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Question Number 46740 by math khazana by abdo last updated on 30/Oct/18
find ∫ x(√((1−(√x))/(1+(√x))))dx
$${find}\:\int\:\:{x}\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}{dx} \\ $$
Commented by behi83417@gmail.com last updated on 31/Oct/18
=∫x.(√(((1−(√x))(1−(√x)))/((1+(√x))(1−(√x)))))dx= =∫x.((1−(√x))/( (√(1−x))))dx=∫[(x/( (√(1−x))))−((x(√x))/( (√(1−x))))]dx= [x=cos^2 t⇒dx=−2cost.sintdt] =∫[((cos^2 t)/(sint))−((cos^3 t)/(sint))](−2cost.sint)dt= =2∫(cos^4 t−cos^3 t)dt= =2∫[(((1+cos2t)/2))^2 −((1/4)cos3t+(3/4)cost)]dt= =(1/4)∫[3+4cos2t+cos4t−2cos3t+6cost]dt= =(3/4)t+(1/2)sin2t+(1/(16))sin4t−(1/6)sin3t+(3/2)sint+C= =(3/4)cos^(−1) (√x)+(√x).(√(1−x))+(1/4)(√x).(√(1−x))(2x−1) −(1/6)[3(√(1−x))−4(1−x)(√(1−x))]+(3/2)(√(1−x))+C. [sin4t=2sin2tcos2t=4sintcost.(2cos^2 t−1)= =4(√x).(√(1−x)).(2x−1) sin3t=3sint−4sin^3 t=3(√(1−x))−4(1−x)(√(1−x))
$$=\int{x}.\sqrt{\frac{\left(\mathrm{1}−\sqrt{{x}}\right)\left(\mathrm{1}−\sqrt{{x}}\right)}{\left(\mathrm{1}+\sqrt{{x}}\right)\left(\mathrm{1}−\sqrt{{x}}\right)}}{dx}= \\ $$$$=\int{x}.\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}{dx}=\int\left[\frac{{x}}{\:\sqrt{\mathrm{1}−{x}}}−\frac{{x}\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}\right]{dx}= \\ $$$$\left[{x}={cos}^{\mathrm{2}} {t}\Rightarrow{dx}=−\mathrm{2}{cost}.{sintdt}\right] \\ $$$$=\int\left[\frac{{cos}^{\mathrm{2}} {t}}{{sint}}−\frac{{cos}^{\mathrm{3}} {t}}{{sint}}\right]\left(−\mathrm{2}{cost}.{sint}\right){dt}= \\ $$$$=\mathrm{2}\int\left({cos}^{\mathrm{4}} {t}−{cos}^{\mathrm{3}} {t}\right){dt}= \\ $$$$=\mathrm{2}\int\left[\left(\frac{\mathrm{1}+{cos}\mathrm{2}{t}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}{cos}\mathrm{3}{t}+\frac{\mathrm{3}}{\mathrm{4}}{cost}\right)\right]{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left[\mathrm{3}+\mathrm{4}{cos}\mathrm{2}{t}+{cos}\mathrm{4}{t}−\mathrm{2}{cos}\mathrm{3}{t}+\mathrm{6}{cost}\right]{dt}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{t}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{t}+\frac{\mathrm{1}}{\mathrm{16}}{sin}\mathrm{4}{t}−\frac{\mathrm{1}}{\mathrm{6}}{sin}\mathrm{3}{t}+\frac{\mathrm{3}}{\mathrm{2}}{sint}+{C}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{cos}^{−\mathrm{1}} \sqrt{{x}}+\sqrt{{x}}.