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

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

Commented by Spillover last updated on 10/Oct/24

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

Commented by Frix last updated on 10/Oct/24

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

Answered by Spillover last updated on 10/Oct/24

$$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\ln \left(\frac{1+\sin x}{1-\sin x}\right)$$$$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\ln \left(\frac{1+\sin x}{1-\sin x}\times \frac{1+\sin x}{1+\sin x}\right)=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\ln \frac{{(1+\sin x)}^2}{\cos {}^{2}x}$$$$\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\ln \frac{{(1+\sin x)}^2}{\cos {}^{2}x}=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\ln {(1+\sin x)}^2-\ln \cos {}^{2}x$$$$\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\ln {(1+\sin x)}^2-\ln \cos {}^{2}x$$$$\frac{1}{4}{\int }_0^{\frac{\pi }{2}}[2\ln (1+\sin x)-2\ln \cos x]dx$$$$\left[\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln (1+\sin x)dx\right]-\left[\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \cos x\right]dx$$$$\left[\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \cos x\right]dx=-\frac{\pi }{2}\ln 2$$$$\left[\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln (1+\sin x)dx\right]$$$$....$$

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

$$\mathrm{\Omega }=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\ln \left(\frac{1+\sin x}{1-\sin x}\right)dx=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)dx$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\frac{\sin (\frac{x}{2}+\frac{\pi }{4})}{\cos (\frac{x}{2}+\frac{\pi }{4})}\right)dx=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right)dx$$$$={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{\cos x}{\sin x}\right)dx=-{\int }_0^{\frac{\pi }{4}}\ln \left(\tan x\right)dx=G$$$$G=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^2},{\mathrm{Catalan}}^{\prime }\mathrm{s}\mathrm{constant}.$$

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

$$HowdidyouknowifsirMJs.username$$$$isdifferent$$

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

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

Commented by Ghisom last updated on 13/Oct/24

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

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

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

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