1-x-1-x-dx-

Question Number 154641 by cesarL last updated on 20/Sep/21

$$\int\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}{dx} \\ $$

Answered by puissant last updated on 20/Sep/21

u=(√(1−x))→ u^2 =1−x → x=1−u^2 ⇒ dx=−2udu I=∫((1−(√(1−u^2 )))/u)(−2udu)=2∫((√(1−u^2 ))−1)du =2∫(√(1−u^2 ))du−2u u=sint→du=costdt ⇒ I=2∫cos^2 tdt−2u = ∫(1+cos2t)dt−2u =t+(1/2)sin2t−2u+C ⇒ I=arcsinu+(1/2)sin(2arcsinu)−2(√(1−x))+C ∴∵ I=arcsin(√(1−x))+(√(x−x^2 ))−2(√(1−x))+C.

$${u}=\sqrt{\mathrm{1}−{x}}\rightarrow\:{u}^{\mathrm{2}} =\mathrm{1}−{x}\:\rightarrow\:{x}=\mathrm{1}−{u}^{\mathrm{2}} \\ $$$$\Rightarrow\:{dx}=−\mathrm{2}{udu}\: \\ $$$${I}=\int\frac{\mathrm{1}−\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{{u}}\left(−\mathrm{2}{udu}\right)=\mathrm{2}\int\left(\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }−\mathrm{1}\right){du} \\ $$$$=\mathrm{2}\int\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }{du}−\mathrm{2}{u} \\ $$$${u}={sint}\rightarrow{du}={costdt} \\ $$$$\Rightarrow\:{I}=\mathrm{2}\int{cos}^{\mathrm{2}} {tdt}−\mathrm{2}{u}\:=\:\int\left(\mathrm{1}+{cos}\mathrm{2}{t}\right){dt}−\mathrm{2}{u} \\ $$$$={t}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{t}−\mathrm{2}{u}+{C} \\ $$$$\Rightarrow\:{I}={arcsinu}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{arcsinu}\right)−\mathrm{2}\sqrt{\mathrm{1}−{x}}+{C} \\ $$$$\therefore\because\:{I}={arcsin}\sqrt{\mathrm{1}−{x}}+\sqrt{{x}−{x}^{\mathrm{2}} }−\mathrm{2}\sqrt{\mathrm{1}−{x}}+{C}. \\ $$

Answered by MJS_new last updated on 20/Sep/21

∫((1−(√x))/( (√(1−x))))dx= [t=((√(1−x))/(1−(√x))) → dx=2(√x)(1−(√x))(√(1−x))dt] =2∫((t^2 −1)/((t^2 +1)^3 ))dt= [Ostrogradski′s Method] =−((2t(t^2 +3))/((t^2 +1)^2 ))−2∫(dt/(t^2 +1))= =−((2t(t^2 +3))/((t^2 +1)^2 ))−2arctan t = ...t =((√x)−2)(√(1−x))+2arctan ((1−(√x))/( (√(1−x)))) +C

$$\int\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}\left(\mathrm{1}−\sqrt{{x}}\right)\sqrt{\mathrm{1}−{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=−\frac{\mathrm{2}{t}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=−\frac{\mathrm{2}{t}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2arctan}\:{t}\:= \\ $$$$…{t} \\ $$$$=\left(\sqrt{{x}}−\mathrm{2}\right)\sqrt{\mathrm{1}−{x}}+\mathrm{2arctan}\:\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:+{C} \\ $$

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