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Question Number 84386 by M±th+et£s last updated on 12/Mar/20
∫(√(x−(√(4−x^2 )))) dx
$$\int\sqrt{{x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:{dx} \\ $$
Answered by mind is power last updated on 13/Mar/20
x=2sin(t)⇒ =∫2cos(t)(√(2sin(t)−2cos(t)))dt let f(t)=∫cos(t)(√(sin(t)−cos(t)))dt by part=sin(t)(√(sin(t)−cos(t)))dt−∫((sin(t)(sin(t)+cos(t)))/(2(√(sin(t)−cos(t)))))dt =sin(t)(√(sin(t)−cos(t)))−(1/2).∫((sin^2 (t)+sin(t)cos(t))/( (√(sin(t)−cos(t)))))dt sin^2 (t)=1−cos^2 (t)⇒ =sin(t)(√(sin(t)−cos(t)))−(1/2)∫(dt/( (√(sin(t)−cos(t))) ))−(1/2)∫((cos(t)(sin(t)−cos(t))/( (√(sin(t)−cos(t)))))dt ⇒(3/2)f(t)=sin(t)(√(sin(t)−cos(t)))−(1/2).∫(dt/( (√(sin(t)−cos(t))))) ,let g(t)=∫(dt/( (√(sin(t)−cos(t)))))=(1/( (√(√2))))∫(dt/( (√(−cos(t+(π/4)))))) sin(t)−cos(t)=−(√2)cos(t+(π/4)) let u=(√(−cos(t+(π/4))))⇒t=arcos(−u^2 )−(π/4) g(t)⇔(1/2^(1/4) )∫((2udu)/( (√(1−u^4 )).u))=2^(3/4) ∫(du/( (√(1−u^4 )))) (1/( (√(1−x))))=1+Σ_(n≥1) (((2n−1)!!)/(2^n .n!)).x^n (1/( (√(1−u^4 ))))=1+Σ_(n≥1) (((2n−1)!!)/(2^n .n!)).x^(4n) ⇒∫(du/( (√(1−u^4 ))))=∫(1+Σ_(n≥1) (((2n−1)!!)/(2^n .n!))u^(4n) )du =u+Σ_(n≥1) (((2n−1)!!u^(4n+1) )/(2^n .n!.(4n+1)))=u(1+Σ_(n≥1) (((2n−1)!!)/(2^n .(4n+1))).(((u^4 )^n )/(n!)))..E (((2n−1)!!)/2^n )=Π_(k=0) ^(n−1) ((1/2)+k) (1/(4n+1))=((Π_(k=0) ^(n−1) ((1/4)+k))/(Π_(k=0) ^(n−1) ((5/4)+k−1))) E⇔u(1+Σ_(n≥1) ((Π_(k=0) ^(n−1) ((1/2)+k).Π_(k=0) ^(n−1) ((1/4)+k))/(Π_(k=0) ^(n−1) ((5/4)+k))).(((u^4 )^n )/(n!))) =u _2 F_1 ((1/2),(1/4);(5/4);u^4 ),u=(√(−cos(t+(π/4)))) g(t)=2^(3/4) (√(−cos(t+(π/4)))) _2 F_1 ((1/2),(1/4);(5/4);cos^2 (t+(π/4))) f(t)=(2/3)(sin(t)(√(sin(t)−cos(t)))−(1/2^(1/4) ).(√(−cos(t+(π/4)))) _2 F_1 ((1/2),(1/4);(5/4);cos^2 (t+(π/4)))) get our answer t=arcsin((x/2)) ∫(√(x−(√(4−x^2 ))))dx=g(x) g(x)=2f(arcsin((x/2))).
