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Question Number 151220 by peter frank last updated on 19/Aug/21
show that ∫_0 ^(π/2) ((cos x)/(cos x+sin x+1))dx=(1/4)(π−2ln 2)
$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}+\mathrm{1}}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\left(\pi−\mathrm{2ln}\:\mathrm{2}\right) \\ $$
Answered by ArielVyny last updated on 19/Aug/21
t=tan((x/2))→dt=(1/2)(1+t^2 )dx ∫_0 ^1 (((1−t^2 )/(1+t^2 ))/(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 ))+1))×(2/(1+t^2 ))dt=2∫_0 ^1 (((1−t^2 )/(1+t^2 ))/((1−t^2 )+(2t)+(1+t^2 )))dt =2∫_0 ^1 ((1−t^2 )/(1+t^2 ))×(1/(2+2t))dt=∫_0 ^1 (((1−t))/((1+t^2 ))) dt=∫_0 ^1 (1/(1+t^2 ))dt−∫_0 ^1 (t/(1+t^2 ))dt =∫_0 ^1 (1/(1+t^2 ))dt−∫_0 ^1 (t/(1+t^2 ))dt=(π/4)−Σ(−1)^n ∫_0 ^1 t^(2n+1) (π/4)−Σ(−1)^n (1/(2n+2))=(π/4)−(1/2)ln(2) ∫_0 ^(π/2) ((cosx)/(cosx+sinx+1))=(π/4)−(1/2)ln(2+
$${t}={tan}\left(\frac{{x}}{\mathrm{2}}\right)\rightarrow{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{1}}×\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\left(\mathrm{2}{t}\right)+\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}+\mathrm{2}{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}\right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\frac{\pi}{\mathrm{4}}−\Sigma\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\frac{\pi}{\mathrm{4}}−\Sigma\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cosx}}{{cosx}+{sinx}+\mathrm{1}}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}+\right. \\ $$
Commented by peter frank last updated on 19/Aug/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Lordose last updated on 19/Aug/21
Ω = ∫_0 ^(𝛑/2) ((cos(x))/(cos(x)+sin(x)+1))dx =^(t=tan((x/2))) ∫_0 ^( 1) (((1−t^2 )/(1+t^2 ))/((1+t^2 )(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 ))+1)))dt Ω = 2∫_0 ^( 1) (((1−t^2 ))/((1+t^2 )(1−t^2 +2t+1+t^2 ))) = ∫_0 ^( 1) ((1−t)/(1+t^2 ))dt Ω = ∫_0 ^( 1) (1/(1+t^2 ))dt − (1/2)∫_0 ^( 1) ((2t)/(1+t^2 ))dt 𝛀 = (𝛑/4) − (1/2)ln(2)
$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{1}}\mathrm{dx}\:\overset{\mathrm{t}=\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\mathrm{1}\right)}\mathrm{dt} \\ $$$$\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\boldsymbol{\Omega}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right) \\ $$
Commented by peter frank last updated on 19/Aug/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Lordose last updated on 19/Aug/21
Ω = ∫_0 ^(𝛑/2) ((cos(x))/(cos(x)+sin(x)+1))dx =^(t=tan((x/2))) ∫_0 ^( 1) (((1−t^2 )/(1+t^2 ))/((1+t^2 )(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 ))+1)))dt Ω = 2∫_0 ^( 1) (((1−t^2 ))/((1+t^2 )(1−t^2 +2t+1+t^2 ))) = ∫_0 ^( 1) ((1−t)/(1+t^2 ))dt Ω = ∫_0 ^( 1) (1/(1+t^2 ))dt − (1/2)∫_0 ^( 1) ((2t)/(1+t^2 ))dt 𝛀 = (𝛑/4) − (1/2)ln(2)
$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{1}}\mathrm{dx}\:\overset{\mathrm{t}=\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\mathrm{1}\right)}\mathrm{dt} \\ $$$$\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\boldsymbol{\Omega}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right) \\ $$

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