0-pi-2-log-sin-d-

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Question Number 313 by Vishal Bhardwaj last updated on 25/Jan/15

$${\int }_0^{\frac{\pi }{2}}logsin\theta d\theta$$

Answered by prakash jain last updated on 20/Dec/14

$$I={\int }_0^{\pi /2}\ln \sin \theta d\theta \dots ..\left(i\right)$$$$\mathrm{put}\theta =\frac{\pi }{2}-\alpha ,d\theta =-d\alpha$$$$I={\int }_{\pi /2}^0\ln \sin (\frac{\pi }{2}-\alpha )(-d\alpha )$$$$=-{\int }_{\pi /2}^0\ln \cos \alpha d\alpha ={\int }_0^{\pi /2}\ln \cos \alpha d\alpha$$$$={\int }_0^{\pi /2}\ln \cos \theta d\theta \dots .\left(ii\right)$$$$\mathrm{add}\left(i\right)\mathrm{and}\left(ii\right)$$$$2I={\int }_0^{\pi /2}(\ln \sin \theta +\ln \cos \theta )d\theta$$$$2\mathrm{I}={\int }_0^{\pi /2}\ln \frac{\sin 2\theta }{2}d\theta$$$$2I={\int }_0^{\pi /2}(\ln \sin 2\theta -\ln 2)d\theta$$$$2I={\int }_0^{\pi /2}\ln \sin 2\theta d\theta -{\int }_0^{\pi /2}\ln 2d\theta$$$$\mathrm{In}\mathrm{first}\mathrm{integral}\mathrm{put}2\theta =\beta ,2d\theta =d\beta \mathrm{so}$$$$\mathrm{the}\mathrm{equation}\mathrm{one}\mathrm{now}\mathrm{with}\mathrm{updated}\mathrm{limits}$$$$2I={\int }_0^{\pi }\ln \sin \beta \frac{d\beta }{2}-{\int }_0^{\pi /2}\ln 2d\theta$$$$2I=\frac{1}{2}{\int }_0^{\pi }\ln \sin \beta d\beta -{\int }_0^{\pi /2}\ln 2d\theta \dots .\left(iii\right)$$$$\mathrm{Now}\mathrm{we}\mathrm{will}\mathrm{evaluate}\mathrm{first}\mathrm{integral}$$$${\int }_0^{\pi }\ln \sin \beta d\beta ={\int }_0^{\pi /2}\ln \sin \beta d\beta +{\int }_{\pi /2}^{\pi }\ln \sin \beta d\beta$$$${\int }_0^{\pi }\ln \sin \beta d\beta ==I+{\int }_{\pi /2}^{\pi }\ln \sin \beta d\beta$$$$\mathrm{put}\beta =\pi -td\beta =-dt$$$${\int }_0^{\pi }\ln \sin \beta d\beta ==I+{\int }_{\pi /2}^0\ln \sin (\pi -t)(-dt)$$$${\int }_0^{\pi }\ln \sin \beta d\beta ==I-{\int }_{\pi /2}^0\ln \sin tdt$$$${\int }_0^{\pi }\ln \sin \beta d\beta ==I+{\int }_0^{\pi /2}\ln \sin tdt$$$${\int }_0^{\pi }\ln \sin \beta d\beta ==I+I=2I\dots .\left(iv\right)$$$$\mathrm{putting}\mathrm{this}\mathrm{value}\mathrm{in}\left(iii\right)$$$$2I=\frac{1}{2}\left(2I\right)-{\int }_0^{\pi /2}\ln 2d\theta$$$$2I=I-\frac{\pi }{2}\ln 2$$$$I=-\frac{\pi }{2}\ln 2$$

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