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Question-39222

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Question Number 39222 by behi83417@gmail.com last updated on 04/Jul/18
Commented by math khazana by abdo last updated on 04/Jul/18
1) f is defined on [0,1[ f^′ (x)=((((−1)/(2(√(1−x))))(1−(√x)) −(√(1−x))(((−1)/(2(√x)))))/((1−(√x))^2 )) =((−2(√x)(1−(√x))+2(1−x))/(4(√x)(√(1−x))(1−(√x))^2 )) =((−2(√x) +2x +2−2x)/(4(√x)(√(1−x))(1−(√x))^2 )) = (2/(4(√x)(√(1−x))(1−(√x)))) >0 ⇒ f is increasing on[0,1[ f(o)= 1 and lim _(x→1^− ) f(x) =lim_(x→1^− ) ((√(1−x))/(1−(√x))) =lim_(x→1^− ) (((−1)/(2(√(1−x))))/(−(1/(2(√x))))) = lim_(x→1^− ) ((2(√x))/(2(√(1−x)))) =lim_(x→1^− ) ((√x)/( (√(1−x)))) = +∞ ⇒ f([0,1[)=[1,+∞[ . 2)f(x)=y ⇔ x=f^(−1) (y) ⇒ ((√(1−x))/(1−(√x)))=y ⇒ (√( ((1−x)/((1−(√x))^2 )))) =y ⇒ ((1−x)/(1+x−2(√x))) =y^2 let (√x)=t ⇒ ((1−t^2 )/(1+t^2 −2t))=y^2 ⇒ 1−t^2 =y^2 +y^2 t^2 −2y^2 t ⇒ (1+y^2 )t^2 −2y^2 t +y^2 −1=0 Δ^′ =y^4 −(y^2 +1)(y^2 −1)=y^4 −(y^4 −1)=1 t_1^ =((y^(2 ) +1)/(1+y^2 )) =1 and t_2 = ((y^2 −1)/(y^2 +1)) ⇒x =1 or x=(((y^2 −1)/(y^2 +1)))^2 f^(−1) is not constant ⇒ f^(−1) (x) = {((x^2 −1)/(x^2 +1))}^2 .
$$\left.\mathrm{1}\right)\:\:{f}\:{is}\:{defined}\:{on}\:\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$$${f}^{'} \left({x}\right)=\frac{\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}}}\left(\mathrm{1}−\sqrt{{x}}\right)\:−\sqrt{\mathrm{1}−{x}}\left(\frac{−\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)}{\left(\mathrm{1}−\sqrt{{x}}\right)^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{2}\sqrt{{x}}\left(\mathrm{1}−\sqrt{{x}}\right)+\mathrm{2}\left(\mathrm{1}−{x}\right)}{\mathrm{4}\sqrt{{x}}\sqrt{\mathrm{1}−{x}}\left(\mathrm{1}−\sqrt{{x}}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{2}\sqrt{{x}}\:+\mathrm{2}{x}\:+\mathrm{2}−\mathrm{2}{x}}{\mathrm{4}\sqrt{{x}}\sqrt{\mathrm{1}−{x}}\left(\mathrm{1}−\sqrt{{x}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{4}\sqrt{{x}}\sqrt{\mathrm{1}−{x}}\left(\mathrm{1}−\sqrt{{x}}\right)}\:>\mathrm{0}\:\Rightarrow\:{f}\:{is}\:{increasing}\:{on}\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$$${f}\left({o}\right)=\:\mathrm{1}\:{and}\: \\ $$$${lim}\:_{{x}\rightarrow\mathrm{1}^{−} } {f}\left({x}\right)\:={lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\:\:\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\:\:\:\:\frac{\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{x}}}}{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}\:=\:{lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\:\:\frac{\mathrm{2}\sqrt{{x}}}{\mathrm{2}\sqrt{\mathrm{1}−{x}}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\:\:\:\:\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:=\:+\infty\:\Rightarrow \\ $$$${f}\left(\left[\mathrm{0},\mathrm{1}\left[\right)=\left[\mathrm{1},+\infty\left[\:.