\left({sin}\left({x}\right)\right){dx} \\ $$$${Ln}\left({sin}\left({x}\right)\right)=β{Ln}\left(\mathrm{2}\right)β\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}},\:\:\:\mathrm{0}<{x}\leqslant\frac{\pi}{\mathrm{2}}. \\ $$$$\therefore\:\Omega=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)+\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} {cos}\left(\mathrm{2}{nx}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}\pi}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left(\frac{\left(β\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}\pi}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)β\left(\zeta\left(\mathrm{3}\right)β\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right)\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)β\frac{\mathrm{9}\pi}{\mathrm{16}}=\frac{\pi^{\mathrm{3}} {Ln}\left(\mathrm{4}\right)β\mathrm{9}\pi}{\mathrm{16}} \\ $$
Commented by mnjuly1970 last updated on 25/Aug/21
$${excellent}… \\ $$