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Question Number 162068 by HongKing last updated on 25/Dec/21

Answered by Lordose last updated on 26/Dec/21

$$$$$$\mathrm{\Omega }=\underset{ϵ\to 0}{\lim }{\int }_{{\sin }^{-1}\left(ϵ\right)}^{{\sin }^{-1}(1-ϵ)}\log ({\left(\cos \left(\mathrm{x}\right)\right)}^{\cot \left(\mathrm{x}\right)}\cdot {\left(\sin \left(\mathrm{x}\right)\right)}^{\frac{\cos \left(\mathrm{x}\right)}{1+\sin \left(\mathrm{x}\right)}})\mathrm{dx}$$$$\mathrm{\Omega }={\int }_0^{\frac{\mathit{\pi }}{2}}\cot \left(\mathrm{x}\right)\log \left(\cos \left(\mathrm{x}\right)\right)\mathrm{dx}+{\int }_0^{\frac{\mathit{\pi }}{2}}\frac{\cos \left(\mathrm{x}\right)}{1+\sin \left(\mathrm{x}\right)}\log \left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx}$$$$\mathrm{\Omega }=\mathrm{A}+\mathrm{B}$$$$\mathrm{A}\stackrel{\mathrm{u}=\cos \left(\mathrm{x}\right)}{=}{\int }_0^1\frac{\mathrm{u}}{1-{\mathrm{u}}^2}\log \left(\mathrm{u}\right)\mathrm{du}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{u}}^{2\mathrm{n}+1}\log \left(\mathrm{u}\right)\mathrm{du}$$$$\mathrm{A}\stackrel{\mathbf{IBP}}{=}-\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}+1}{\int }_0^1{\mathrm{u}}^{2\mathrm{n}+1}\mathrm{du}=-\frac{1}{4}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+1)}^2}$$$$\mathrm{A}=-\frac{1}{4}\mathit{\zeta }\left(2\right)=-\frac{{\mathit{\pi }}^2}{24}$$$$\mathrm{B}\stackrel{\mathrm{u}=\sin \left(\mathrm{x}\right)}{=}{\int }_0^1\frac{\log \left(\mathrm{u}\right)}{1+\mathrm{u}}\mathrm{du}={\mathbf{Li}}_2(-1)=-\frac{{\mathit{\pi }}^2}{12}$$$$\mathrm{\Omega }=\mathrm{A}+\mathrm{B}$$$$\mathbf{\Omega }=-\frac{{\mathit{\pi }}^2}{8}$$$$\mathit{\varnothing }\mathbf{sE}$$

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