Question Number 36943 by maxmathsup by imad last updated on 07/Jun/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{lnx}}{\:\sqrt{{x}}\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx} \\ $$
Commented by math khazana by abdo last updated on 10/Jun/18
![changement x=sin^2 t give I = ∫_0 ^(π/2) ((2ln(sint))/(sint cos^3 t)) 2sint cost dt = ∫_0 ^(π/2) ((4ln(sint))/(cos^2 t))dt by parts I = 4 { [tant ln(sint)]_0 ^(π/2) −∫_0 ^(π/2) tant ((cost)/(sint)) dt} =−4 ∫_0 ^(π/2) dt = −4.(π/2) =−2π now let prove?that lim_(t→((π )/2)) tan(t)ln(sint)=0 t =(π/2) −x ⇒ tan(t)ln(sint) =(1/(tan(x)))ln(cosx) ∼ (1/x)ln(1 −(x^2 /2))∼(1/x)(−(x^2 /2))=−(x/2) →0(x→0) let prove that lim_(t→0) tan(t)ln(sint) =0 tant ln(sint) ∼t ln(t)→0(t→0) so ★ I =−2π★](https://www.tinkutara.com/question/Q37172.png)
$${changement}\:{x}={sin}^{\mathrm{2}} {t}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{\mathrm{2}{ln}\left({sint}\right)}{{sint}\:{cos}^{\mathrm{3}} {t}}\:\mathrm{2}{sint}\:{cost}\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{4}{ln}\left({sint}\right)}{{cos}^{\mathrm{2}} {t}}{dt}\:\:{by}\:{parts} \\ $$$${I}\:=\:\mathrm{4}\:\left\{\:\:\left[{tant}\:{ln}\left({sint}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{tant}\:\frac{{cost}}{{sint}}\:{dt}\right\} \\ $$$$=−\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{dt}\:=\:−\mathrm{4}.\frac{\pi}{\mathrm{2}}\:=−\mathrm{2}\pi\:{now}\:{let}\:{prove}?{that} \\ $$$${lim}_{{t}\rightarrow\frac{\pi\:\:}{\mathrm{2}}} \:\:\:{tan}\left({t}\right){ln}\left({sint}\right)=\mathrm{0} \\ $$$${t}\:=\frac{\pi}{\mathrm{2}}\:−{x}\:\Rightarrow\:{tan}\left({t}\right){ln}\left({sint}\right)\:=\frac{\mathrm{1}}{{tan}\left({x}\right)}{ln}\left({cosx}\right) \\ $$$$\sim\:\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sim\frac{\mathrm{1}}{{x}}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)=−\frac{{x}}{\mathrm{2}}\:\rightarrow\mathrm{0}\left({x}\rightarrow\mathrm{0}\right) \\ $$$${let}\:{prove}\:{that}\:{lim}_{{t}\rightarrow\mathrm{0}} \:{tan}\left({t}\right){ln}\left({sint}\right)\:=\mathrm{0} \\ $$$${tant}\:{ln}\left({sint}\right)\:\sim{t}\:{ln}\left({t}\right)\rightarrow\mathrm{0}\left({t}\rightarrow\mathrm{0}\right)\:\:{so} \\ $$$$\bigstar\:{I}\:=−\mathrm{2}\pi\bigstar \\ $$$$ \\ $$