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Integral-I-0-pi-ln-sin-x-tan-1-cot-x-dx-0-proof-I-0-pi-ln-sin-x-tan-1-tan-pi-2-x-dx-0-pi-pi-2-x-

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Question Number 152201 by mnjuly1970 last updated on 26/Aug/21
...Integral... I := ∫_0 ^( π) ln (sin(x) ).tan^( −1) (cot(x))dx=^? 0 proof :: .... I := ∫_0 ^( π) ln (sin(x) ). tan^( −1) ( tan((π/2) −x ))dx := ∫_0 ^( π) ((π/2) −x ).ln(sin(x))dx := (π/2) ∫_0 ^( π) ln(sin(x))dx−∫_0 ^( π) xln(sin(x))dx := (π/2) (−π ln (2 )) −J ......( 1 ) J : = ∫_0 ^( π) (π − x) ln (sin(x))dx := π (−π ln(2))−J ∴ J :=((−π^( 2) )/2) ln( 2 ) .......(2) (2) ⇛ (1 ) : I = 0 .........■
$$ \\ $$$$\:\:\:…\mathrm{Integral}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\pi} {ln}\:\left({sin}\left({x}\right)\:\right).{tan}^{\:−\mathrm{1}} \left({cot}\left({x}\right)\right){dx}\overset{?} {=}\:\mathrm{0} \\ $$$$\:\:\:\:\:{proof}\:::\:…. \\ $$$$\:\:\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\pi} {ln}\:\left({sin}\left({x}\right)\:\right).\:{tan}^{\:−\mathrm{1}} \left(\:{tan}\left(\frac{\pi}{\mathrm{2}}\:−{x}\:\right)\right){dx} \\ $$$$\:\:\:\:\:\:\::=\:\int_{\mathrm{0}} ^{\:\pi} \left(\frac{\pi}{\mathrm{2}}\:−{x}\:\right).{ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\::=\:\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\pi} {ln}\left({sin}\left({x}\right)\right){dx}−\int_{\mathrm{0}} ^{\:\pi} {xln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\::=\:\frac{\pi}{\mathrm{2}}\:\left(−\pi\:{ln}\:\left(\mathrm{2}\:\right)\right)\:−\mathrm{J}\:\:\:……\left(\:\mathrm{1}\:\right)\:\: \\ $$$$\:\:\:\:\:\:\mathrm{J}\::\:=\:\int_{\mathrm{0}} ^{\:\pi} \left(\pi\:−\:{x}\right)\:{ln}\:\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\::=\:\pi\:\left(−\pi\:{ln}\left(\mathrm{2}\right)\right)−\mathrm{J} \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\mathrm{J}\::=\frac{−\pi^{\:\mathrm{2}} }{\mathrm{2}}\:{ln}\left(\:\mathrm{2}\:\right)\:…….\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\left(\mathrm{2}\right)\:\Rrightarrow\:\left(\mathrm{1}\:\right)\::\:\:\:\:\:\mathrm{I}\:=\:\mathrm{0}\:………\blacksquare \\ $$$$ \\ $$

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