The value of ∫$$01$$ log⁡$$sin$$⁡π$$x2dx$$ is equal to

Question

The value of ∫$$01$$ log⁡$$sin$$⁡π$$x2dx$$ is equal to

Infinity Learn · Accepted Answer

Let, $$I=$$∫$$01$$ log⁡$$sin$$⁡π$$x2dxput$$, π$$x2=t$$⇒$$dx=2$$$$π$$$$dtI=2$$π∫$$0$$π$$/2$$ log⁡$$sin$$⁡$$tdt=2$$$$π$$$$I1-----iwhere$$,$$I1=$$∫$$0$$π$$/2$$ log⁡$$sin$$⁡$$tdt=$$∫$$0$$π$$/2$$ log⁡$$sin$$⁡π$$2$$−$$tdt=$$∫$$0$$π$$/2$$ log⁡$$cos$$⁡tdt∴ $$2I1=$$∫$$0$$π$$/2$$ (log$$⁡$$sin$$⁡$$t+log$$⁡$$cos$$⁡$$t)dt=$$∫$$0$$π$$/2$$ log⁡$$($$$$sin$$⁡tcos⁡$$t)dt=$$∫$$0$$π$$/2$$ log⁡$$sin$$⁡$$2t2dt=$$∫$$0$$π$$/2$$ $$(log$$⁡$$sin$$⁡$$2t$$−log⁡$$2)dt=12$$∫$$0$$π log⁡$$sin$$⁡zdz−log⁡$$2$$∫$$0$$π$$/2$$ $$dt=12$$⋅$$2$$∫$$0$$π$$/2$$ log⁡$$sin$$⁡zdz−π$$2log$$⁡$$2$$⇒ $$2I1=I1$$−π$$2log$$⁡$$2$$⇒$$I1=$$−π$$2log$$⁡$$2From$$ Eq. $$(i), $$I=2$$π−π$$2log$$⁡$$2=$$−log⁡$$2$$

The value of $\int_{0}^{1} \log \sin (\frac{πx}{2}) dx$ is equal to

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The value of ∫01 log⁡sin⁡πx2dx is equal to

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The value of $\int_{0}^{1} \log \sin (\frac{πx}{2}) dx$ is equal to