Integration Questions

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

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

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

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

Commented by bemath last updated on 28/Jul/20

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

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

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

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

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

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

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

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

Answered by bemath last updated on 29/Jul/20

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

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