Question Number 161945 by Ahmed777hamouda last updated on 24/Dec/21
$$\:\:\boldsymbol{{P}\mathrm{rove}}\:\boldsymbol{\mathrm{that}}\:\int_{\mathrm{0}} ^{\infty} \frac{\left(\boldsymbol{{xcos}}\left(\boldsymbol{{x}}\right)β\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\right)^{\mathrm{2}} }{\boldsymbol{{x}}^{\mathrm{6}} }\boldsymbol{{dx}}\:=\frac{\boldsymbol{\pi}}{\mathrm{15}} \\ $$
Answered by Ar Brandon last updated on 24/Dec/21
![I=β«_0 ^β (((xcosxβsinx)^2 )/x^6 )dx=β«_0 ^β (((cos^2 x)/x^4 )β((sin2x)/x^5 )+((sin^2 x)/x^6 ))dx =[β((cos^2 x)/(3x^3 ))]_0 ^β β(1/3)β«((sin2x)/x^3 )dxβ16β«_0 ^β ((sinu)/u^5 )duβ[((sin^2 x)/(5x^5 ))]_0 ^β +(1/5)β«_0 ^β ((sin2x)/x^5 )dx =β(4/3)β«_0 ^β ((sint)/t^3 )dtβ16β«_0 ^β ((sinu)/u^5 )du+((16)/5)β«_0 ^β ((sint)/t^5 )dt =β(4/3)β(Ο/(2Ξ(3)sin(((3Ο)/2))))β((64)/5)β(Ο/(2Ξ(5)sin(((5Ο)/2))))=(Ο/3)β((4Ο)/(15))=(Ο/(15))](https://www.tinkutara.com/question/Q161949.png)
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\left({x}\mathrm{cos}{x}β\mathrm{sin}{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{6}} }{dx}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{cos}^{\mathrm{2}} {x}}{{x}^{\mathrm{4}} }β\frac{\mathrm{sin2}{x}}{{x}^{\mathrm{5}} }+\frac{\mathrm{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{6}} }\right){dx} \\ $$$$\:\:\:=\left[β\frac{\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{3}{x}^{\mathrm{3}} }\right]_{\mathrm{0}} ^{\infty} β\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{sin2}{x}}{{x}^{\mathrm{3}} }{dx}β\mathrm{16}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{u}}{{u}^{\mathrm{5}} }{du}β\left[\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{5}{x}^{\mathrm{5}} }\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin2}{x}}{{x}^{\mathrm{5}} }{dx} \\ $$$$\:\:\:=β\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{t}}{{t}^{\mathrm{3}} }{dt}β\mathrm{16}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{u}}{{u}^{\mathrm{5}} }{du}+\frac{\mathrm{16}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{t}}{{t}^{\mathrm{5}} }{dt} \\ $$$$\:\:\:=β\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{3}\right)\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}β\frac{\mathrm{64}}{\mathrm{5}}\centerdot\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{5}\right)\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{2}}\right)}=\frac{\pi}{\mathrm{3}}β\frac{\mathrm{4}\pi}{\mathrm{15}}=\frac{\pi}{\mathrm{15}} \\ $$