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Question Number 26357 by abdo imad last updated on 24/Dec/17

find the value of ∫_0 ^(π/2) ln(cosθ)dθ and ∫_0 ^(π/2) ln(sinθ)dθ .

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\theta\right){d}\theta\:\:{and}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sin}\theta\right){d}\theta\:\:\:. \\ $$

Commented by prakash jain last updated on 24/Dec/17

∫_0 ^(π/2) ln (cos x)dx u=(π/2)−x du=−dx I=∫_0 ^(π/2) ln (sin x)dx=−∫_(π/2) ^0 ln (cos u)du =∫_0 ^(π/2) ln (cos x)dx 2I=∫_0 ^(π/2) ln (cos xsin x)dx =∫_0 ^(π/2) ln sin 2xdx−∫_0 ^(π/2) ln 2dx =I−(π/2)ln 2 I=−(π/2)ln 2

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{cos}\:{x}\right){dx} \\ $$$${u}=\frac{\pi}{\mathrm{2}}−{x} \\ $$$${du}=−{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{sin}\:{x}\right){dx}=−\int_{\pi/\mathrm{2}} ^{\mathrm{0}} \mathrm{ln}\:\left(\mathrm{cos}\:{u}\right){du} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{cos}\:{x}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\left(\mathrm{cos}\:{x}\mathrm{sin}\:{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\mathrm{sin}\:\mathrm{2}{xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\:\mathrm{2}{dx} \\ $$$$={I}−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${I}=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$

Answered by prakash jain last updated on 24/Dec/17

−(π/2)ln 2

$$−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$

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