\sqrt{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{x}}.\sqrt{\mathrm{1}−\mathrm{x}}\left(\mathrm{2x}−\mathrm{1}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}\left[\mathrm{3}\sqrt{\mathrm{1}−\mathrm{x}}−\mathrm{4}\left(\mathrm{1}−\mathrm{x}\right)\sqrt{\mathrm{1}−\mathrm{x}}\right]+\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{x}}+\mathrm{C}. \\ $$$$\left[{sin}\mathrm{4}{t}=\mathrm{2sin2tcos2t}=\mathrm{4sintcost}.\left(\mathrm{2cos}^{\mathrm{2}} \mathrm{t}−\mathrm{1}\right)=\right. \\ $$$$=\mathrm{4}\sqrt{\mathrm{x}}.\sqrt{\mathrm{1}−\mathrm{x}}.\left(\mathrm{2x}−\mathrm{1}\right) \\ $$$$\mathrm{sin3t}=\mathrm{3sint}−\mathrm{4sin}^{\mathrm{3}} \mathrm{t}=\mathrm{3}\sqrt{\mathrm{1}−\mathrm{x}}−\mathrm{4}\left(\mathrm{1}−\mathrm{x}\right)\sqrt{\mathrm{1}−\mathrm{x}} \\ $$
Commented by maxmathsup by imad last updated on 31/Oct/18
we have x(√((1−(√x))/(1+(√x))))=x(√(((1−(√x))^2 )/(1−x)))=x((1−(√x))/( (√(1−x)))) =(x/( (√(1−x)))) −((x(√x))/( (√(1−x)))) ⇒ I = ∫ ((xdx)/( (√(1−x)))) −∫ ((x(√x))/( (√(1−x)))) changement x=sin^2 θ give ∫ (x/( (√(1−x))))dx = ∫ ((sin^2 θ)/(cosθ)) 2sinθ cosθ = 2 ∫ sin^3 θ dθ but sin^3 θ =(((e^(iθ ) −e^(−iθ) )/(2i)))^3 =−(1/(8i)) Σ_(k=0) ^3 C_3 ^k e^(ikθ) e^(−(3−k)iθ) =(i/8)Σ_(k=0) ^3 C_3 ^k (−1)^(3−k) e^(i(2k−3)θ) =(1/8){− e^(−i3θ) +3 e^(−iθ) −3 e^(iθ) + e^(i3θ) } =(i/8){2isin(3θ) −6i sin(θ)} =−(1/4)sin(3θ) +(3/4)sin(θ) ⇒ ∫ (x/( (√(1−x))))dx =−(1/2) ∫ sin(3θ)dθ +(3/2) ∫ sinθ dθ +c_1 =(1/6) cos(3θ)−(3/2)cosθ +c_1 also the same changeent x=sin^2 θ give ∫ ((x(√x))/( (√(1−x)))) dx = ∫ ((sin^2 θ sinθ)/(cosθ)) 2sinθ cosθ dθ = 2 ∫ sin^4 θ dθ =2 ∫ (((1−cos(2θ))/2))^2 dθ =(1/2) ∫ (1−2cos(2θ) +((1+cos(4θ))/2))dθ =(θ/2) −(1/2)sin(2θ) +(θ/4) +(1/8)sin(4θ) +c_2 =((3θ)/4) −((sin(2θ))/2) +((sin(4θ))/8) +c_2 ⇒ I =(1/6)cos(3θ)−(3/2) cosθ +((3θ)/4) −((sin(2θ))/2) +((sin(4θ))/8) +C =(1/6) cos(3 arcsin((√x)))−(3/2) cos(arcsin((√x)))+(3/4) arcsin((√x))−(1/2)sin(2arcsin(√x)) +(1/8) sin(4 arcsin(√x)) +C .
