Question Number 33628 by naka3546 last updated on 20/Apr/18
Commented by MJS last updated on 20/Apr/18
$$\mathrm{this}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:=\mathrm{0} \\ $$
Commented by naka3546 last updated on 21/Apr/18
$${how}\:\:{to}\:{get}\:\:\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 21/Apr/18
![let put I = ∫_0 ^4 ((ln(x))/( (√(4x−x^2 )))) dx we have −x^2 +4x =−(x^2 −4x) =−(x^2 −4x +4 −4) =−(x−2)^2 +4 =4 −(x−2)^2 ch. x−2 =2 sint give I = ∫_(−(π/2)) ^(π/2) ((ln(2+2sint))/(2(√(1−sin^2 t)))) 2 cost dt = ∫_(−(π/2)) ^(π/2) ln(2+2sint)dt =π ln(2) + ∫_(−(π/2)) ^(π/2) ln(1+sint)dt let find the value of this integral let f(x)=∫_(−(π/2)) ^(π/2) ln(1+xsint)dt with ∣x∣≤1 and x≠0 f^′ (x) = ∫_(−(π/2)) ^(π/2) ((sint)/(1+xsint))dt =(1/x)∫_(−(π/2)) ^(π/2) ((1+xsint −1)/(1+xsint))dt =(π/x) −(1/x) ∫_(−(π/2)) ^(π/2) (dt/(1+x sint)) ch tan((t/2)) =u give ∫_(−(π/2)) ^(π/2) (dt/(1+x sint)) = ∫_(−1) ^1 (1/(1+x ((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) = 2 ∫_(−1) ^1 (du/(1+u^2 +2xu)) = 2∫_(−1) ^1 (du/(u^2 +2xu +1)) =2 ∫_(−1) ^1 (du/((u+x)^2 +1−x^2 )) (ch. u+x=(√(1−x^2 )) α) = 2 ∫_((x−1)/( (√(1−x^2 )))) ^((x+1)/( (√(1−x^2 )))) ((√(1−x^2 ))/((1−x^2 )( 1+α^2 )))dα = (2/( (√(1−x^2 )))) ∫_(−(1/( (√(1+x))))) ^(1/( (√(1−x)))) (dα/(1+α^2 )) = (2/( (√(1−x^2 ))))[arctan(α)]_(−(1/( (√(1+x))))) ^(1/( (√(1−x)))) = (2/( (√(1−x^2 )))) (arctan((1/( (√(1−x))))) +arctan((1/( (√(1+x))))))) = (2/( (√(1−x^2 ))))( π −arctan((√(1−x))) −arctan((√(1+x)))) = ((2π)/( (√(1−x^2 )))) −(2/( (√(1−x^2 ))))( arctan((√(1−x))) +arctan((√(1+x)))) ...be continued....](https://www.tinkutara.com/question/Q33638.png)
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{4}} \:\:\frac{{ln}\left({x}\right)}{\:\sqrt{\mathrm{4}{x}−{x}^{\mathrm{2}} }}\:{dx}\:\:{we}\:{have} \\ $$$$−{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:=−\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)\:=−\left({x}^{\mathrm{2}} \:−\mathrm{4}{x}\:+\mathrm{4}\:−\mathrm{4}\right) \\ $$$$=−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}\:=\mathrm{4}\:−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:\:{ch}.\:{x}−\mathrm{2}\:=\mathrm{2}\:{sint}\:{give} \\ $$$${I}\:=\:\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{ln}\left(\mathrm{2}+\mathrm{2}{sint}\right)}{\mathrm{2}\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}}\:\mathrm{2}\:{cost}\:{dt} \\ $$$$=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{2}+\mathrm{2}{sint}\right){dt} \\ $$$$=\pi\:{ln}\left(\mathrm{2}\right)\:+\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{1}+{sint}\right){dt}\:\:{let}\:{find}\:{the}\:{value} \\ $$$${of}\:{this}\:{integral}\:{let}\:\:{f}\left({x}\right)=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{1}+{xsint}\right){dt} \\ $$$${with}\:\mid{x}\mid\leqslant\mathrm{1}\:{and}\:{x}\neq\mathrm{0} \\ $$$${f}^{'} \left({x}\right)\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sint}}{\mathrm{1}+{xsint}}{dt} \\ $$$$=\frac{\mathrm{1}}{{x}}\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{xsint}\:−\mathrm{1}}{\mathrm{1}+{xsint}}{dt}\:=\frac{\pi}{{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{1}+{x}\:{sint}} \\ $$$${ch}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{sint}}\:=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{2}{xu}}\:=\:\mathrm{2}\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}{xu}\:+\mathrm{1}} \\ $$$$=\mathrm{2}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\:\frac{{du}}{\left({u}+{x}\right)^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:\left({ch}.\:{u}+{x}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:\alpha\right) \\ $$$$=\:\mathrm{2}\:\int_{\frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} \:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\:\mathrm{1}+\alpha^{\mathrm{2}} \right)}{d}\alpha \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:\:\int_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}} \:\:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left[{arctan}\left(\alpha\right)\right]_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\left({arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}\right)\:+{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\left.\mathrm{1}+{x}\right)}}\right)\right) \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\:\pi\:−{arctan}\left(\sqrt{\left.\mathrm{1}−{x}\right)}\:−{arctan}\left(\sqrt{\mathrm{1}+{x}}\right)\right)\right. \\ $$$$=\:\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\:{arctan}\left(\sqrt{\left.\mathrm{1}−{x}\right)}\:+{arctan}\left(\sqrt{\mathrm{1}+{x}}\right)\right)\right. \\ $$$$…{be}\:{continued}…. \\ $$
Commented by prof Abdo imad last updated on 21/Apr/18
$${let}\:{put}\:{I}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{sinx}\right){dx} \\ $$$${J}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{1}−{sinx}\right){dx}\:\:.{ch}.{x}=−{t}\:\Rightarrow \\ $$$${J}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{sint}\right){dt}\:={I}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right){dx}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{2}{ln}\left({cosx}\right){dx} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cosx}\right){dx}\:=\:\mathrm{4}\left(−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}\right) \\ $$$$=\:−\mathrm{2}\pi\:{ln}\left(\mathrm{2}\right)\:\Rightarrow\:{I}\:=−\pi\:{ln}\left(\mathrm{2}\right)\:{so} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \:\:\:\:\frac{{ln}\left({x}\right)}{\:\sqrt{−{x}^{\mathrm{2}} \:+\mathrm{4}{x}}}\:{dx}\:\:=\:\pi\:{ln}\left(\mathrm{2}\right)\:−\pi{ln}\left(\mathrm{2}\right)\:=\mathrm{0}\:. \\ $$