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Question Number 24852 by nnnavendu last updated on 27/Nov/17

$$\mathrm{please}\mathrm{prove}\mathrm{that}{\int }_0^{\frac{\pi }{2}}\log \left(\mathrm{sinx}\right)\mathrm{dx}=-\frac{\pi }{2}\log 2$$$$\mathrm{or}$$$${\int }_0^{\frac{\pi }{2}}\log \left(\mathrm{cosx}\right)\mathrm{dx}=-\frac{\pi }{2}\log 2$$

Commented by Tinku Tara last updated on 28/Nov/17

$$\sin \left(x\right)=\cos (\frac{\pi }{2}-x)$$$$\Rightarrow {\int }_0^{\pi /2}\log \left(\cos x\right)$$$$\mathrm{substitute}$$$$u=\frac{\pi }{2}-x\Rightarrow x=0,u=\frac{\pi }{2},x=\frac{\pi }{2},u=0$$$$du=-dx$$$${\int }_0^{\pi /2}\log \left(\cos x\right)=-{\int }_{\pi /2}^0\log \left(\sin u\right)du$$$$={\int }_0^{\pi /2}\log \left(\sin u\right)du$$$${\int }_0^{\pi /2}\log \left(\sin x\right)dx={\int }_0^{\pi /2}\log \left(\cos x\right)dx=I$$$$2I={\int }_0^{\pi /2}[\log \left(\sin x\right)+\log \left(\cos x\right)]dx$$$$={\int }_0^{\pi /2}\log \frac{\sin 2x}{2}dx$$$$={\int }_0^{\pi /2}\log \left(\sin 2x\right)dx-{\int }_0^{\pi /2}\ln 2dx$$$${\int }_0^{\pi /2}\log \left(\sin 2x\right)dx$$$$2x=u\Rightarrow dx=du/2$$$$limits\mathrm{change}\mathrm{to}0\mathrm{to}\pi$$$$=\frac{1}{2}{\int }_0^{\pi }\log \left(\sin u\right)du-\frac{\pi }{2}\ln 2$$$$=\frac{1}{2}[{\int }_0^{\pi /2}\log \left(\sin u\right)du+{\int }_{\pi /2}^{\pi }\log \left(\sin u\right)du]-\frac{\pi }{2}\ln 2$$$$=\frac{1}{2}[I+{\int }_{\pi /2}^{\pi }\log \left(\sin u\right)du]-\frac{\pi }{2}\ln 2$$$$t=u-\frac{\pi }{2}\Rightarrow u=\frac{\pi }{2}+t\Rightarrow \sin u=\cos t$$$$dt=du,\mathrm{limits}\mathrm{change}\mathrm{to}0\mathrm{to}\frac{\pi }{2}$$$$=\frac{1}{2}[I+{\int }_0^{\pi /2}\log \left(\cos t\right)dt]-\frac{\pi }{2}\ln 2$$$$2I=\frac{1}{2}[I+I]-\frac{\pi }{2}\ln 2$$$$I=-\frac{\pi }{2}\ln 2$$

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