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

∫_0 ^(π/2) sin 2xlog( tan x)dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:\mathrm{2xlog}\left(\:\mathrm{tan}\:\mathrm{x}\right)\mathrm{dx} \\ $$

Answered by qaz last updated on 27/Aug/21

∫_0 ^(π/2) sin 2x∙lntan xdx  =2∫_0 ^(π/2) sin xcos x(lnsin x−lncos x)dx  ={∫lnsin xd(sin^2 x)+∫lncos xd(cos^2 x)}_0 ^(π/2)   =sin^2 xlnsin x+cos^2 xlncos x ∣_0 ^(π/2)   =0

$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:\mathrm{2x}\centerdot\mathrm{lntan}\:\mathrm{xdx} \\ $$ $$=\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}\:\mathrm{xcos}\:\mathrm{x}\left(\mathrm{lnsin}\:\mathrm{x}−\mathrm{lncos}\:\mathrm{x}\right)\mathrm{dx} \\ $$ $$=\left\{\int\mathrm{lnsin}\:\mathrm{xd}\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)+\int\mathrm{lncos}\:\mathrm{xd}\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)\right\}_{\mathrm{0}} ^{\pi/\mathrm{2}} \\ $$ $$=\mathrm{sin}\:^{\mathrm{2}} \mathrm{xlnsin}\:\mathrm{x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{xlncos}\:\mathrm{x}\:\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} \\ $$ $$=\mathrm{0} \\ $$

Answered by mnjuly1970 last updated on 27/Aug/21

   I=∫_0 ^( (π/2))  sin(2x)ln(tan(x))dx ...(1)     I= ∫_0 ^( (π/2)) sin(2x).ln(cot(x))dx ...(2)   (1)+(2):  2I= ∫_0 ^( (π/2)) sin(2x)ln(1)=0      I=0...

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

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