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Question Number 137277 by mnjuly1970 last updated on 31/Mar/21

$$$$$$\mathit{\varphi }={\int }_0^{\frac{\pi }{2}}{sin}^2\left(x\right).ln\left(sin\left(x\right)\right)dx=?$$$$$$

Answered by Dwaipayan Shikari last updated on 31/Mar/21

$${\int }_0^{\frac{\pi }{2}}{sin}^a\left(x\right)dx=\frac{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }(\frac{a}{2}+1)}$$$${\int }_0^{\frac{\pi }{2}}log\left(sinx\right){sin}^{a}xdx=\frac{\frac{\mathrm{\Gamma }(\frac{a}{2}+1){\mathrm{\Gamma }}^{\prime }\left(\frac{a+1}{2}\right)\sqrt{\pi }}{2}-\frac{\mathrm{\Gamma }\left(\frac{a+1}{2}\right){\mathrm{\Gamma }}^{\prime }(\frac{a}{2}+1)\sqrt{\pi }}{2}}{2{\mathrm{\Gamma }}^2(\frac{a}{2}+1)}$$$${\int }_0^{\frac{\pi }{2}}{sin}^{2}xlog\left(sinx\right)dx=\frac{\frac{\pi }{4}\psi \left(\frac{3}{2}\right)-\frac{\pi }{4}(1-\gamma )}{2}$$$$=\frac{\pi }{8}(-\gamma +2-2log\left(2\right)-1+\gamma )=\frac{\pi }{8}(1-log\left(4\right))$$

Commented by mnjuly1970 last updated on 31/Mar/21

$$verynice..$$$$thankyoumrDwaipayan..$$

Answered by Ar Brandon last updated on 31/Mar/21

$$\varphi ={\int }_0^{\frac{\pi }{2}}{\sin }^2\left(\mathrm{x}\right)\ln \left(\mathrm{sinx}\right)\mathrm{dx}$$$$=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =2}{\int }_0^{\frac{\pi }{2}}{\sin }^{\alpha }\left(\mathrm{x}\right)\mathrm{dx}=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =2}\frac{\mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }\left(\frac{\alpha +2}{2}\right)}$$$$=\frac{\sqrt{\pi }}{2}{\mid }_{\alpha =2}\frac{\mathrm{\Gamma }(\frac{\alpha }{2}+1){\mathrm{\Gamma }}^{\prime }(\frac{\alpha }{2}+\frac{1}{2})-\mathrm{\Gamma }(\frac{\alpha }{2}+\frac{1}{2}){\mathrm{\Gamma }}^{\prime }(\frac{\alpha }{2}+1)}{{\mathrm{\Gamma }}^2(\frac{\alpha }{2}+1)}$$$$=\frac{\sqrt{\pi }}{2}\cdot \frac{1}{2}\cdot \frac{\mathrm{\Gamma }\left(2\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)[\psi \left(\frac{3}{2}\right)-\psi \left(2\right)]}{{\mathrm{\Gamma }}^2\left(2\right)}$$$$=\frac{\pi }{8}(2-\gamma -2\ln 2-1+\gamma )=\frac{\pi }{8}(1-2\ln 2)$$

Answered by Ñï= last updated on 31/Mar/21

$${\int }_0^{\pi /2}\sin {}^{2}xln\sin xdx={\int }_0^{\pi /2}\cos {}^{2}xln\sin xdx+\frac{\pi }{4}.........\left(1\right)...byparts$$$${\int }_0^{\pi /2}\sin {}^{2}xln\sin xdx+{\int }_0^{\pi /2}\cos {}^{2}xln\sin xdx=-\frac{\pi }{2}ln2....\left(2\right)$$$$\Rightarrow {\int }_0^{\pi /2}\sin {}^{2}xln\sin xdx=\frac{1}{2}(\frac{\pi }{4}-\frac{\pi }{2}ln2)=\frac{\pi }{8}(1-2ln2)$$

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