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What is the value of $$\int_0^{\Large \frac{\pi}2}\ln(\sin x)dx=\text{?}$$ Can this be done using by parts?

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marked as duplicate by heropup, Daniel Fischer Feb 26 '16 at 19:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Notice, let $$I=\int_0^{\pi/2}\ln(\sin x)\ dx\tag 1$$ $$I=\int_0^{\pi/2}\ln\left(\sin \left(\frac{\pi}{2}-x\right)\right)\ dx$$ $$I=\int_0^{\pi/2}\ln\left(\cos x\right)\ dx\tag 2$$ adding (1) & (2), $$I+I=\int_0^{\pi/2}\ln\left(\sin x\right)\ dx+\int_0^{\pi/2}\ln\left(\cos x\right)\ dx$$ $$2I=\int_0^{\pi/2}\ln\left(\sin x\cos x\right)\ dx$$ $$I=\frac 12\int_0^{\pi/2}\ln\left(\frac{2 \sin x\cos x}{2}\right)\ dx$$ $$I=\frac 12\int_0^{\pi/2}\ln\left(\sin 2x\right)\ dx-\frac 12\int_0^{\pi/2}\ln\left(2\right)\ dx$$ $$I=\frac 14\int_0^{\pi}\ln\left(\sin x\right)\ dx-\frac 12\ln\left(2\right)\int_0^{\pi/2}\ dx$$ $$I=\frac 12\int_0^{\pi/2}\ln\left(\sin x\right)\ dx-\frac 12\ln\left(2\right)\cdot \frac \pi2$$ $$I=\frac I2-\frac{\pi}{4}\ln(2)$$ $$\frac I2=\color{}{-\frac{\pi}{4}\ln(2)}$$ $$\color{red}{I=-\frac{\pi}{2}\ln(2)}$$

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    $\begingroup$ Good lord, why do u answer this question which has already been answerd hundreds of times? $\endgroup$ – tired Feb 26 '16 at 20:40
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Let $$I=\int_0^{\frac{\pi}2}\ln(\sin x)dx$$ Using $$\int _a^b f(x)dx=\int _a^b f(a+b-x) dx$$ We get $$ I= \int_0^{\frac{\pi}2}\ln(\cos x)dx$$ Adding, $$\implies 2I= \int_0^{\frac{\pi}2}[\ln(\sin x)+\ln(\cos x)]dx$$ $$\implies 2I= \int_0^{\frac{\pi}2}\ln(\frac{\sin 2x}{2})dx$$ $$\implies 2I= \int_0^{\frac{\pi}2}\ln(\sin 2x)dx-\int_0^{\frac{\pi}2}\ln 2dx$$ $$\implies 2I= \frac{1}{2}\int_0^{\pi}\ln(\sin t)dt-\frac{\pi}{2}\ln2$$ Using $$\int_0^{\pi}\ln(\sin x)dx=2\int_0^{\frac{\pi}{2}}\ln(\sin x)dx$$ We get $$2I=I-\frac{\pi}{2}\ln2$$ $$\implies I= -\frac{\pi}{2}\ln2$$

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