Menu Close

nice-calculus-0-pi-2-x-3-cot-x-dx-a-16-a-m-n-




Question Number 151954 by mnjuly1970 last updated on 24/Aug/21
      nice...calculus          𝛗 := ∫_0 ^( (Ο€/2)) x^( 3) . cot (x )dx =(a/(16))                     a :=?  m.n...
$$ \\ $$$$\:\:\:\:{nice}…{calculus} \\ $$$$\: \\ $$$$\:\:\:\:\:\boldsymbol{\phi}\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}^{\:\mathrm{3}} .\:{cot}\:\left({x}\:\right){dx}\:=\frac{{a}}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\::=? \\ $$$${m}.{n}… \\ $$
Answered by Kamel last updated on 24/Aug/21
  Ξ©=∫_0 ^(Ο€/2) x^3 cot(x)dx=^(IBP) βˆ’3∫_0 ^(Ο€/2) x^2 Ln(sin(x))dx  Ln(sin(x))=βˆ’Ln(2)βˆ’Ξ£_(n=1) ^(+∞) ((cos(2nx))/n),   0<x≀(Ο€/2).  ∴ Ξ©=(Ο€^3 /8)Ln(2)+3Ξ£_(n=1) ^(+∞) (1/n)∫_0 ^(Ο€/2) x^2 cos(2nx)dx          =(Ο€^3 /8)Ln(2)+((3Ο€)/2)Ξ£_(n=1) ^(+∞) (1/n)((((βˆ’1)^n )/(2n^2 )))         =(Ο€^3 /8)Ln(2)+((3Ο€)/4)((1/8)ΞΆ(3)βˆ’(ΞΆ(3)βˆ’(1/8)ΞΆ(3)))         =(Ο€^3 /8)Ln(2)βˆ’((9Ο€)/(16))=((Ο€^3 Ln(4)βˆ’9Ο€)/(16))
$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{3}} {cot}\left({x}\right){dx}\overset{{IBP}} {=}βˆ’\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} {Ln}\left({sin}\left({x}\right)\right){dx} \\ $$$${Ln}\left({sin}\left({x}\right)\right)=βˆ’{Ln}\left(\mathrm{2}\right)βˆ’\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}},\:\:\:\mathrm{0}<{x}\leqslant\frac{\pi}{\mathrm{2}}. \\ $$$$\therefore\:\Omega=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)+\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} {cos}\left(\mathrm{2}{nx}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}\pi}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left(\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}\pi}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)βˆ’\left(\zeta\left(\mathrm{3}\right)βˆ’\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right)\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{Ln}\left(\mathrm{2}\right)βˆ’\frac{\mathrm{9}\pi}{\mathrm{16}}=\frac{\pi^{\mathrm{3}} {Ln}\left(\mathrm{4}\right)βˆ’\mathrm{9}\pi}{\mathrm{16}} \\ $$
Commented by mnjuly1970 last updated on 25/Aug/21
excellent...
$${excellent}… \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *