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Question Number 114094 by Lordose last updated on 17/Sep/20
∫ln(x)sin^(−1) (x)dx
$$\int\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}} \\ $$
Answered by Olaf last updated on 17/Sep/20
u′ = lnx, u = xlnx−x v = arcsinx, v′ = (1/( (√(1−x^2 )))) (xlnx−x)arcsinx−∫((xlnx−x)/( (√(1−x^2 ))))dx xln((x/e))arcsinx+∫ln((x/e))((−2x)/( 2(√(1−x^2 ))))dx u = ln((x/e)), u′ = (1/x) v′ = ((−2x)/(2(√(1−x^2 )))), v = (√(1−x^2 )) xln((x/e))arcsinx+ln((x/e))(√(1−x^2 ))−∫(1/x)(√(1−x^2 ))dx ln((x/e))(xarcsinx+(√(1−x^2 )))−∫(1/x)(√(1−x^2 ))dx x = sinu ∫(1/x)(√(1−x^2 ))dx = ∫(1/(sinu))(√(1−sin^2 u)).cosudu = ∫ε((cos^2 u)/(sinu))du = ε∫((1−sin^2 u)/(sinu))du (ε=±1) = ε∫(cscu−sinu)du = ε((1/2)ln∣((1−cosu)/(1+cosu))∣+cosu) = ε((1/2)ln∣((1−(√(1−x^2 )))/(1+(√(1−x^2 ))))∣+(√(1−x^2 )))...
$${u}'\:=\:\mathrm{ln}{x},\:{u}\:=\:{x}\mathrm{ln}{x}−{x} \\ $$$${v}\:=\:\mathrm{arcsin}{x},\:{v}'\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\left({x}\mathrm{ln}{x}−{x}\right)\mathrm{arcsin}{x}−\int\frac{{x}\mathrm{ln}{x}−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${x}\mathrm{ln}\left(\frac{{x}}{{e}}\right)\mathrm{arcsin}{x}+\int\mathrm{ln}\left(\frac{{x}}{{e}}\right)\frac{−\mathrm{2}{x}}{\:\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${u}\:=\:\mathrm{ln}\left(\frac{{x}}{{e}}\right),\:{u}'\:=\:\frac{\mathrm{1}}{{x}} \\ $$$${v}'\:=\:\frac{−\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }},\:{v}\:=\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${x}\mathrm{ln}\left(\frac{{x}}{{e}}\right)\mathrm{arcsin}{x}+\mathrm{ln}\left(\frac{{x}}{{e}}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\int\frac{\mathrm{1}}{{x}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$\mathrm{ln}\left(\frac{{x}}{{e}}\right)\left({x}\mathrm{arcsin}{x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\int\frac{\mathrm{1}}{{x}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$${x}\:=\:\mathrm{sinu} \\ $$$$\int\frac{\mathrm{1}}{{x}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=\: \\ $$$$\int\frac{\mathrm{1}}{\mathrm{sin}{u}}\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {u}}.\mathrm{cos}{udu}\:= \\ $$$$\int\varepsilon\frac{\mathrm{cos}^{\mathrm{2}} {u}}{\mathrm{sin}{u}}{du}\:=\:\varepsilon\int\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {u}}{\mathrm{sin}{u}}{du}\:\left(\varepsilon=\pm\mathrm{1}\right) \\ $$$$=\:\varepsilon\int\left(\mathrm{csc}{u}−\mathrm{sin}{u}\right){du}\:= \\ $$$$\varepsilon\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\mathrm{1}−\mathrm{cos}{u}}{\mathrm{1}+\mathrm{cos}{u}}\mid+\mathrm{cos}{u}\right)\:=\: \\ $$$$\varepsilon\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mid+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)…\: \\ $$$$ \\ $$
Commented by Lordose last updated on 17/Sep/20
Complete it sir
$$\boldsymbol{\mathrm{Complete}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{sir}} \\ $$
Answered by 1549442205PVT last updated on 17/Sep/20
Integrating by part we have I=∫ln(x)sin^(−1) (x)dx=∫(lnx−1)sin^(−1) (x)dx+∫sin^(−1) (x)dx =(xlnx−x)sin^(−1) (x)−∫x[(lnx−1)sin^(−1) (x)]′dx+∫sin^(−1) xdx =(xlnx−x)−∫((xlnx)/( (√(1−x^2 ))))dx+∫(x/( (√(1−x^2 ))))dx(1) Since ∫(x/( (√(1−x^2 ))))dx=(√(1−x^2 )) , We just need must find∫ ((xlnx)/( (√(1−x^2 ))))dx Put x=sinθ⇒dx=cosθdθ ∫((xlnx)/( (√(1−x^2 ))))dx=∫sinθlnsinθdθ=−cosθlnsinθ −∫(−cosθ)(lnsinθ)′dθ=−cosθlnsinθ+∫((cos^2 θ)/(sinθ))dθ =−cosθlnsinθ+∫((1−sin^2 θ)/(sinθ))dθ= =−cosθlnsinθ+∫(cosecθ−sinθ)dθ =−cosθlnsinθ+ln(cosecθ−cosθ)+cosθ =−(√(1−x^2 ))lnx+ln(((1−(√(1−x^2 )))/x))+(√(1−x^2 )) From (1)we get I=(xlnx−x)sin^(−1) x+(√(1−x^2 )) lnx−ln(((1−(√(1−x^2 )))/x))+C
