Question Number 113188 by mathdave last updated on 11/Sep/20
$${proporsed}\:{by}\:{m}.{n}\:{july}\:\mathrm{1790} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx} \\ $$
Answered by mathdave last updated on 11/Sep/20
![my solution to I=∫_0 ^π ln(1−(1/2)sinx)dx=2∫_0 ^(π/2) ln(1−(1/2)sinx)dx.....(1) and J=2∫_0 ^(π/2) ln(1+(1/2)sinx)dx...........(2) adding (1) and (2) I+J=2∫_0 ^(π/2) ln(1−((sin^2 x)/4))dx=2∫_0 ^(π/2) ln(cos^2 x+sin^2 x−((sin^2 x)/4))dx I+J=2∫_0 ^(π/2) ln(cos^2 x+(3/4)sin^2 x)dx recall that ∫_0 ^(π/2) ln(a^2 cos^2 x+b^2 sin^2 x)dx=πln(((a+b)/2)) I+J=2πln((1/2)−((√3)/4))=2πln(((2+(√3))/4))......(x) subtracting (1) from (2) J−I=2∫_0 ^(π/2) ln(((1+(1/2)sinx)/(1−(1/2)sinx)))dx (apply faynmann trick) J−I=2∫_0 ^(π/2) dx∫_(−(1/2)) ^(1/2) (((sinx)/(1+ysinx)))dx J−I=2∫_0 ^(π/2) dx[∫_(−(1/2)) ^0 (((sinx)/(1+ysinx)))dy+∫_0 ^(1/2) (((sinx)/(1+ysinx)))dy] put y=−y J−I=2∫_0 ^(π/2) dx[∫_0 ^(1/2) (((sinx)/(1−ysinx)))dy+∫_0 ^(π/2) (((sinx)/(1+ysinx)))dy] J−I=2∫_0 ^(π/2) dx[∫_0 ^(1/2) (((sinx(1+ysinx+1−ysinx))/((1−ysinx)(1+ysinx))))dy] J−I=4∫_0 ^(π/2) dx∫_0 ^(1/2) (((sinx)/(1−y^2 sin^2 x)))dy (but sin^2 x=1−cos^2 x) J−I=4∫_0 ^(π/2) dx∫_0 ^(1/2) (((sinx)/(1−y^2 +y^2 cos^2 x)))dy J−I=4∫_0 ^(1/2) dy∫_0 ^(π/2) (((sinx)/( (√(1−y^2 ))+(ycosx)^2 )))dx let Z=ycosx ,A=(√(1−y^2 )),but ∫(1/(A^2 +Z^2 ))dZ=(1/A)tan^(−1) ((Z/A)) J−I=−4∫_0 ^(1/2) dy[(1/(y(√(1−y^2 ))))tan^(−1) (((ycosx)/( (√(1−y^2 )))))_0 ^(π/2) ] J−I=4∫_0 ^(1/2) [((tan^(−1) ((y/( (√(1−y^2 ))))))/(y(√(1−y^2 ))))] dy ( let y=siny) J−I=4∫_0 ^(π/6) (y/(siny))dy (using IBP) note ∫(1/(siny))dy=ln(tan(y/2)) J−I=4[yln(tan(y/2))]_0 ^(π/6) −4∫_0 ^(π/6) ln(tan(y/2))dy J−I=4•(π/6)ln(tan(π/(12)))−8∫_0 ^(π/(12)) ln(tany)dy notes tan(A/2)=(√((1−cos2A)/(1+cos2A))) and from lemma (2) ∫_0 ^(π/(12)) ln(tany)dy=−(2/3)G ∵J−I=(π/3)ln[((1−cos(π/6))/(1+cos(π/6)))]−8[−(2/3)G] J−I=(π/3)[((2−(√3))/(2+(√3)))]+((16)/3)G=((16)/3)−((2π)/3)ln(2+(√3)).....(xx) ∵I=∫_0 ^π ln(1−(1/2)sinx)dx=((J+I−(J−I))/2) I=((2πln(((2+(√3))/4))−((16)/3)G+((2π)/3)ln(2+(√3)))/2) I=πln(((2+(√3))/4))−(8/3)G+(π/3)ln(2+(√3)) ∵∫_0 ^π (1−(1/2)sinx)dx=((4π)/3)ln(2+(√3))−(8/3)G−2πln2 by mathdave(12/09/2020)](https://www.tinkutara.com/question/Q113207.png)
$${my}\:{solution}\:{to} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx}…..\left(\mathrm{1}\right) \\ $$$${and}\:{J}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx}………..\left(\mathrm{2}\right) \\ $$$${adding}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$${I}+{J}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{4}}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} {x}−\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{4}}\right){dx} \\ $$$${I}+{J}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}^{\mathrm{2}} {x}+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} {x}\right){dx} \\ $$$${recall}\:{that}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left({a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}\right){dx}=\pi\mathrm{ln}\left(\frac{{a}+{b}}{\mathrm{2}}\right) \\ $$$${I}+{J}=\mathrm{2}\pi\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)=\mathrm{2}\pi\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\right)……\left({x}\right) \\ $$$${subtracting}\:\left(\mathrm{1}\right)\:{from}\:\left(\mathrm{2}\right) \\ $$$${J}−{I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}}\right){dx}\:\:\:\:\left({apply}\:{faynmann}\:{trick}\right) \\ $$$${J}−{I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}}{\mathrm{1}+{y}\mathrm{sin}{x}}\right){dx} \\ $$$${J}−{I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\left[\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{0}} \left(\frac{\mathrm{sin}{x}}{\mathrm{1}+{y}\mathrm{sin}{x}}\right){dy}+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}}{\mathrm{1}+{y}\mathrm{sin}{x}}\right){dy}\right] \\ $$$${put}\:{y}=−{y} \\ $$$${J}−{I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\left[\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}}{\mathrm{1}−{y}\mathrm{sin}{x}}\right){dy}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}}{\mathrm{1}+{y}\mathrm{sin}{x}}\right){dy}\right] \\ $$$${J}−{I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\left[\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}\left(\mathrm{1}+{y}\mathrm{sin}{x}+\mathrm{1}−{y}\mathrm{sin}{x}\right)}{\left(\mathrm{1}−{y}\mathrm{sin}{x}\right)\left(\mathrm{1}+{y}\mathrm{sin}{x}\right)}\right){dy}\right] \\ $$$${J}−{I}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}}{\mathrm{1}−{y}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}\right){dy}\:\:\:\:\:\left({but}\:\:\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right) \\ $$$${J}−{I}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}}{\mathrm{1}−{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}}\right){dy} \\ $$$${J}−{I}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {dy}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{sin}{x}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }+\left({y}\mathrm{cos}{x}\right)^{\mathrm{2}} }\right){dx} \\ $$$${let}\:\:{Z}={y}\mathrm{cos}{x}\:\:,{A}=\sqrt{\mathrm{1}−{y}^{\mathrm{2}} },{but}\:\:\int\frac{\mathrm{1}}{{A}^{\mathrm{2}} +{Z}^{\mathrm{2}} }{dZ}=\frac{\mathrm{1}}{{A}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{Z}}{{A}}\right) \\ $$$${J}−{I}=−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {dy}\left[\frac{\mathrm{1}}{{y}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{y}\mathrm{cos}{x}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\right)_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \right] \\ $$$${J}−{I}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\right)}{{y}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\right]\:{dy}\:\:\:\:\left(\:{let}\:\:{y}=\mathrm{sin}{y}\right) \\ $$$${J}−{I}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \frac{{y}}{\mathrm{sin}{y}}{dy}\:\:\:\:\:\left({using}\:{IBP}\right)\:{note}\:\int\frac{\mathrm{1}}{\mathrm{sin}{y}}{dy}=\mathrm{ln}\left(\mathrm{tan}\frac{{y}}{\mathrm{2}}\right) \\ $$$${J}−{I}=\mathrm{4}\left[{y}\mathrm{ln}\left(\mathrm{tan}\frac{{y}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} −\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{ln}\left(\mathrm{tan}\frac{{y}}{\mathrm{2}}\right){dy} \\ $$$${J}−{I}=\mathrm{4}\bullet\frac{\pi}{\mathrm{6}}\mathrm{ln}\left(\mathrm{tan}\frac{\pi}{\mathrm{12}}\right)−\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \mathrm{ln}\left(\mathrm{tan}{y}\right){dy} \\ $$$${notes}\:\mathrm{tan}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}−\mathrm{cos2}{A}}{\mathrm{1}+\mathrm{cos2}{A}}}\:{and}\:{from} \\ $$$${lemma}\:\left(\mathrm{2}\right)\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \mathrm{ln}\left(\mathrm{tan}{y}\right){dy}=−\frac{\mathrm{2}}{\mathrm{3}}{G} \\ $$$$\because{J}−{I}=\frac{\pi}{\mathrm{3}}\mathrm{ln}\left[\frac{\mathrm{1}−\mathrm{cos}\frac{\pi}{\mathrm{6}}}{\mathrm{1}+\mathrm{cos}\frac{\pi}{\mathrm{6}}}\right]−\mathrm{8}\left[−\frac{\mathrm{2}}{\mathrm{3}}{G}\right] \\ $$$${J}−{I}=\frac{\pi}{\mathrm{3}}\left[\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right]+\frac{\mathrm{16}}{\mathrm{3}}{G}=\frac{\mathrm{16}}{\mathrm{3}}−\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)…..\left({xx}\right) \\ $$$$\because{I}=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx}=\frac{{J}+{I}−\left({J}−{I}\right)}{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{2}\pi\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\right)−\frac{\mathrm{16}}{\mathrm{3}}{G}+\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$${I}=\pi\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\right)−\frac{\mathrm{8}}{\mathrm{3}}{G}+\frac{\pi}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$\because\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{x}\right){dx}=\frac{\mathrm{4}\pi}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)−\frac{\mathrm{8}}{\mathrm{3}}{G}−\mathrm{2}\pi\mathrm{ln2} \\ $$$${by}\:{mathdave}\left(\mathrm{12}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 11/Sep/20
$${exellent}\:.{very}\:{nice} \\ $$$${grateful}\:{mr}\:{dave}.. \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$