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Question Number 152275 by peter frank last updated on 27/Aug/21

$${\int }_0^{\frac{\pi }{2}}\sin 2\mathrm{xlog}\left(\tan \mathrm{x}\right)\mathrm{dx}$$

Answered by qaz last updated on 27/Aug/21

$${\int }_0^{\pi /2}\sin 2\mathrm{x}\cdot \mathrm{lntan}\mathrm{xdx}$$$$=2{\int }_0^{\pi /2}\sin \mathrm{xcos}\mathrm{x}(\mathrm{lnsin}\mathrm{x}-\mathrm{lncos}\mathrm{x})\mathrm{dx}$$$$={\{\int \mathrm{lnsin}\mathrm{xd}\left(\sin {}^2\mathrm{x}\right)+\int \mathrm{lncos}\mathrm{xd}\left(\cos {}^2\mathrm{x}\right)\}}_0^{\pi /2}$$$$=\sin {}^2\mathrm{xlnsin}\mathrm{x}+\cos {}^2\mathrm{xlncos}\mathrm{x}{\mid }_0^{\pi /2}$$$$=0$$

Answered by mnjuly1970 last updated on 27/Aug/21

$$I={\int }_0^{\frac{\pi }{2}}sin\left(2x\right)ln\left(tan\left(x\right)\right)dx...\left(1\right)$$$$I={\int }_0^{\frac{\pi }{2}}sin\left(2x\right).ln\left(cot\left(x\right)\right)dx...\left(2\right)$$$$\left(1\right)+\left(2\right):2I={\int }_0^{\frac{\pi }{2}}sin\left(2x\right)ln\left(1\right)=0$$$$I=0...$$

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