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Question Number 122631 by MJS_new last updated on 18/Nov/20

solve ∫_((a−1)^2 ) ^a^2 cosh^(−1) (1/( (√(a−(√x))))) dx with a>0

$$\mathrm{solve}\:\underset{\left({a}−\mathrm{1}\right)^{\mathrm{2}} } {\overset{{a}^{\mathrm{2}} } {\int}}\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\sqrt{{x}}}}\:{dx}\:\mathrm{with}\:{a}>\mathrm{0} \\ $$

Commented byliberty last updated on 18/Nov/20

waw..nice question

$${waw}..{nice}\:{question} \\ $$

Commented byMJS_new last updated on 19/Nov/20

f(a)=∫_((a−1)^2 ) ^a^2 cosh^(−1) (1/( (√(a−(√x))))) dx is linear for a≥1 I found a nice path to solve it, will post it later

$${f}\left({a}\right)=\underset{\left({a}−\mathrm{1}\right)^{\mathrm{2}} } {\overset{{a}^{\mathrm{2}} } {\int}}\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\sqrt{{x}}}}\:{dx}\:\mathrm{is}\:\mathrm{linear}\:\mathrm{for}\:{a}\geqslant\mathrm{1} \\ $$ $$\mathrm{I}\:\mathrm{found}\:\mathrm{a}\:\mathrm{nice}\:\mathrm{path}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it},\:\mathrm{will}\:\mathrm{post}\:\mathrm{it}\:\mathrm{later} \\ $$

Answered by mathmax by abdo last updated on 18/Nov/20

I =∫_((a−1)^2 ) ^a^2 argch((1/( (√(a−(√x))))))dx chamgement (√(a−(√x)))=t give a+(√x)=t^2 ⇒(√x)=t^2 −a ⇒x=(t^2 −a)^2 ⇒ I = ∫_1 ^o argch((1/t))4t(t^2 −a)dt =−4 ∫_o ^1 t(t^2 −a)ln((1/t)+(√((1/t^2 )−1)))dt =−4 ∫_0 ^1 t(t^2 −a)ln(((1+(√(1−t^2 )))/t))dt =_(t=sinθ) −4 ∫_0 ^(π/2) sinθ(sin^2 θ−a)ln(((1+cosθ)/(sinθ)))cosθ dθ =−4 ∫_0 ^(π/2) sinθ(sin^2 θ−a)ln(((2cos^2 ((θ/2)))/(2cos((θ/2))sin((θ/2)))))cosθ dθ =4 ∫_0 ^(π/2) sinθ cosθ(sin^2 θ−a)ln(tan((θ/2)))dθ =4 ∫_0 ^(π/2) (sin^3 θcosθ −sinθ cosθ)ln(tan((θ/2)))dθ this integral can be solved by parts if we put u^′ =sin^3 cosθ−sinθ cosθ and v=ln(tan((θ/2))) ...be continued...

