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Question Number 162894 by mathacek last updated on 02/Jan/22

Answered by Ar Brandon last updated on 02/Jan/22

$${\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)dx-{\int }_0^{\frac{\pi }{2}}\ln \left(\cos x\right)dx$$$$={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)dx-{\int }_0^{\frac{\pi }{2}}\ln \left(\sin u\right)du,u=\frac{\pi }{2}-x$$$$=0$$

Commented by mathacek last updated on 02/Jan/22

$$Nice!$$

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