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Question Number 122697 by john santu last updated on 19/Nov/20

∫ (dx/( (√((x−a)(b−x))))) ?

$$\:\:\int\:\frac{{dx}}{\:\sqrt{\left({x}−{a}\right)\left({b}−{x}\right)}}\:? \\ $$

Answered by liberty last updated on 19/Nov/20

Solve ϕ(x)=∫ (dx/( (√((x−a)(b−x))))) ? Solution : letting x = acos^2 q + b sin^2 q where a<x<b ,0<q<(π/2) we get { ((x−a=(b−a)sin^2 q)),((b−x=(b−a)cos^2 q )) :} and tan^2 q = ((x−a)/(b−x)) or tan q = (√((x−a)/(b−x))) dx = 2(b−a)sin qcos q dq implying that ϕ(x) = ∫ (dx/( (√((x−a)(b−x))))) = ∫ 2dq ϕ(x) = 2q + c = 2arctan ((√((x−a)/(b−x)))) + c. ✓

$$\:\:{Solve}\:\varphi\left({x}\right)=\int\:\frac{{dx}}{\:\sqrt{\left({x}−{a}\right)\left({b}−{x}\right)}}\:? \\ $$$$\:\:\:{Solution}\::\: \\ $$$$\:{letting}\:{x}\:=\:{a}\mathrm{cos}\:^{\mathrm{2}} {q}\:+\:{b}\:\mathrm{sin}\:^{\mathrm{2}} {q}\: \\ $$$${where}\:{a}<{x}<{b}\:,\mathrm{0}<{q}<\frac{\pi}{\mathrm{2}} \\ $$$${we}\:{get}\:\begin{cases}{{x}−{a}=\left({b}−{a}\right)\mathrm{sin}\:^{\mathrm{2}} {q}}\\{{b}−{x}=\left({b}−{a}\right)\mathrm{cos}\:^{\mathrm{2}} {q}\:}\end{cases} \\ $$$$\:{and}\:\mathrm{tan}\:^{\mathrm{2}} {q}\:=\:\frac{{x}−{a}}{{b}−{x}}\:{or}\:\mathrm{tan}\:{q}\:=\:\sqrt{\frac{{x}−{a}}{{b}−{x}}} \\ $$$${dx}\:=\:\mathrm{2}\left({b}−{a}\right)\mathrm{sin}\:{q}\mathrm{cos}\:{q}\:{dq} \\ $$$${implying}\:{that}\: \\ $$$$\varphi\left({x}\right)\:=\:\int\:\frac{{dx}}{\:\sqrt{\left({x}−{a}\right)\left({b}−{x}\right)}}\:=\:\int\:\mathrm{2}{dq} \\ $$$$\:\varphi\left({x}\right)\:=\:\mathrm{2}{q}\:+\:{c}\:=\:\mathrm{2arctan}\:\left(\sqrt{\frac{{x}−{a}}{{b}−{x}}}\right)\:+\:{c}.\:\checkmark \\ $$

Commented by john santu last updated on 19/Nov/20

nice

$${nice} \\ $$$$ \\ $$

Answered by som(math1967) last updated on 19/Nov/20

let x−a=z^2 dx=2zdz ∫((2zdz)/(z(√(b−a−z^2 )))) 2∫(dz/( (√(b−a−z^2 )))) 2sin^(−1) (z/( (√(b−a)))) +c 2sin^(−1) ((√((x−a)/(b−a))))+c

$$\mathrm{let}\:\mathrm{x}−\mathrm{a}=\mathrm{z}^{\mathrm{2}} \\ $$$$\mathrm{dx}=\mathrm{2zdz} \\ $$$$\int\frac{\mathrm{2zdz}}{\mathrm{z}\sqrt{\mathrm{b}−\mathrm{a}−\mathrm{z}^{\mathrm{2}} }} \\ $$$$\mathrm{2}\int\frac{\mathrm{dz}}{\:\sqrt{\mathrm{b}−\mathrm{a}−\mathrm{z}^{\mathrm{2}} }} \\ $$$$\mathrm{2sin}^{−\mathrm{1}} \frac{\mathrm{z}}{\:\sqrt{\mathrm{b}−\mathrm{a}}}\:+\mathrm{c} \\ $$$$\mathrm{2sin}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{x}−\mathrm{a}}{\mathrm{b}−\mathrm{a}}}\right)+\mathrm{c} \\ $$

Answered by MJS_new last updated on 19/Nov/20

∫(dx/( (√((x−a)(b−x)))))= [t=((√(x−a))/( (√(b−x)))) → dx=(2/(b−a))(√(x−a))(√((b−x)^3 ))dt] =2∫(dt/(t^2 +1))=2arctan t =2arctan ((√(x−a))/( (√(b−x)))) +C

$$\int\frac{{dx}}{\:\sqrt{\left({x}−{a}\right)\left({b}−{x}\right)}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{{x}−{a}}}{\:\sqrt{{b}−{x}}}\:\rightarrow\:{dx}=\frac{\mathrm{2}}{{b}−{a}}\sqrt{{x}−{a}}\sqrt{\left({b}−{x}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2arctan}\:{t}\:=\mathrm{2arctan}\:\frac{\sqrt{{x}−{a}}}{\:\sqrt{{b}−{x}}}\:+{C} \\ $$

Commented by liberty last updated on 19/Nov/20

amazing...your method very fastest sir

$${amazing}...{your}\:{method}\:{very}\:{fastest} \\ $$$${sir}\: \\ $$

Commented by MJS_new last updated on 19/Nov/20

I′m just an old integral lover...

$$\mathrm{I}'\mathrm{m}\:\mathrm{just}\:\mathrm{an}\:\mathrm{old}\:\mathrm{integral}\:\mathrm{lover}... \\ $$

Commented by liberty last updated on 19/Nov/20

hahahaha....great

$${hahahaha}....{great}\: \\ $$

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