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Question Number 157096 by amin96 last updated on 19/Oct/21

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

Answered by mindispower last updated on 19/Oct/21

$${(-xcos\left(x\right)+sin\left(x\right))ln\left(sin\left(x\right)\right)]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}\frac{{xcos}^2\left(x\right)}{sin\left(x\right)}-cosxdx$$$$={\int }_0^{\frac{\pi }{2}}\frac{x(1-{sin}^2\left(x\right))}{sin\left(x\right)}dx-1$$$$={\int }_0^{\frac{\pi }{2}}\frac{x}{sin\left(x\right)}dx-{\int }_0^{\frac{\pi }{2}}xsin\left(x\right)dx-1$$$$={\int }_0^{\frac{\pi }{2}}\frac{x}{sin\left(x\right)}dx+{[xcos\left(x\right)-sin\left(x\right)]}_0^{\frac{\pi }{2}}-1$$$$={\int }_0^{\frac{\pi }{2}}\frac{x}{sin\left(x\right)}dx-2,tg\left(\frac{x}{2}\right)=u$$$$={\int }_0^1\frac{2arctan\left(u\right)}{u}du=2{\int }_0^1\sum _{k⩾0}\frac{{(-1)}^k}{2k+1}{u}^{2k}-2$$$$=2\sum _{k⩾0}\frac{{(-1)}^k}{{(2k+1)}^2}-2=2(G-1)$$$$$$

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