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Question Number 105412 by john santu last updated on 28/Jul/20

∫_0 ^(π/2) ln (cos x) dx

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)\:{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 28/Jul/20

∫_0 ^(π/2) log(cosx)dx=I=∫_0 ^(π/2) log(sinx)dx 2I=∫_0 ^(π/2) log(sinxcosx)dx 2I=∫_0 ^(π/4) log(sin2x)dx−∫_0 ^(π/2) log(2) { ∫_0 ^(π/4) log(sin2x)=∫_0 ^(π/2) log(sint)} 2I=I−∫_0 ^(π/2) log(2) I=−(π/2)log(2)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({cosx}\right){dx}={I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinxcosx}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({sin}\mathrm{2}{x}\right){dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\mathrm{2}\right)\:\:\:\:\:\left\{\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({sin}\mathrm{2}{x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sint}\right)\right\} \\ $$$$\mathrm{2}{I}={I}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\mathrm{2}\right) \\ $$$${I}=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$

Commented by bemath last updated on 28/Jul/20

∫_0 ^(π/2) log (2) dx ?

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{log}\:\left(\mathrm{2}\right)\:{dx}\:?\: \\ $$

Commented by Dwaipayan Shikari last updated on 28/Jul/20

[xlog(2)]_0 ^(π/2) =((π/2)−0)log2=(π/2)log2

$$\left[{xlog}\left(\mathrm{2}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\left(\frac{\pi}{\mathrm{2}}−\mathrm{0}\right){log}\mathrm{2}=\frac{\pi}{\mathrm{2}}{log}\mathrm{2} \\ $$

Commented by Dwaipayan Shikari last updated on 28/Jul/20

∫_0 ^(π/2) log(cosx)dx =∫_0 ^(π/2) log(((e^(ix) +e^(−ix) )/2))dx =∫_0 ^(π/2) log e^(−ix) +log(e^(2ix) +1)−∫_0 ^(π/2) log2 =∫_0 ^(π/2) −ix+∫log(e^(2ix) +1)−(π/2)log(2) =−((iπ^2 )/8)−(π/2)log(2)+∫_0 ^(π/2) log(e^(2ix) +1) =((−iπ^2 )/8)−(π/2)log(2)+((iπ^2 )/8)=−(π/2)log(2)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({cosx}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\:{e}^{−{ix}} +{log}\left({e}^{\mathrm{2}{ix}} +\mathrm{1}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\mathrm{2} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −{ix}+\int{log}\left({e}^{\mathrm{2}{ix}} +\mathrm{1}\right)−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$$=−\frac{{i}\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({e}^{\mathrm{2}{ix}} +\mathrm{1}\right) \\ $$$$=\frac{−{i}\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)+\frac{{i}\pi^{\mathrm{2}} }{\mathrm{8}}=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 28/Jul/20

Commented by Dwaipayan Shikari last updated on 28/Jul/20

From wolfram alpha

$${From}\:{wolfram}\:{alpha} \\ $$

Answered by bemath last updated on 29/Jul/20

Commented by mathmax by abdo last updated on 29/Jul/20

no sir the correct answer is ∫_0 ^(π/2) ln(sinu)du =−(π/2)ln(2)

$$\mathrm{no}\:\mathrm{sir}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{answer}\:\mathrm{is}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sinu}\right)\mathrm{du}\:=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$

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