$$\mathrm{I}\:=\int_{\left(\mathrm{a}−\mathrm{1}\right)^{\mathrm{2}} } ^{\mathrm{a}^{\mathrm{2}} } \:\:\mathrm{argch}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}−\sqrt{\mathrm{x}}}}\right)\mathrm{dx}\:\:\mathrm{chamgement}\:\sqrt{\mathrm{a}−\sqrt{\mathrm{x}}}=\mathrm{t}\:\mathrm{give} \\ $$ $$\mathrm{a}+\sqrt{\mathrm{x}}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\sqrt{\mathrm{x}}=\mathrm{t}^{\mathrm{2}} −\mathrm{a}\:\Rightarrow\mathrm{x}=\left(\mathrm{t}^{\mathrm{2}} −\mathrm{a}\right)^{\mathrm{2}} \:\Rightarrow \\ $$ $$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{o}} \:\mathrm{argch}\left(\frac{\mathrm{1}}{\mathrm{t}}\right)\mathrm{4t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{a}\right)\mathrm{dt}\:=−\mathrm{4}\:\int_{\mathrm{o}} ^{\mathrm{1}} \mathrm{t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{a}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{t}}+\sqrt{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }−\mathrm{1}}\right)\mathrm{dt} \\ $$ $$=−\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{a}\right)\mathrm{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}}\right)\mathrm{dt} \\ $$ $$=_{\mathrm{t}=\mathrm{sin}\theta} \:\:\:−\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\theta\left(\mathrm{sin}^{\mathrm{2}} \theta−\mathrm{a}\right)\mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{sin}\theta}\right)\mathrm{cos}\theta\:\mathrm{d}\theta \\ $$ $$=−\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\theta\left(\mathrm{sin}^{\mathrm{2}} \theta−\mathrm{a}\right)\mathrm{ln}\left(\frac{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2cos}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\theta}{\mathrm{2}}\right)}\right)\mathrm{cos}\theta\:\mathrm{d}\theta \\ $$ $$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\theta\:\mathrm{cos}\theta\left(\mathrm{sin}^{\mathrm{2}} \theta−\mathrm{a}\right)\mathrm{ln}\left(\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)\right)\mathrm{d}\theta \\ $$ $$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}^{\mathrm{3}} \theta\mathrm{cos}\theta\:−\mathrm{sin}\theta\:\mathrm{cos}\theta\right)\mathrm{ln}\left(\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)\right)\mathrm{d}\theta\:\:\mathrm{this}\:\mathrm{integral}\:\mathrm{can}\:\mathrm{be} \\ $$ $$\mathrm{solved}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{if}\:\mathrm{we}\:\mathrm{put}\:\mathrm{u}^{'} \:=\mathrm{sin}^{\mathrm{3}} \mathrm{cos}\theta−\mathrm{sin}\theta\:\mathrm{cos}\theta\:\mathrm{and}\:\mathrm{v}=\mathrm{ln}\left(\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)\right) \\ $$ $$...\mathrm{be}\:\mathrm{continued}... \\ $$

Commented byMJS_new last updated on 19/Nov/20

thank you for trying

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{trying} \\ $$

Answered by MJS_new last updated on 19/Nov/20

∫cosh^(−1) (1/( (√(a−(√x))))) dx= by parts but with a simple trick: u^′ =1 → u=x−a^2 v=cosh^(−1) (1/( (√(a−(√x))))) → v′=−(1/(4(√x)((√x)−a)(√((√x)−a+1)))) =(x−a^2 )cosh^(−1) (1/( (√(a−(√x)))))+(1/4)∫((x−a^2 )/( (√x)((√x)−a)(√((√x)−a+1))))dx= (1/4)∫((x−a^2 )/( (√x)((√x)−a)(√((√x)−a+1))))dx= [t=(√((√x)−a+1)) → dx=4(√x)(√((√x)−a+1))dt] =∫(t^2 +2a−1)dt=(1/3)t(t^2 +3(2a−1))= =(1/3)((√x)+5a−2)(√((√x)−a+1)) =(x−a^2 )cosh^(−1) (1/( (√(a−(√x))))) +(1/3)((√x)+5a−2)(√((√x)−a+1))+C inserting the borders we get f(a)=(2a−1)cosh^(−1) (1/( (√(a−∣a−1∣)))) −(((5a+∣a−1∣−2)(√(1−a+∣a−1∣)))/3)+(((5a+∣a∣−2)(√(1−a+∣a∣)))/3) for 0<a<1 we get f(a)=(2a−1)cosh^(−1) (1/( (√(2a−1)))) −(((4a−1)(√(2(1−a))))/3)+((2(3a−1))/3) for a≥1 we get f(a)=((2(3a−1))/3)

