The value of ∫$$0$$π xlog⁡$$($$$$sin$$⁡$$x)dx$$ is $$(given$$ that ∫$$0$$π$$/2$$ log⁡$$sin$$⁡$$xdx=$$−π$$2log$$⁡$$2$$

Correct option is 1−π22log⁡2

Questions

The value of $\int_{0}^{π} x \log (\sin x) d x$ is (given that $\int_{0}^{π / 2} \log \sin x d x = - \frac{π}{2} \log 2)$

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−π22log⁡2

−π24log⁡2

−π28log⁡2

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detailed solution

Correct option is A

I=∫0π xlog⁡(sin⁡x)dx=∫0π (π−x)log⁡[sin⁡(π−x)]dx⇒2I=π∫0π log⁡sin⁡xdx=2π∫0π/2 log⁡sin⁡xdx⇒I=π∫0π/2 log⁡(sin)dx=-π22log⁡2

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The value of ∫0π xlog⁡(sin⁡x)dx is (given that ∫0π/2 log⁡sin⁡xdx=−π2log⁡2

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The value of $\int_{0}^{π} x \log (\sin x) d x$ is (given that $\int_{0}^{π / 2} \log \sin x d x = - \frac{π}{2} \log 2)$