find-G-1-4-0-pi-2-ln-1-sin-x-1-sin-x-dx-

Question Number 212319 by Ghisom last updated on 09/Oct/24

find G=(1/4)∫_0 ^(π/2) ln ((1+sin x)/(1−sin x)) dx

$$\mathrm{find} \\ $$$${G}=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\:{dx} \\ $$

Commented by Spillover last updated on 10/Oct/24

$$\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}\:\:{right}? \\ $$

Commented by Frix last updated on 10/Oct/24

$$\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}\approx.\mathrm{544} \\ $$$${G}\approx.\mathrm{916} \\ $$

Answered by Spillover last updated on 10/Oct/24

=(1/4)∫_0 ^(π/2) ln (((1+sin x)/(1−sin x))) =(1/4)∫_0 ^(π/2) ln (((1+sin x)/(1−sin x))×((1+sin x)/(1+sin x)))=(1/4)∫_0 ^(π/2) ln (((1+sin x)^2 )/(cos^2 x)) (1/4)∫_0 ^(π/2) ln (((1+sin x)^2 )/(cos^2 x))=(1/4)∫_0 ^(π/2) ln (1+sin x)^2 −ln cos^2 x (1/4)∫_0 ^(π/2) ln (1+sin x)^2 −ln cos^2 x (1/4)∫_0 ^(π/2) [2ln (1+sin x)−2ln cosx]dx [(1/2)∫_0 ^(π/2) ln (1+sin x)dx]−[(1/2)∫_0 ^(π/2) ln cosx]dx [(1/2)∫_0 ^(π/2) ln cosx]dx=−(π/2)ln 2 [(1/2)∫_0 ^(π/2) ln (1+sin x)dx] ....

$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}×\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\frac{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\frac{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} −\mathrm{ln}\:\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} −\mathrm{ln}\:\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{2ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)−\mathrm{2ln}\:\mathrm{cos}{x}\right]{dx} \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right){dx}\right]−\left[\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\mathrm{cos}{x}\right]{dx} \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\mathrm{cos}{x}\right]{dx}=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right){dx}\right] \\ $$$$…. \\ $$

Answered by Ar Brandon last updated on 11/Oct/24

Ω=(1/4)∫_0 ^(π/2) ln(((1+sinx)/(1−sinx)))dx=(1/2)∫_0 ^(π/2) ln(((cos(x/2)+sin(x/2))/(cos(x/2)−sin(x/2))))dx =(1/2)∫_0 ^(π/2) ln(((sin((x/2)+(π/4)))/(cos((x/2)+(π/4)))))dx=(1/2)∫_0 ^(π/2) ln(((cos(x/2))/(sin(x/2))))dx =∫_0 ^(π/4) ln(((cosx)/(sinx)))dx=−∫_0 ^(π/4) ln(tanx)dx=G G=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) , Catalan′s constant.

$$\Omega=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{sin}{x}}{\mathrm{1}−\mathrm{sin}{x}}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{cos}\frac{{x}}{\mathrm{2}}+\mathrm{sin}\frac{{x}}{\mathrm{2}}}{\mathrm{cos}\frac{{x}}{\mathrm{2}}−\mathrm{sin}\frac{{x}}{\mathrm{2}}}\right){dx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{sin}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{cos}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{sin}\frac{{x}}{\mathrm{2}}}\right){dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\frac{\mathrm{cos}{x}}{\mathrm{sin}{x}}\right){dx}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx}={G} \\ $$$$\:\:\:\:{G}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:,\:\mathrm{Catalan}'\mathrm{s}\:\mathrm{constant}. \\ $$

Commented by Ghisom last updated on 12/Oct/24

$$\mathrm{yes} \\ $$

Commented by Ar Brandon last updated on 12/Oct/24

Yep, Sir MJS ��

Commented by Spillover last updated on 12/Oct/24

$${How}\:{did}\:{you}\:{know}\:{if}\:{sir}\:{MJs}.{user}\:{name} \\ $$$${is}\:{different} \\ $$

Commented by Ar Brandon last updated on 12/Oct/24

I can tell from his writings.

Commented by Frix last updated on 12/Oct/24

I′m confused. I thought it was me? So if it′s not me, then who am I? (I′m not Ghisom, Scout′s honor!)

$$\mathrm{I}'\mathrm{m}\:\mathrm{confused}.\:\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{was}\:\mathrm{me}?\:\mathrm{So}\:\mathrm{if}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{not}\:\mathrm{me},\:\mathrm{then}\:\mathrm{who}\:\mathrm{am}\:\mathrm{I}? \\ $$$$\left(\mathrm{I}'\mathrm{m}\:{not}\:\mathrm{Ghisom},\:\mathrm{Scout}'\mathrm{s}\:\mathrm{honor}!\right) \\ $$

Commented by Ghisom last updated on 13/Oct/24

if you were me I would see you in my mirror and v/v. but I see me, at least I see the same guy who′s on the photographs with my wife.

$$\mathrm{if}\:\mathrm{you}\:\mathrm{were}\:\mathrm{me}\:\mathrm{I}\:\mathrm{would}\:\mathrm{see}\:\mathrm{you}\:\mathrm{in}\:\mathrm{my} \\ $$$$\mathrm{mirror}\:\mathrm{and}\:\mathrm{v}/\mathrm{v}.\:\mathrm{but}\:\mathrm{I}\:\mathrm{see}\:\mathrm{me},\:\mathrm{at}\:\mathrm{least}\:\mathrm{I} \\ $$$$\mathrm{see}\:\mathrm{the}\:\mathrm{same}\:\mathrm{guy}\:\mathrm{who}'\mathrm{s}\:\mathrm{on}\:\mathrm{the}\:\mathrm{photographs} \\ $$$$\mathrm{with}\:\mathrm{my}\:\mathrm{wife}. \\ $$

Commented by Ar Brandon last updated on 13/Oct/24

You're the long-white-bearded old man. Haha!

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