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Question Number 151246 by mathmax by abdo last updated on 19/Aug/21

$$\mathrm{find}\mathrm{I}={\int }_0^{\frac{\pi }{4}}\ln \left(\mathrm{cosx}\right)\mathrm{dx}\mathrm{and}\mathrm{J}={\int }_0^{\frac{\pi }{4}}\ln \left(\mathrm{sinx}\right)\mathrm{dx}$$

Answered by qaz last updated on 19/Aug/21

$${\int }_0^{\pi /4}\mathrm{lnsin}\mathrm{xdx}$$$$=\frac{1}{2}{\int }_0^{\pi /2}\mathrm{lnsin}\frac{\mathrm{x}}{2}\mathrm{dx}$$$$=\frac{1}{4}{\int }_0^{\pi /2}\ln \frac{1-\cos \mathrm{x}}{2}\mathrm{dx}$$$$=\frac{1}{4}{\int }_0^{\pi /2}\ln \left(\sin \mathrm{xtan}\frac{\mathrm{x}}{2}\right)\mathrm{dx}-\frac{1}{4}{\int }_0^{\pi /2}\ln 2\mathrm{dx}$$$$=-\frac{\pi }{8}\ln 2+\frac{1}{4}{\int }_0^{\pi /2}\mathrm{lntan}\frac{\mathrm{x}}{2}\mathrm{dx}-\frac{\pi }{8}\ln 2$$$$=-\frac{\pi }{4}\ln 2+\frac{1}{2}{\int }_0^{\pi /4}\mathrm{lntan}\mathrm{xdx}$$$$=-\frac{\pi }{4}\ln 2+\frac{1}{2}{\int }_0^1\frac{\mathrm{lnx}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=-\frac{\pi }{4}\ln 2+\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}}\mathrm{lnxdx}$$$$=-\frac{\pi }{4}\ln 2+\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{(2\mathrm{n}+1)}^2}$$$$=-\frac{\pi }{4}\ln 2-\frac{1}{2}\mathrm{G}$$$$---------------$$$${\int }_0^{\pi /4}\mathrm{lncos}\mathrm{xdx}+{\int }_0^{\pi /4}\mathrm{lnsin}\mathrm{xdx}$$$$={\int }_0^{\pi /4}\ln \left(\frac{1}{2}\sin 2\mathrm{x}\right)\mathrm{dx}$$$$={\int }_0^{\pi /4}\ln \frac{1}{2}\mathrm{dx}+\frac{1}{2}{\int }_0^{\pi /2}\mathrm{lnsin}\mathrm{xdx}$$$$=-\frac{\pi }{2}\ln 2$$$$\Rightarrow {\int }_0^{\pi /4}\mathrm{lncos}\mathrm{xdx}=-\frac{\pi }{2}\ln 2-(-\frac{\pi }{4}\ln 2-\frac{1}{2}\mathrm{G})=\frac{1}{2}\mathrm{G}-\frac{\pi }{4}\ln 2$$

Commented by peter frank last updated on 19/Aug/21

$$\mathrm{thank}\mathrm{you}$$

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