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Question Number 46842 by maxmathsup by imad last updated on 01/Nov/18
find ∫ (dx/(x(√(x−x^2 ))))
$${find}\:\:\int\:\:\:\:\:\frac{{dx}}{{x}\sqrt{{x}−{x}^{\mathrm{2}} }} \\ $$
Commented by maxmathsup by imad last updated on 01/Nov/18
we have x(√(x−x^2 ))=x(√(−x^2 +x))=x(√(−(x^2 −x)))=x(√(−(x^2 −2(x/2)+(1/4)−(1/4)))) =x(√((1/4)−(x−(1/2))^2 )) so changement x−(1/2)=(1/2)sint give ∫ (dx/(x(√(x−x^2 )))) = ∫ (1/(2((1/2)+(1/2)sint)(1/2)cost)) cost dt = ∫ ((2dt)/(1+sint)) =_(tan((t/2))=u) ∫ (2/(1+((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =4 ∫ (du/(1+u^2 +2u)) =4 ∫ (du/((u+1)^2 )) =−(4/(u+1)) +c =((−4)/(tan((t/2))+1)) +c but sin(t)=2x−1 ⇒t =arcsin(2x−1) ⇒ I = ((−4)/(tan(((arcsin(2x−1))/2))+1)) +c .
$${we}\:{have}\:{x}\sqrt{{x}−{x}^{\mathrm{2}} }={x}\sqrt{−{x}^{\mathrm{2}} \:+{x}}={x}\sqrt{−\left({x}^{\mathrm{2}} −{x}\right)}={x}\sqrt{−\left({x}^{\mathrm{2}} −\mathrm{2}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$={x}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:\:{so}\:{changement}\:\:{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{sint}\:{give} \\ $$$$\int\:\:\:\frac{{dx}}{{x}\sqrt{{x}−{x}^{\mathrm{2}} }}\:=\:\int\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{sint}\right)\frac{\mathrm{1}}{\mathrm{2}}{cost}}\:{cost}\:{dt} \\ $$$$=\:\int\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{sint}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\:\int\:\:\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int\:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{2}{u}}\:=\mathrm{4}\:\int\:\:\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{4}}{{u}+\mathrm{1}}\:+{c}\:=\frac{−\mathrm{4}}{{tan}\left(\frac{{t}}{\mathrm{2}}\right)+\mathrm{1}}\:+{c}\:{but} \\ $$$${sin}\left({t}\right)=\mathrm{2}{x}−\mathrm{1}\:\Rightarrow{t}\:={arcsin}\left(\mathrm{2}{x}−\mathrm{1}\right)\:\Rightarrow \\ $$$${I}\:=\:\frac{−\mathrm{4}}{{tan}\left(\frac{{arcsin}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{2}}\right)+\mathrm{1}}\:+{c}\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
t=(1/x) dx=((−1)/t^2 )dt ∫((−dt)/(t^2 ×(1/t)×(√((1/t)−(1/t^2 ))))) ∫((−dt)/(t×(√((t−1)/t^2 )))) ∫((−dt)/( (√(t−1)) )) =−1×(((t−1)^(1/2) )/(1/2))+c =−2((1/x)−1)^(1/2) +c
$${t}=\frac{\mathrm{1}}{{x}}\:\:\:{dx}=\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{−{dt}}{{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}}×\sqrt{\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}} \\ $$$$\int\frac{−{dt}}{{t}×\sqrt{\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} }}} \\ $$$$\int\frac{−{dt}}{\:\sqrt{{t}−\mathrm{1}}\:} \\ $$$$=−\mathrm{1}×\frac{\left({t}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}+{c} \\ $$$$=−\mathrm{2}\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{c} \\ $$

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