$$\int\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\sqrt{{x}}}}\:{dx}= \\ $$ $$\:\:\:\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{but}\:\mathrm{with}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{trick}: \\ $$ $$\:\:\:\:\:{u}^{'} =\mathrm{1}\:\rightarrow\:{u}={x}−{a}^{\mathrm{2}} \\ $$ $$\:\:\:\:\:{v}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\sqrt{{x}}}}\:\rightarrow\:{v}'=−\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}\left(\sqrt{{x}}−{a}\right)\sqrt{\sqrt{{x}}−{a}+\mathrm{1}}} \\ $$ $$=\left({x}−{a}^{\mathrm{2}} \right)\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\sqrt{{x}}}}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}−{a}^{\mathrm{2}} }{\:\sqrt{{x}}\left(\sqrt{{x}}−{a}\right)\sqrt{\sqrt{{x}}−{a}+\mathrm{1}}}{dx}= \\ $$ $$ \\ $$ $$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}−{a}^{\mathrm{2}} }{\:\sqrt{{x}}\left(\sqrt{{x}}−{a}\right)\sqrt{\sqrt{{x}}−{a}+\mathrm{1}}}{dx}= \\ $$ $$\:\:\:\:\:\:\:\:\:\:\left[{t}=\sqrt{\sqrt{{x}}−{a}+\mathrm{1}}\:\rightarrow\:{dx}=\mathrm{4}\sqrt{{x}}\sqrt{\sqrt{{x}}−{a}+\mathrm{1}}{dt}\right] \\ $$ $$\:\:\:\:\:=\int\left({t}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{1}\right){dt}=\frac{\mathrm{1}}{\mathrm{3}}{t}\left({t}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}{a}−\mathrm{1}\right)\right)= \\ $$ $$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt{{x}}+\mathrm{5}{a}−\mathrm{2}\right)\sqrt{\sqrt{{x}}−{a}+\mathrm{1}} \\ $$ $$ \\ $$ $$=\left({x}−{a}^{\mathrm{2}} \right)\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\sqrt{{x}}}}\:+\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt{{x}}+\mathrm{5}{a}−\mathrm{2}\right)\sqrt{\sqrt{{x}}−{a}+\mathrm{1}}+{C} \\ $$ $$ \\ $$ $$\mathrm{inserting}\:\mathrm{the}\:\mathrm{borders}\:\mathrm{we}\:\mathrm{get} \\ $$ $${f}\left({a}\right)=\left(\mathrm{2}{a}−\mathrm{1}\right)\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{a}−\mid{a}−\mathrm{1}\mid}}\:−\frac{\left(\mathrm{5}{a}+\mid{a}−\mathrm{1}\mid−\mathrm{2}\right)\sqrt{\mathrm{1}−{a}+\mid{a}−\mathrm{1}\mid}}{\mathrm{3}}+\frac{\left(\mathrm{5}{a}+\mid{a}\mid−\mathrm{2}\right)\sqrt{\mathrm{1}−{a}+\mid{a}\mid}}{\mathrm{3}} \\ $$ $$\mathrm{for}\:\mathrm{0}<{a}<\mathrm{1}\:\mathrm{we}\:\mathrm{get} \\ $$ $${f}\left({a}\right)=\left(\mathrm{2}{a}−\mathrm{1}\right)\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}−\mathrm{1}}}\:−\frac{\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{2}\left(\mathrm{1}−{a}\right)}}{\mathrm{3}}+\frac{\mathrm{2}\left(\mathrm{3}{a}−\mathrm{1}\right)}{\mathrm{3}} \\ $$ $$\mathrm{for}\:{a}\geqslant\mathrm{1}\:\mathrm{we}\:\mathrm{get} \\ $$ $${f}\left({a}\right)=\frac{\mathrm{2}\left(\mathrm{3}{a}−\mathrm{1}\right)}{\mathrm{3}} \\ $$

Commented byliberty last updated on 19/Nov/20

by parts ∫ v du = vu−∫u dv ?

$${by}\:{parts}\:\int\:{v}\:{du}\:=\:{vu}−\int{u}\:{dv}\:?\: \\ $$

Commented byMJS_new last updated on 19/Nov/20

∫u′v=uv−∫uv′

$$\int{u}'{v}={uv}−\int{uv}' \\ $$

Commented byliberty last updated on 19/Nov/20

haha..it′s same sir

$${haha}..{it}'{s}\:{same}\:{sir} \\ $$

Commented byMJS_new last updated on 19/Nov/20

yes

$$\mathrm{yes} \\ $$

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