\right.\right.\right.\right. \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)={y}\:\Leftrightarrow\:{x}={f}^{−\mathrm{1}} \left({y}\right)\:\Rightarrow\:\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}}={y}\:\Rightarrow \\ $$$$\sqrt{\:\frac{\mathrm{1}−{x}}{\left(\mathrm{1}−\sqrt{{x}}\right)^{\mathrm{2}} }}\:\:={y}\:\Rightarrow\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}−\mathrm{2}\sqrt{{x}}}\:={y}^{\mathrm{2}} \\ $$$${let}\:\:\sqrt{{x}}={t}\:\Rightarrow\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} \:−\mathrm{2}{t}}={y}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{1}−{t}^{\mathrm{2}} \:={y}^{\mathrm{2}} \:+{y}^{\mathrm{2}} {t}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} {t}\:\Rightarrow \\ $$$$\left(\mathrm{1}+{y}^{\mathrm{2}} \right){t}^{\mathrm{2}} \:−\mathrm{2}{y}^{\mathrm{2}} {t}\:+{y}^{\mathrm{2}} \:−\mathrm{1}=\mathrm{0} \\ $$$$\Delta^{'} \:={y}^{\mathrm{4}} −\left({y}^{\mathrm{2}} \:+\mathrm{1}\right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)={y}^{\mathrm{4}} \:−\left({y}^{\mathrm{4}} −\mathrm{1}\right)=\mathrm{1} \\ $$$${t}_{\mathrm{1}^{} } \:=\frac{{y}^{\mathrm{2}\:} \:+\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }\:=\mathrm{1}\:{and}\:{t}_{\mathrm{2}} =\:\frac{{y}^{\mathrm{2}} −\mathrm{1}}{{y}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow{x}\:=\mathrm{1}\:\:{or}\:{x}=\left(\frac{{y}^{\mathrm{2}} −\mathrm{1}}{{y}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\mathrm{2}} \:\:{f}^{−\mathrm{1}} {is}\:{not}\:{constant}\:\Rightarrow \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:\left\{\frac{{x}^{\mathrm{2}} \:−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\right\}^{\mathrm{2}} . \\ $$
Commented by maxmathsup by imad last updated on 04/Jul/18
3)let I = ∫_0 ^((√2)/2) ((√(1−x))/(1−(√x)))dx changement (√x)=t give I = ∫_0 ^(1/((^4 (√2)))) ((√(1−t^2 ))/(1−t)) 2t dt I =−2 ∫_0 ^2^(−(1/4)) ((1−t−1)/(1−t))(√(1−t^2 )) dt =−2 ∫_0 ^2^(−(1/4)) (√(1−t^2 )) + 2 ∫_0 ^2^(−(1/4)) ((√(1−t^2 ))/(1−t)) dt ∫_0 ^2^(−(1/4)) (√(1−t^2 ))dt =_(t=sinα) ∫_0 ^(arcsin(2^(−(1/4)) )) cosα cosα dα = (1/2) ∫_0 ^(arcsin(2^(−(1/4)) )) (1+cos(2α))dα=(1/2)arcsin(2^(−(1/4)) ) +(1/4) sin(2arcsin(2^(−(1/4)) )) ∫_0 ^(2^(−(1/4)) ) ((√(1−t^2 ))/(1−t)) =_(t=sinα) ∫_0 ^(arcsin(2^(−(1/4)) )) ((cosα)/(1−sinα)) cosα dα = ∫_0 ^(arcsin(2^(−(1/4)) )) ((1−sin^2 α)/(1−sinα)) dα =∫_0 ^(arcsin(2^(−(1/4)) )) (1+sinα)dα =arcsin(2^(−(1/4)) ) +[−cosα]_0 ^(arsin(2^(−(1/4)) )) =arcsin(2^(−(1/4)) ) +1−cos(arcsin(2^(−(1/4)) )) so the value of I is known.
