Question Number 75960 by Tony Lin last updated on 21/Dec/19
$${prove}\:{that}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx}=−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2} \\ $$
Commented by Kunal12588 last updated on 21/Dec/19
$${now}\:{solve}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\:{x}\right){dx} \\ $$
Commented by Tony Lin last updated on 23/Dec/19
![∫_0 ^(π/2) ln(sinx)dx let θ=(π/2)−x,dθ=−dx −∫_(π/2) ^0 ln[sin((π/2)−x)]dθ =∫_0 ^(π/2) ln(cosx)dx](https://www.tinkutara.com/question/Q76054.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinx}\right){dx} \\ $$$${let}\:\theta=\frac{\pi}{\mathrm{2}}−{x},{d}\theta=−{dx} \\ $$$$−\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {ln}\left[{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\right]{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosx}\right){dx} \\ $$
Answered by Kunal12588 last updated on 21/Dec/19
![I=∫_0 ^(π/2) ln(sin x) dx ⇒I=∫_0 ^(π/2) ln(cos x) dx 2I=∫_0 ^(π/2) ln(sin x cos x)dx ⇒2I=∫_0 ^(π/2) [ln(sin 2x)−ln(2)]dx ⇒2I=∫_0 ^(π/2) ln(sin 2x) dx − [xln2]_0 ^(π/2) let t=2x⇒dx=(1/2)dt x→0⇒t→0 x→(π/2)⇒t→π 2I=(1/2)∫_0 ^π ln(sin t) dt − (π/2)ln2 ⇒2I=(1/2)×2∫_0 ^(π/2) ln(sin x)dx−(π/2)ln2 ⇒2I=I−(π/2)ln2 ⇒I=−(π/2)ln2 proved](https://www.tinkutara.com/question/Q75962.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\:{x}\right)\:{dx} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\:{x}\right)\:{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\:{x}\:{cos}\:{x}\right){dx} \\ $$$$\Rightarrow\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[{ln}\left({sin}\:\mathrm{2}{x}\right)−{ln}\left(\mathrm{2}\right)\right]{dx} \\ $$$$\Rightarrow\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\:\mathrm{2}{x}\right)\:{dx}\:−\:\left[{xln}\mathrm{2}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$${let}\:{t}=\mathrm{2}{x}\Rightarrow{dx}=\frac{\mathrm{1}}{\mathrm{2}}{dt} \\ $$$${x}\rightarrow\mathrm{0}\Rightarrow{t}\rightarrow\mathrm{0} \\ $$$${x}\rightarrow\frac{\pi}{\mathrm{2}}\Rightarrow{t}\rightarrow\pi \\ $$$$\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {ln}\left({sin}\:{t}\right)\:{dt}\:−\:\frac{\pi}{\mathrm{2}}{ln}\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\:{x}\right){dx}−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}{I}={I}−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2} \\ $$$$\Rightarrow{I}=−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}\:\:\:\:{proved} \\ $$
Commented by Tony Lin last updated on 23/Dec/19
$${thanks}\:{sir} \\ $$