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Question-201349




Question Number 201349 by sonukgindia last updated on 05/Dec/23
Answered by Sutrisno last updated on 05/Dec/23
misal :  A = ∫_0 ^(π/2) ln(cosx) dx  B = ∫_0 ^(π/2) ln(sinx) dx  A+B = ∫_0 ^(π/2) ln(cosx) +∫_0 ^(π/2) ln(sinx) dx  A+B = ∫_0 ^(π/2) ln(cosx.sinx) dx  A+B = ∫_0 ^(π/2) ln((2/2)cosx.sinx) dx  A+B = ∫_0 ^(π/2) ln(((sin2x)/2)) dx  A+B = ∫_0 ^(π/2) ln(sin2x) dx −∫_0 ^(π/2)  ln2dx  A+B = (1/2)∫_0 ^π ln(sinx) dx −∫_0 ^(π/2)  ln2dx   (∫_0 ^(π/2) ln(sinx)dx=∫_(π/2) ^π sinxdx)  A+B = (1/2).2B+ −∫_0 ^(π/2)  ln2dx  A = −x. ln2∣_0 ^(π/2)   A = −(π/2). ln2
$${misal}\:: \\ $$$${A}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({cosx}\right)\:{dx} \\ $$$${B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({cosx}\right)\:+\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({cosx}.{sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left(\frac{\mathrm{2}}{\mathrm{2}}{cosx}.{sinx}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left(\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\right)\:{dx} \\ $$$${A}+{B}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sin}\mathrm{2}{x}\right)\:{dx}\:−\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:{ln}\mathrm{2}{dx} \\ $$$${A}+{B}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}{ln}\left({sinx}\right)\:{dx}\:−\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:{ln}\mathrm{2}{dx}\:\:\:\left(\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ln}\left({sinx}\right){dx}=\underset{\frac{\pi}{\mathrm{2}}} {\overset{\pi} {\int}}{sinxdx}\right) \\ $$$${A}+{B}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}{B}+\:−\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:{ln}\mathrm{2}{dx} \\ $$$${A}\:=\:−{x}.\:{ln}\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\mid}} \\ $$$${A}\:=\:−\frac{\pi}{\mathrm{2}}.\:{ln}\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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