$$\left.\mathrm{3}\right){let}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\:\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}}{dx}\:\:{changement}\:\sqrt{{x}}={t}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}} \:\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{1}−{t}}\:\mathrm{2}{t}\:{dt} \\ $$$${I}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} } \:\frac{\mathrm{1}−{t}−\mathrm{1}}{\mathrm{1}−{t}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} } \:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} \:}\:\:+\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} } \:\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{1}−{t}}\:{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} } \:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=_{{t}={sin}\alpha} \:\:\:\int_{\mathrm{0}} ^{{arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)} \:{cos}\alpha\:{cos}\alpha\:{d}\alpha \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)} \:\left(\mathrm{1}+{cos}\left(\mathrm{2}\alpha\right)\right){d}\alpha=\frac{\mathrm{1}}{\mathrm{2}}{arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:{sin}\left(\mathrm{2}{arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:} \:\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{1}−{t}}\:\:=_{{t}={sin}\alpha} \:\:\int_{\mathrm{0}} ^{{arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)} \:\:\:\frac{{cos}\alpha}{\mathrm{1}−{sin}\alpha}\:{cos}\alpha\:{d}\alpha \\ $$$$=\:\int_{\mathrm{0}} ^{{arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)} \:\frac{\mathrm{1}−{sin}^{\mathrm{2}} \alpha}{\mathrm{1}−{sin}\alpha}\:{d}\alpha\:=\int_{\mathrm{0}} ^{{arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)} \left(\mathrm{1}+{sin}\alpha\right){d}\alpha \\ $$$$={arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)\:\:+\left[−{cos}\alpha\right]_{\mathrm{0}} ^{{arsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)} \\ $$$$={arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)\:\:+\mathrm{1}−{cos}\left({arcsin}\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \right)\right)\:{so}\:{the}\:{value}\:{of}\:{I}\:{is}\:{known}. \\ $$
Commented by behi83417@gmail.com last updated on 04/Jul/18
dear pro.abdo!thank you for hard work god bless you sir.what do you think about #4 ?^
$${dear}\:{pro}.{abdo}!{thank}\:{you}\:{for}\:{hard}\:{work} \\ $$$${god}\:{bless}\:{you}\:{sir}.{what}\:{do}\:{you}\:{think}\: \\ $$$${about}\:#\mathrm{4}\:\overset{} {?} \\ $$
Commented by math khazana by abdo last updated on 04/Jul/18
for x=y we get f^2 (4x) =((1−f(2x))/((1−f(x))^2 )) x=(1/8) ⇒f^2 ((1/2))=(((1/( (√2)))/(1−(1/( (√2))))))^2 =((1/( (√2)−1)))^2 =(1/(3−2(√2))) ((1−f(2x))/((1−f(x))^2 )) = ((1−f((1/4)))/((1−f((1/8)))^2 )) butf((1/4))=(((√3)/2)/(1/2)) =(√3) f((1/8))=(((√7)/( (√8)))/(1−(1/( (√8))))) = ((√7)/(2(√2) −1)) ⇒1−f((1/8)) =1−((√7)/(2(√2)−1)) = ((2(√2)−1−(√7))/(2(√2))) ⇒(1−f((1/8)))^2 =(((2(√2)−1−(√7))/(2(√2))))^2 and its clear that f^2 (4x)=((1−f(2x))/((1−f(x))^2 )) is not true withx=(1/8) so tbe equality is not correct.
$${for}\:{x}={y}\:{we}\:{get}\:\:{f}^{\mathrm{2}} \left(\mathrm{4}{x}\right)\:=\frac{\mathrm{1}−{f}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}−{f}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow{f}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}−{f}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}−{f}\left({x}\right)\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}−{f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{1}−{f}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\right)^{\mathrm{2}} }\:\:{butf}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}\:=\sqrt{\mathrm{3}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\frac{\sqrt{\mathrm{7}}}{\:\sqrt{\mathrm{8}}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}}}}\:=\:\frac{\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}\:−\mathrm{1}}\:\Rightarrow\mathrm{1}−{f}\left(\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$=\mathrm{1}−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow\left(\mathrm{1}−{f}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:{and}\:{its}\:{clear}\:{that} \\ $$$${f}^{\mathrm{2}} \left(\mathrm{4}{x}\right)=\frac{\mathrm{1}−{f}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}−{f}\left({x}\right)\right)^{\mathrm{2}} }\:{is}\:{not}\:{true}\:{withx}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${so}\:{tbe}\:{equality}\:{is}\:{not}\:{correct}. \\ $$
Commented by math khazana by abdo last updated on 04/Jul/18
nevermind sir Behi.
