Question Number 158596 by mnjuly1970 last updated on 06/Nov/21
$$ \\ $$$$\:\:\:\:\:{calculate}\:: \\ $$$$\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\:\mathrm{1}+\:{tan}^{\:\mathrm{4}} \left({x}\right)}{{cot}^{\:\mathrm{2}} \left({x}\right)}\:{dx}=? \\ $$$$ \\ $$
Answered by puissant last updated on 06/Nov/21
$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+{tan}^{\mathrm{4}} \left({x}\right)}{{cotan}^{\mathrm{2}} \left({x}\right)}{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{{tan}^{\mathrm{2}} {x}+{tan}^{\mathrm{6}} {x}\right\}{dx} \\ $$$${u}={tanx}\rightarrow{du}=\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}\:\rightarrow\:{dx}=\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{2}} +{u}^{\mathrm{6}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{6}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right\}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \left\{{u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right\}{du} \\ $$$$\Rightarrow\:\Omega\:=\:\mathrm{1}−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}−\frac{\pi}{\mathrm{4}}\:=\:\frac{\mathrm{28}}{\mathrm{15}}−\frac{\pi}{\mathrm{2}}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}{{cotan}^{\mathrm{2}} \left({x}\right)}{dx}\:=\:\frac{\mathrm{28}}{\mathrm{15}}−\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:…………….\mathscr{L}{e}\:{puissant}…………… \\ $$