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Question Number 165380 by mnjuly1970 last updated on 31/Jan/22

$$$$$${\int }_0^{\frac{\pi }{2}}\frac{x^3}{{sin}^2\left(x\right)}dx\stackrel{?}{=}\frac{3}{8}({\pi }^{2}ln\left(4\right)-7\zeta \left(3\right))$$

Answered by Ar Brandon last updated on 31/Jan/22

$$I={\int }_0^{\frac{\pi }{2}}{x}^3{\mathrm{cosec}}^{2}xdx=-{\left[x^3\cot x\right]}_0^{\frac{\pi }{2}}+3{\int }_0^{\frac{\pi }{2}}{x}^2\cot xdx$$$$=3{\left[x^2\ln \left(\sin x\right)\right]}_0^{\frac{\pi }{2}}-6{\int }_0^{\frac{\pi }{2}}x\ln (\sin \left(x\right)dx$$

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