$$\mathrm{Integrating}\:\mathrm{by}\:\mathrm{part}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{I}=\int\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{dx}}=\int\left(\mathrm{lnx}−\mathrm{1}\right)\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx}+\int\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\left(\mathrm{xlnx}−\mathrm{x}\right)\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)−\int\mathrm{x}\left[\left(\mathrm{lnx}−\mathrm{1}\right)\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\right]'\mathrm{dx}+\int\mathrm{sin}^{−\mathrm{1}} \mathrm{xdx} \\ $$$$=\left(\mathrm{xlnx}−\mathrm{x}\right)−\int\frac{\mathrm{xlnx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}+\int\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\left(\mathrm{1}\right) \\ $$$$\mathrm{Since}\:\int\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:, \\ $$$$\mathrm{We}\:\mathrm{just}\:\mathrm{need}\:\mathrm{must}\:\mathrm{find}\int\:\frac{\mathrm{xlnx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\mathrm{Put}\:\mathrm{x}=\mathrm{sin}\theta\Rightarrow\mathrm{dx}=\mathrm{cos}\theta\mathrm{d}\theta \\ $$$$\int\frac{\mathrm{xlnx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\int\mathrm{sin}\theta\mathrm{lnsin}\theta\mathrm{d}\theta=−\mathrm{cos}\theta\mathrm{lnsin}\theta \\ $$$$−\int\left(−\mathrm{cos}\theta\right)\left(\mathrm{lnsin}\theta\right)'\mathrm{d}\theta=−\mathrm{cos}\theta\mathrm{lnsin}\theta+\int\frac{\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{sin}\theta}\mathrm{d}\theta \\ $$$$=−\mathrm{cos}\theta\mathrm{lnsin}\theta+\int\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{sin}\theta}\mathrm{d}\theta= \\ $$$$=−\mathrm{cos}\theta\mathrm{lnsin}\theta+\int\left(\mathrm{cosec}\theta−\mathrm{sin}\theta\right)\mathrm{d}\theta \\ $$$$=−\mathrm{cos}\theta\mathrm{lnsin}\theta+\mathrm{ln}\left(\mathrm{cosec}\theta−\mathrm{cos}\theta\right)+\mathrm{cos}\theta \\ $$$$=−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{lnx}+\mathrm{ln}\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}}\right)+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\boldsymbol{\mathrm{I}}=\left(\boldsymbol{\mathrm{x}}\mathrm{l}\boldsymbol{\mathrm{nx}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \boldsymbol{\mathrm{x}}+\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\boldsymbol{\mathrm{lnx}}−\boldsymbol{\mathrm{ln}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}{\boldsymbol{\mathrm{x}}}\right)+\boldsymbol{\mathrm{C}} \\ $$
Answered by MJS_new last updated on 17/Sep/20
again just because something went wrong above... ∫ln x arcsin x dx= [by parts] =((√(1−x^2 ))+xarcsin x)ln x −∫((√(1−x^2 ))/x)+arcsin x dx ∫((√(1−x^2 ))/x)dx=(√(1−x^2 ))+ln (x/(1+(√(1−x^2 )))) ∫arcsin x dx=(√(1−x^2 ))+xarcsin x ⇒ ∫ln x arcsin x dx= =ln ((1+(√(1−x^2 )))/x) +((√(1−x^2 ))+xarcsin x)ln x −xarcsin x −2(√(1−x^2 )) +C
$$\mathrm{again}\:\mathrm{just}\:\mathrm{because}\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}\:\mathrm{above}… \\ $$$$\int\mathrm{ln}\:{x}\:\mathrm{arcsin}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}\mathrm{arcsin}\:{x}\right)\mathrm{ln}\:{x}\:−\int\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}+\mathrm{arcsin}\:{x}\:{dx} \\ $$$$\int\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}{dx}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\mathrm{ln}\:\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\int\mathrm{arcsin}\:{x}\:{dx}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}\mathrm{arcsin}\:{x} \\ $$$$\Rightarrow \\ $$$$\int\mathrm{ln}\:{x}\:\mathrm{arcsin}\:{x}\:{dx}= \\ $$$$=\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:+\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}\mathrm{arcsin}\:{x}\right)\mathrm{ln}\:{x}\:−{x}\mathrm{arcsin}\:{x}\:−\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{C} \\ $$

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