1-x-1-x-dx-

Question Number 122274 by benjo_mathlover last updated on 15/Nov/20

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

Answered by liberty last updated on 15/Nov/20

let (√x) = cos θ ⇒ x=cos^2 θ ⇒ dx =−2sin θ cos θ dθ I =∫ ((√(1−cos θ))/( (√(1+cos θ)))) (−2sin θ cos θ )dθ I = ∫ ((−2(1−cos θ)(sin θ cos θ)dθ)/(sin θ)) I= −2∫(cos θ−cos^2 θ)dθ I=−2∫ (cos θ−((1/2)+(1/2)cos 2θ))dθ I= −2sin θ+θ+sin θ cos θ + c I=−2(√(1−x)) + arc cos ((√x))+(√(x−x^2 )) + c .▲

$$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\mathrm{cos}\:\theta\:\Rightarrow\:\mathrm{x}=\mathrm{cos}\:^{\mathrm{2}} \theta\: \\ $$$$\Rightarrow\:\mathrm{dx}\:=−\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\:\mathrm{d}\theta\: \\ $$$$\mathrm{I}\:=\int\:\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:\theta}}{\:\sqrt{\mathrm{1}+\mathrm{cos}\:\theta}}\:\left(−\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\:\right)\mathrm{d}\theta \\ $$$$\mathrm{I}\:=\:\int\:\frac{−\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left(\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right)\mathrm{d}\theta}{\mathrm{sin}\:\theta} \\ $$$$\mathrm{I}=\:−\mathrm{2}\int\left(\mathrm{cos}\:\theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\mathrm{d}\theta \\ $$$$\mathrm{I}=−\mathrm{2}\int\:\left(\mathrm{cos}\:\theta−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}\theta\right)\right)\mathrm{d}\theta \\ $$$$\mathrm{I}=\:−\mathrm{2sin}\:\theta+\theta+\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:+\:\mathrm{c} \\ $$$$\mathrm{I}=−\mathrm{2}\sqrt{\mathrm{1}−\mathrm{x}}\:+\:\mathrm{arc}\:\mathrm{cos}\:\left(\sqrt{\mathrm{x}}\right)+\sqrt{\mathrm{x}−\mathrm{x}^{\mathrm{2}} }\:+\:\mathrm{c}\:.\blacktriangle \\ $$

Commented by benjo_mathlover last updated on 15/Nov/20

$${thank}\:{you} \\ $$

Answered by MJS_new last updated on 15/Nov/20

∫((√(1−(√x)))/( (√(1+(√x)))))dx= [t=((√(1−(√x)))/( (√(1+(√x))))) → dx=−2(√(1−(√x)))(√((1+(√x))^3 ))(√x)]] =8∫((t^2 (t^2 −1))/((t^2 +1)^3 ))dt= [Ostrogradski′s Method] =−((2t(3t^2 +1))/((t^2 +1)^2 ))+2∫(dt/(t^2 +1))= =−((2t(3t^2 +1))/((t^2 +1)^2 ))+2arctan t = =−(2−(√x))(√(1−(√x)))(√(1+(√x)))+2arctan ((√(1−(√x)))/( (√(1+(√x))))) +C

$$\int\frac{\sqrt{\mathrm{1}−\sqrt{{x}}}}{\:\sqrt{\mathrm{1}+\sqrt{{x}}}}{dx}= \\ $$$$\left.\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{1}−\sqrt{{x}}}}{\:\sqrt{\mathrm{1}+\sqrt{{x}}}}\:\rightarrow\:{dx}=−\mathrm{2}\sqrt{\mathrm{1}−\sqrt{{x}}}\sqrt{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{3}} }\sqrt{{x}}\right]\right] \\ $$$$=\mathrm{8}\int\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=−\frac{\mathrm{2}{t}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=−\frac{\mathrm{2}{t}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{2arctan}\:{t}\:= \\ $$$$=−\left(\mathrm{2}−\sqrt{{x}}\right)\sqrt{\mathrm{1}−\sqrt{{x}}}\sqrt{\mathrm{1}+\sqrt{{x}}}+\mathrm{2arctan}\:\frac{\sqrt{\mathrm{1}−\sqrt{{x}}}}{\:\sqrt{\mathrm{1}+\sqrt{{x}}}}\:+{C} \\ $$

Commented by MJS_new last updated on 15/Nov/20

$$=−\left(\mathrm{2}−\sqrt{{x}}\right)\sqrt{\mathrm{1}−{x}}+\mathrm{arccos}\:\sqrt{{x}}\:+{C} \\ $$

Commented by benjo_mathlover last updated on 15/Nov/20

$${thank}\:{you}\:{sir} \\ $$

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