Question Number 24852 by nnnavendu last updated on 27/Nov/17
$$\mathrm{please}\:\mathrm{prove}\:\mathrm{that}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sinx}\right)\mathrm{dx}=−\frac{\pi}{\mathrm{2}}\mathrm{log2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{or} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi\:\:}{\mathrm{2}}} \mathrm{log}\left(\mathrm{cosx}\right)\mathrm{dx}=−\frac{\pi}{\mathrm{2}}\mathrm{log2} \\ $$
Commented by Tinku Tara last updated on 28/Nov/17
![sin(x)=cos ((π/2)−x) ⇒∫_0 ^( π/2) log (cos x) substitute u=(π/2)−x⇒x=0,u=(π/2),x=(π/2),u=0 du=−dx ∫_0 ^( π/2) log (cos x)=−∫_(π/2) ^0 log (sin u)du =∫_0 ^( π/2) log (sin u)du ∫_0 ^( π/2) log (sinx) dx=∫_0 ^(π/2) log (cos x) dx=I 2I=∫_0 ^(π/2) [log (sin x)+log (cos x)]dx =∫_0 ^( π/2) log ((sin 2x)/2)dx =∫_0 ^(π/2) log (sin 2x)dx−∫_0 ^(π/2) ln 2dx ∫_0 ^(π/2) log (sin 2x)dx 2x=u⇒dx=du/2 limits change to 0 to π =(1/2)∫_0 ^π log (sin u)du−(π/2)ln 2 =(1/2)[∫_0 ^(π/2) log (sin u)du+∫_(π/2) ^π log (sin u)du]−(π/2)ln 2 =(1/2)[I+∫_(π/2) ^π log (sin u)du]−(π/2)ln 2 t=u−(π/2)⇒u=(π/2)+t⇒sin u=cos t dt=du, limits change to 0 to (π/2) =(1/2)[I+∫_0 ^(π/2) log (cos t)dt]−(π/2)ln 2 2I=(1/2)[I+I]−(π/2)ln 2 I=−(π/2)ln 2](https://www.tinkutara.com/question/Q24891.png)
$$\mathrm{sin}\left({x}\right)=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{substitute} \\ $$$${u}=\frac{\pi}{\mathrm{2}}−{x}\Rightarrow{x}=\mathrm{0},{u}=\frac{\pi}{\mathrm{2}},{x}=\frac{\pi}{\mathrm{2}},{u}=\mathrm{0} \\ $$$${du}=−{dx} \\ $$$$\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{cos}\:{x}\right)=−\int_{\pi/\mathrm{2}} ^{\mathrm{0}} \mathrm{log}\:\left(\mathrm{sin}\:{u}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{sin}\:{u}\right){du} \\ $$$$\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{sin}{x}\right)\:{dx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{cos}\:{x}\right)\:{dx}={I} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left[\mathrm{log}\:\left(\mathrm{sin}\:{x}\right)+\mathrm{log}\:\left(\mathrm{cos}\:{x}\right)\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{log}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{sin}\:\mathrm{2}{x}\right){dx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\mathrm{2}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{sin}\:\mathrm{2}{x}\right){dx} \\ $$$$\mathrm{2}{x}={u}\Rightarrow{dx}={du}/\mathrm{2} \\ $$$${limits}\:\mathrm{change}\:\mathrm{to}\:\mathrm{0}\:\mathrm{to}\:\pi \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{log}\:\left(\mathrm{sin}\:{u}\right){du}−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{sin}\:{u}\right){du}+\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{log}\:\left(\mathrm{sin}\:{u}\right){du}\right]−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{I}+\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{log}\:\left(\mathrm{sin}\:{u}\right){du}\right]−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${t}={u}−\frac{\pi}{\mathrm{2}}\Rightarrow{u}=\frac{\pi}{\mathrm{2}}+{t}\Rightarrow\mathrm{sin}\:{u}=\mathrm{cos}\:{t} \\ $$$${dt}={du},\:\mathrm{limits}\:\mathrm{change}\:\mathrm{to}\:\mathrm{0}\:\mathrm{to}\:\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{I}+\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{log}\:\left(\mathrm{cos}\:{t}\right){dt}\right]−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}\left[{I}+{I}\right]−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${I}=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$