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Question Number 158596 by mnjuly1970 last updated on 06/Nov/21
calculate : Ω = ∫_0 ^( (π/4)) (( 1+ tan^( 4) (x))/(cot^( 2) (x))) dx=?
$$ \\ $$$$\:\:\:\:\:{calculate}\:: \\ $$$$\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\:\mathrm{1}+\:{tan}^{\:\mathrm{4}} \left({x}\right)}{{cot}^{\:\mathrm{2}} \left({x}\right)}\:{dx}=? \\ $$$$ \\ $$
Answered by puissant last updated on 06/Nov/21
Ω=∫_0 ^(π/4) ((1+tan^4 (x))/(cotan^2 (x)))dx = ∫_0 ^(π/4) {tan^2 x+tan^6 x}dx u=tanx→du=(1+tan^2 x)dx → dx=(du/(1+u^2 )) ⇒ Ω=∫_0 ^1 ((u^2 +u^6 )/(1+u^2 ))du=∫_0 ^1 ((u^2 +1−1)/(1+u^2 ))du+∫_0 ^1 (u^6 /(1+u^2 ))du Ω=∫_0 ^1 {1−(1/(1+u^2 ))}dx+∫_0 ^1 {u^4 −u^2 +1−(1/(1+u^2 ))}du ⇒ Ω = 1−(π/4)+(1/5)−(1/3)+1−(π/4) = ((28)/(15))−(π/2). Ω = ∫_0 ^1 ((1+tan^2 (x))/(cotan^2 (x)))dx = ((28)/(15))−(π/2) ................Le puissant...............
$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+{tan}^{\mathrm{4}} \left({x}\right)}{{cotan}^{\mathrm{2}} \left({x}\right)}{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{{tan}^{\mathrm{2}} {x}+{tan}^{\mathrm{6}} {x}\right\}{dx} \\ $$$${u}={tanx}\rightarrow{du}=\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}\:\rightarrow\:{dx}=\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{2}} +{u}^{\mathrm{6}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{6}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right\}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \left\{{u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right\}{du} \\ $$$$\Rightarrow\:\Omega\:=\:\mathrm{1}−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}−\frac{\pi}{\mathrm{4}}\:=\:\frac{\mathrm{28}}{\mathrm{15}}−\frac{\pi}{\mathrm{2}}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}{{cotan}^{\mathrm{2}} \left({x}\right)}{dx}\:=\:\frac{\mathrm{28}}{\mathrm{15}}−\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:…………….\mathscr{L}{e}\:{puissant}…………… \\ $$

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