$${nevermind}\:{sir}\:{Behi}. \\ $$
Answered by MJS last updated on 04/Jul/18
y=((√(1−x))/(1−(√x))) ⇒ x∈[0; 1[ ∧ y∈[1; +∞[ f^(−1) (x)=(((x^2 −1)/(x^2 +1)))^2 with x∈[1; +∞[ ∧ y∈[0; 1[ ∫((√(1−x))/(1−(√x)))dx=∫(((1+(√x))(√(1−x)))/((1+(√x))(1−(√x))))dx= =∫(((√(1−x))+(√(x−x^2 )))/(1−x))dx=∫(dx/( (√(1−x))))+∫(√(x/(1−x)))dx= [t=(√x) → dx=2(√x)dt] 2∫(t/( (√(1−t^2 ))))dt+2∫(t^2 /( (√(1−t^2 ))))dt= 2∫(t/( (√(1−t^2 ))))dt=−2(√(1−t^2 ))=−2(√(1−x)) 2∫(t^2 /( (√(1−t^2 ))))dt= [u=arcsin t → dt=(√(1−t^2 ))du] =2∫sin^2 u du=u−sin u cos u = =arcsin t −t(√(1−t^2 ))=arcsin (√(x ))−(√x)(√(1−x)) =arcsin (√(x ))−(2+(√x))(√(1−x))+C ∫_0 ^((√2)/2) ((√(1−x))/(1−(√x)))dx≈1.46146 but ∫_0 ^1 ((√(1−x))/(1−(√x)))dx=2+(π/2) (4) is not true for the given function
$${y}=\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}}\:\Rightarrow\:{x}\in\left[\mathrm{0};\:\mathrm{1}\left[\:\wedge\:{y}\in\left[\mathrm{1};\:+\infty\left[\right.\right.\right.\right. \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} \mathrm{with}\:{x}\in\left[\mathrm{1};\:+\infty\left[\:\wedge\:{y}\in\left[\mathrm{0};\:\mathrm{1}\left[\right.\right.\right.\right. \\ $$$$\int\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}}{dx}=\int\frac{\left(\mathrm{1}+\sqrt{{x}}\right)\sqrt{\mathrm{1}−{x}}}{\left(\mathrm{1}+\sqrt{{x}}\right)\left(\mathrm{1}−\sqrt{{x}}\right)}{dx}= \\ $$$$=\int\frac{\sqrt{\mathrm{1}−{x}}+\sqrt{{x}−{x}^{\mathrm{2}} }}{\mathrm{1}−{x}}{dx}=\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}}}+\int\sqrt{\frac{{x}}{\mathrm{1}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$\mathrm{2}\int\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}+\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}= \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{2}\int\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=−\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=−\mathrm{2}\sqrt{\mathrm{1}−{x}} \\ $$$$\:\:\:\:\:\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\mathrm{arcsin}\:{t}\:\rightarrow\:{dt}=\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{du}\right] \\ $$$$\:\:\:\:\:=\mathrm{2}\int\mathrm{sin}^{\mathrm{2}} \:{u}\:{du}={u}−\mathrm{sin}\:{u}\:\mathrm{cos}\:{u}\:= \\ $$$$\:\:\:\:\:=\mathrm{arcsin}\:{t}\:−{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=\mathrm{arcsin}\:\sqrt{{x}\:}−\sqrt{{x}}\sqrt{\mathrm{1}−{x}} \\ $$$$ \\ $$$$=\mathrm{arcsin}\:\sqrt{{x}\:}−\left(\mathrm{2}+\sqrt{{x}}\right)\sqrt{\mathrm{1}−{x}}+{C} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\int}}\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}}{dx}\approx\mathrm{1}.\mathrm{46146} \\ $$$$\mathrm{but}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\sqrt{\mathrm{1}−{x}}}{\mathrm{1}−\sqrt{{x}}}{dx}=\mathrm{2}+\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{true}\:\mathrm{for}\:\mathrm{the}\:\mathrm{given}\:\mathrm{function} \\ $$
Commented by behi83417@gmail.com last updated on 04/Jul/18
thank you so much dear MJS! god bless you sir. #4,for what?please!
$${thank}\:{you}\:{so}\:{much}\:{dear}\:{MJS}! \\ $$$${god}\:{bless}\:{you}\:{sir}. \\ $$$$#\mathrm{4},{for}\:{what}?{please}! \\ $$

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