An integral of Romanian Mathematical Magazine

Find a closed form: $$\mathrm{\Omega }={\int }_1^{e^{\pi }}{\int }_0^1\frac{\arcsin (x^a)\arccos (x^a)}{x}dxda$$ Proposed by Ankush Kumar Parcha - India 

Solution by Ose Favour - Nigeria $$\mathrm{\Omega }={\int }_0^1\frac{\arccos (x)\arcsin (x)}{x}dx={\int }_0^1\frac{\arcsin (x)}{x}(\frac{\pi }{2}-\arcsin (x))dx=$$ $$=\frac{\pi }{2}{\int }_0^1\frac{\arcsin (x)}{x}dx-{\int }_0^1\frac{(\arcsin (x){)}^2}{x}dx\cong$$ $$=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}y\cot (y)dy-{\int }_0^{\frac{\pi }{2}}{y}^2\cot (y)dy=\frac{{\pi }^2}{4}\ln (2)+2{\int }_0^{\frac{\pi }{2}}y\ln (\sin (y))dy=$$ $$=\frac{{\pi }^2}{4}\ln (2)-\frac{{\pi }^2}{8}\ln (2)-\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}{\int }_0^{\frac{\pi }{2}}y\cos (2ky)dy=\frac{7}{8}\zeta (3).$$