Question Number 201349 by sonukgindia last updated on 05/Dec/23
Answered by Sutrisno last updated on 05/Dec/23
$${misal}\:: \\ $$$${A}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({cosx}\right)\:{dx} \\ $$$${B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({cosx}\right)\:+\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({cosx}.{sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left(\frac{\mathrm{2}}{\mathrm{2}}{cosx}.{sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left(\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sin}\mathrm{2}{x}\right)\:{dx}\:−\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:{ln}\mathrm{2}{dx} \\ $$$${A}+{B}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}{ln}\left({sinx}\right)\:{dx}\:−\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:{ln}\mathrm{2}{dx}\:\:\:\left(\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sinx}\right){dx}=\underset{\frac{\pi}{\mathrm{2}}} {\overset{\pi} {\int}}{sinxdx}\right) \\ $$$${A}+{B}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}{B}+\:−\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:{ln}\mathrm{2}{dx} \\ $$$${A}\:=\:−{x}.\:{ln}\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\mid}} \\ $$$${A}\:=\:−\frac{\pi}{\mathrm{2}}.\:{ln}\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$