Integration Questions

Question Number 161212 by cortano last updated on 14/Dec/21

$$\:\:\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{{x}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{cos}\:^{\mathrm{4}} {x}\:+\mathrm{sin}\:^{\mathrm{4}} {x}}\:{dx}\:=? \\$$

Commented by puissant last updated on 15/Dec/21

Commented by Tawa11 last updated on 15/Dec/21

$$\mathrm{Nice}\:\mathrm{sirs} \\$$

Answered by chhaythean last updated on 14/Dec/21

$$\mathrm{let}\:\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\mathrm{sin}{x}\mathrm{cos}{x}}{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{4}} {x}}\mathrm{d}{x}\:\:\left(\ast\right) \\$$$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{x}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{\mathrm{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)+\mathrm{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}\mathrm{d}{x} \\$$$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{x}\right)\mathrm{sin}{x}\mathrm{cos}{x}}{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{4}} {x}}\mathrm{d}{x}\:\:\left(\ast\ast\right) \\$$$$\mathrm{take}\:\left(\ast\right)+\left(\ast\ast\right)\:\mathrm{we}\:\mathrm{get}: \\$$$$\mathrm{2I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}{x}\mathrm{cos}{x}}{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{4}} {x}}\mathrm{d}{x} \\$$$$\mathrm{let}\:\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{t}\Rightarrow\mathrm{dt}=\mathrm{2sin}{x}\mathrm{cos}{x}\mathrm{d}{x} \\$$$$\mathrm{2I}=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} } \\$$$$\mathrm{2I}=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{1}−\mathrm{2t}+\mathrm{2t}^{\mathrm{2}} }=\frac{\pi}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}} \\$$$$\mathrm{2I}=\frac{\pi}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}} \\$$$$\mathrm{2I}=\frac{\pi}{\mathrm{8}}×\mathrm{2arctan}\left(\mathrm{2t}−\mathrm{1}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\$$$$\mathrm{2I}=\frac{\pi}{\mathrm{8}}\left(\mathrm{2}×\mathrm{2arctan}\left(\mathrm{1}\right)\right) \\$$$$\mathrm{2I}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\Rightarrow\mathrm{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$

Commented by cortano last updated on 14/Dec/21

$${yes}.\:{thanks} \\$$

Answered by Ar Brandon last updated on 14/Dec/21

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\mathrm{sin}{x}\mathrm{cos}{x}}{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{4}} {x}}{dx}=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}{x}\mathrm{cos}{x}}{\mathrm{cos}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{4}} {x}}{dx} \\$$$$\:\:\:=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{tan}{x}\centerdot\mathrm{sec}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{tan}^{\mathrm{4}} {x}}{dx}=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\$$$$\:\:\:=\frac{\pi}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\pi}{\mathrm{8}}\left[\mathrm{tan}^{−\mathrm{1}} \left({u}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$

Commented by Ar Brandon last updated on 14/Dec/21

$$\:\:\:\:\:{u}={t}^{\mathrm{2}} ,\:\:{t}=\mathrm{tan}{x} \\$$