$${we}\:{have}\:{x}\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}={x}\sqrt{\frac{\left(\mathrm{1}−\sqrt{{x}}\right)^{\mathrm{2}} }{\mathrm{1}−{x}}}={x}\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:=\frac{{x}}{\:\sqrt{\mathrm{1}−{x}}}\:−\frac{{x}\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:\Rightarrow \\ $$$${I}\:=\:\int\:\:\frac{{xdx}}{\:\sqrt{\mathrm{1}−{x}}}\:−\int\:\:\frac{{x}\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:\:{changement}\:{x}={sin}^{\mathrm{2}} \theta\:{give} \\ $$$$\int\:\:\frac{{x}}{\:\sqrt{\mathrm{1}−{x}}}{dx}\:=\:\int\:\:\frac{{sin}^{\mathrm{2}} \theta}{{cos}\theta}\:\mathrm{2}{sin}\theta\:{cos}\theta\:=\:\mathrm{2}\:\int\:{sin}^{\mathrm{3}} \theta\:{d}\theta\:{but} \\ $$$${sin}^{\mathrm{3}} \theta\:=\left(\frac{{e}^{{i}\theta\:} \:−{e}^{−{i}\theta} }{\mathrm{2}{i}}\right)^{\mathrm{3}} \:=−\frac{\mathrm{1}}{\mathrm{8}{i}}\:\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \:{C}_{\mathrm{3}} ^{{k}} \:{e}^{{ik}\theta} \:{e}^{−\left(\mathrm{3}−{k}\right){i}\theta} \\ $$$$=\frac{{i}}{\mathrm{8}}\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \:\:\:{C}_{\mathrm{3}} ^{{k}} \:\:\left(−\mathrm{1}\right)^{\mathrm{3}−{k}} {e}^{{i}\left(\mathrm{2}{k}−\mathrm{3}\right)\theta} \:=\frac{\mathrm{1}}{\mathrm{8}}\left\{−\:{e}^{−{i}\mathrm{3}\theta} \:\:+\mathrm{3}\:{e}^{−{i}\theta} \:−\mathrm{3}\:{e}^{{i}\theta} \:+\:{e}^{{i}\mathrm{3}\theta} \right\} \\ $$$$=\frac{{i}}{\mathrm{8}}\left\{\mathrm{2}{isin}\left(\mathrm{3}\theta\right)\:−\mathrm{6}{i}\:{sin}\left(\theta\right)\right\}\:=−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}\theta\right)\:+\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\theta\right)\:\Rightarrow \\ $$$$\int\:\:\frac{{x}}{\:\sqrt{\mathrm{1}−{x}}}{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:{sin}\left(\mathrm{3}\theta\right){d}\theta\:+\frac{\mathrm{3}}{\mathrm{2}}\:\int\:{sin}\theta\:{d}\theta\:+{c}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\:{cos}\left(\mathrm{3}\theta\right)−\frac{\mathrm{3}}{\mathrm{2}}{cos}\theta\:+{c}_{\mathrm{1}} \:\:\:\:{also}\:{the}\:{same}\:{changeent}\:{x}={sin}^{\mathrm{2}} \theta\:{give} \\ $$$$\int\:\:\frac{{x}\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:{dx}\:=\:\int\:\:\frac{{sin}^{\mathrm{2}} \theta\:{sin}\theta}{{cos}\theta}\:\mathrm{2}{sin}\theta\:{cos}\theta\:{d}\theta\:=\:\mathrm{2}\:\int\:\:{sin}^{\mathrm{4}} \theta\:{d}\theta \\ $$$$=\mathrm{2}\:\int\:\:\left(\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right)^{\mathrm{2}} {d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\left(\mathrm{1}−\mathrm{2}{cos}\left(\mathrm{2}\theta\right)\:+\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}\right){d}\theta \\ $$$$=\frac{\theta}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\theta\right)\:+\frac{\theta}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{4}\theta\right)\:+{c}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}\theta}{\mathrm{4}}\:−\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:+\frac{{sin}\left(\mathrm{4}\theta\right)}{\mathrm{8}}\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{6}}{cos}\left(\mathrm{3}\theta\right)−\frac{\mathrm{3}}{\mathrm{2}}\:{cos}\theta\:\:+\frac{\mathrm{3}\theta}{\mathrm{4}}\:−\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:+\frac{{sin}\left(\mathrm{4}\theta\right)}{\mathrm{8}}\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\:{cos}\left(\mathrm{3}\:{arcsin}\left(\sqrt{{x}}\right)\right)−\frac{\mathrm{3}}{\mathrm{2}}\:{cos}\left({arcsin}\left(\sqrt{{x}}\right)\right)+\frac{\mathrm{3}}{\mathrm{4}}\:{arcsin}\left(\sqrt{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{arcsin}\sqrt{{x}}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{8}}\:{sin}\left(\mathrm{4}\:{arcsin}\sqrt{{x}}\right)\:+{C}\:. \\ $$
Answered by MJS last updated on 31/Oct/18
∫x(√((1−(√x))/(1+(√x))))dx= [t=(√(1−(√x))) → dx=−4(√(x(1−(√x))))dt] =4∫((t^2 (1−t^2 )^3 )/( (√(2−t^2 ))))dt= [t=(√2)sin u → u=arcsin ((t(√2))/2); dt=du(√2)cos u] =4∫(((√8)cos u sin^2 u (2sin^2 u −1)^3 )/( (√(2−2sin^2 u))))du= =−8∫sin^2 u (1−2sin^2 u)^3 du= =−8∫(−8sin^8 u +12sin^6 u −6sin^4 u +sin^2 u)du= =(1/2)∫cos 8u du−∫cos 6u du+2∫cos 4u du−3∫cos 2u du+(3/2)∫du= =(1/(16))sin 8u −(1/6)sin 6u +(1/2)sin 4u −(3/2)sin 2u +(3/2)u= =...hard but possible, leads to...= =(3/2)arcsin (√((1−(√x))/2)) +((1/2)x^(3/2) −(2/3)x+(3/4)x^(1/2) −(4/3))(√(1−x))+C
$$\int{x}\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}−\sqrt{{x}}}\:\rightarrow\:{dx}=−\mathrm{4}\sqrt{{x}\left(\mathrm{1}−\sqrt{{x}}\right)}{dt}\right] \\ $$$$=\mathrm{4}\int\frac{{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{\:\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}{dt}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{2}}\mathrm{sin}\:{u}\:\rightarrow\:{u}=\mathrm{arcsin}\:\frac{{t}\sqrt{\mathrm{2}}}{\mathrm{2}};\:{dt}={du}\sqrt{\mathrm{2}}\mathrm{cos}\:{u}\right] \\ $$$$=\mathrm{4}\int\frac{\sqrt{\mathrm{8}}\mathrm{cos}\:{u}\:\mathrm{sin}^{\mathrm{2}} \:{u}\:\left(\mathrm{2sin}^{\mathrm{2}} \:{u}\:−\mathrm{1}\right)^{\mathrm{3}} }{\:\sqrt{\mathrm{2}−\mathrm{2sin}^{\mathrm{2}} \:{u}}}{du}= \\ $$$$=−\mathrm{8}\int\mathrm{sin}^{\mathrm{2}} \:{u}\:\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{u}\right)^{\mathrm{3}} {du}= \\ $$$$=−\mathrm{8}\int\left(−\mathrm{8sin}^{\mathrm{8}} \:{u}\:+\mathrm{12sin}^{\mathrm{6}} \:{u}\:−\mathrm{6sin}^{\mathrm{4}} \:{u}\:+\mathrm{sin}^{\mathrm{2}} \:{u}\right){du}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{cos}\:\mathrm{8}{u}\:{du}−\int\mathrm{cos}\:\mathrm{6}{u}\:{du}+\mathrm{2}\int\mathrm{cos}\:\mathrm{4}{u}\:{du}−\mathrm{3}\int\mathrm{cos}\:\mathrm{2}{u}\:{du}+\frac{\mathrm{3}}{\mathrm{2}}\int{du}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{8}{u}\:−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{sin}\:\mathrm{6}{u}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{4}{u}\:−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{u}\:+\frac{\mathrm{3}}{\mathrm{2}}{u}= \\ $$$$=…\mathrm{hard}\:\mathrm{but}\:\mathrm{possible},\:\mathrm{leads}\:\mathrm{to}…= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{arcsin}\:\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{2}}}\:+\left(\frac{\mathrm{1}}{\mathrm{2}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{2}}{\mathrm{3}}{x}+\frac{\mathrm{3}}{\mathrm{4}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{\mathrm{4}}{\mathrm{3}}\right)\sqrt{\mathrm{1}−{x}}+{C} \\ $$
Commented by maxmathsup by imad last updated on 31/Oct/18
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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