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Question Number 122528 by mnjuly1970 last updated on 17/Nov/20

$$...nicecalculus...$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}xsin\left(x\right).cos\left(x\right)ln(sin\left(x\right).ln\left(cos\left(x\right)\right)dx$$$$\stackrel{???}{=}\frac{\pi }{16}-\frac{{\pi }^3}{192}✓$$

Answered by mindispower last updated on 18/Nov/20

$${\int }_a^{b}f\left(x\right)dx=\frac{1}{2}{\int }_a^b(f(a+b-x)+f\left(x\right))dx$$$$\iff =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}.\frac{\pi }{2}sin\left(x\right)cos\left(x\right)ln\left(sin\left(x\right)\right)ln\left(cos\left(x\right)\right)dx$$$$\mathrm{\Omega }=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}sin\left(x\right)cos\left(x\right)ln\left(sin\left(x\right)\right)ln\left(cos\left(x\right)\right)dx$$$$f(a,b)=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}sin\left(x\right)cos\left(x\right){sin}^a\left(x\right){cos}^b\left(x\right)dx$$$$\mathrm{\Omega }=\frac{{\partial }^2}{\partial a\partial b}f(0,0)$$$$f=\frac{\pi }{8}.2{\int }_0^{\frac{\pi }{2}}{sin}^{2(1+\frac{a}{2})-1}\left(x\right).{cos}^{2(1+\frac{b}{2})-1}\left(x\right)dx$$$$=\frac{\pi }{8}\beta (1+a,1+b)$$$$\frac{\partial }{\partial b}\left(\frac{\partial }{\partial a}\right)f=\frac{\pi }{8}.\frac{\partial }{\partial b}.\frac{1}{2}\beta (\frac{a}{2}+1,\frac{b}{2}+1)(\mathrm{\Psi }(\frac{a}{2}+1)-\mathrm{\Psi }(\frac{1}{2}(a+b)+2)$$$$=\frac{\pi }{32}\beta (\frac{a}{2}+1,\frac{b}{2}+1)(\mathrm{\Psi }(\frac{b}{2}+1)-\mathrm{\Psi }(\frac{1}{2}(a+b)+2))(\mathrm{\Psi }(\frac{a}{2}+1)-\mathrm{\Psi }(\frac{1}{2}(a+b)+2))$$$$-\frac{\pi }{32}{\mathrm{\Psi }}^1(\frac{1}{2}(a+b)+2)\beta (\frac{a}{2}+1,\frac{b}{2}+1)$$$$\mathrm{\Omega }=\frac{\pi }{32}\beta (1,1){(\mathrm{\Psi }\left(1\right)-\mathrm{\Psi }\left(2\right))}^2-\frac{\pi }{32}{\mathrm{\Psi }}^1\left(2\right)\beta (1,1)$$$$=\frac{\pi }{32}-\frac{\pi }{32}{\mathrm{\Psi }}^{\prime }\left(2\right)$$$${\mathrm{\Psi }}^1\left(z\right)=\sum _{n⩾0}\frac{1}{{(n+z)}^2}$$$${\mathrm{\Psi }}^1\left(2\right)=\sum _{n⩾0}\frac{1}{{(n+2)}^2}=\sum _{n⩾1}\left(\frac{1}{n^2}\right)-1=\frac{{\pi }^2}{6}-1$$$$=\frac{\pi }{32}-\frac{\pi }{32}(\frac{{\pi }^2}{6}-1)$$$$=\frac{\mathit{\pi }}{16}-\frac{{\pi }^3}{192}$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 18/Nov/20

$$thankyousirpower...$$$$byusingtheeulerbetafunction$$$$.grateful.sincerelyyours.m.n$$

Commented by mindispower last updated on 18/Nov/20

$$withepleasurniceday$$

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