$${x}=\mathrm{2}{sin}\left({t}\right)\Rightarrow \\ $$$$=\int\mathrm{2}{cos}\left({t}\right)\sqrt{\mathrm{2}{sin}\left({t}\right)−\mathrm{2}{cos}\left({t}\right)}{dt}\:\:{let} \\ $$$${f}\left({t}\right)=\int{cos}\left({t}\right)\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}{dt} \\ $$$${by}\:{part}={sin}\left({t}\right)\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}{dt}−\int\frac{{sin}\left({t}\right)\left({sin}\left({t}\right)+{cos}\left({t}\right)\right)}{\mathrm{2}\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}}{dt} \\ $$$$={sin}\left({t}\right)\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}−\frac{\mathrm{1}}{\mathrm{2}}.\int\frac{{sin}^{\mathrm{2}} \left({t}\right)+{sin}\left({t}\right){cos}\left({t}\right)}{\:\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}}{dt} \\ $$$${sin}^{\mathrm{2}} \left({t}\right)=\mathrm{1}−{cos}^{\mathrm{2}} \left({t}\right)\Rightarrow \\ $$$$={sin}\left({t}\right)\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\:\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}\:\:}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{cos}\left({t}\right)\left({sin}\left({t}\right)−{cos}\left({t}\right)\right.}{\:\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}}{dt} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{2}}{f}\left({t}\right)={sin}\left({t}\right)\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}−\frac{\mathrm{1}}{\mathrm{2}}.\int\frac{{dt}}{\:\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}} \\ $$$$,{let}\:{g}\left({t}\right)=\int\frac{{dt}}{\:\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}}=\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}}\int\frac{{dt}}{\:\sqrt{−{cos}\left({t}+\frac{\pi}{\mathrm{4}}\right)}} \\ $$$${sin}\left({t}\right)−{cos}\left({t}\right)=−\sqrt{\mathrm{2}}{cos}\left({t}+\frac{\pi}{\mathrm{4}}\right) \\ $$$${let}\:{u}=\sqrt{−{cos}\left({t}+\frac{\pi}{\mathrm{4}}\right)}\Rightarrow{t}={arcos}\left(−{u}^{\mathrm{2}} \right)−\frac{\pi}{\mathrm{4}} \\ $$$${g}\left({t}\right)\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }\int\frac{\mathrm{2}{udu}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{4}} }.{u}}=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \int\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{4}} }} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} .{n}!}.{x}^{{n}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{4}} }}=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} .{n}!}.{x}^{\mathrm{4}{n}} \\ $$$$\Rightarrow\int\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{4}} }}=\int\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} .{n}!}{u}^{\mathrm{4}{n}} \right){du} \\ $$$$={u}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!{u}^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{2}^{{n}} .{n}!.\left(\mathrm{4}{n}+\mathrm{1}\right)}={u}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} .\left(\mathrm{4}{n}+\mathrm{1}\right)}.\frac{\left({u}^{\mathrm{4}} \right)^{{n}} }{{n}!}\right)..{E} \\ $$$$\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} }=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right)\: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{1}}=\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{4}}+{k}\right)}{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{5}}{\mathrm{4}}+{k}−\mathrm{1}\right)} \\ $$$${E}\Leftrightarrow{u}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{4}}+{k}\right)}{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{5}}{\mathrm{4}}+{k}\right)}.\frac{\left({u}^{\mathrm{4}} \right)^{{n}} }{{n}!}\right) \\ $$$$={u}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}};{u}^{\mathrm{4}} \right),{u}=\sqrt{−{cos}\left({t}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$${g}\left({t}\right)=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{−{cos}\left({t}+\frac{\pi}{\mathrm{4}}\right)}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}};{cos}^{\mathrm{2}} \left({t}+\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$${f}\left({t}\right)=\frac{\mathrm{2}}{\mathrm{3}}\left({sin}\left({t}\right)\sqrt{{sin}\left({t}\right)−{cos}\left({t}\right)}−\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }.\sqrt{−{cos}\left({t}+\frac{\pi}{\mathrm{4}}\right)}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}};{cos}^{\mathrm{2}} \left({t}+\frac{\pi}{\mathrm{4}}\right)\right)\right) \\ $$$${get}\:{our}\:{answer}\:{t}={arcsin}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\int\sqrt{{x}−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{dx}={g}\left({x}\right) \\ $$$${g}\left({x}\right)=\mathrm{2}{f}\left({arcsin}\left(\frac{{x}}{\mathrm{2}}\right)\right). \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 13/Mar/20
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by mind is power last updated on 13/Mar/20
withe pleasur
$${withe}\:{pleasur} \\ $$

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