1-1-x-4-dx-

Question Number 82891 by jagoll last updated on 25/Feb/20

$$\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\:\mathrm{dx}\:=\:? \\ $$

Answered by mind is power last updated on 25/Feb/20

(1/(.(√(1−t^2 ))))=Σ_(n≥ 0) (((2n)!t^(2n) )/(2^(2n) (n!)^2 )) (1/( (√(1−t^4 ))))=Σ_(n≥0) (((2n)!t^(4n) )/(2^(2n) (n!)^2 )) ∫(dt/( (√(1−t^4 ))))=Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 ))∫t^(4n) dt =Σ_(n≥0) (((2n)!x^(4n+1) )/(2^(2n) (n!)^2 (4n+1)))=xΣ_(n≥0) (((2n)!x^(4n) )/(2^(2n) (n!)^2 (4n+1))) xΣ_(n≥0) ((2^n .n!.(2n+1)!!)/(2^(2n) (n!)^2 (4n+1)))x^(4n) =xΣ_(n≥0) (((2n+1)!!)/(2^n (4n+1))).(((x^4 )^n )/(n!))+c (2n+1)!!=2^n Π_(k=0) ^(n−1) (k+(3/2))=2^n ((3/2))_n =xΣ_(n≥0) ((((3/2))_n .2^n ((1/4)))/(2^n .(n+(1/4)))) .x^(4n) +c =xΣ_(n≥0) ((((3/2))_n .((1/4))_n )/(((5/4))_n )).x^(4n) +c =x _2 F_1 ((3/2),(1/4);(5/5);x^4 )+c

$$\frac{\mathrm{1}}{.\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\underset{{n}\geqslant\:\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{t}^{\mathrm{2}{n}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{t}^{\mathrm{4}{n}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} } \\ $$$$\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\int{t}^{\mathrm{4}{n}} {dt} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{x}^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{4}{n}+\mathrm{1}\right)}={x}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{x}^{\mathrm{4}{n}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{4}{n}+\mathrm{1}\right)} \\ $$$${x}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{n}} .{n}!.\left(\mathrm{2}{n}+\mathrm{1}\right)!!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{4}{n}+\mathrm{1}\right)}{x}^{\mathrm{4}{n}} \\ $$$$={x}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}{\mathrm{2}^{{n}} \left(\mathrm{4}{n}+\mathrm{1}\right)}.\frac{\left({x}^{\mathrm{4}} \right)^{{n}} }{{n}!}+{c} \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!!=\mathrm{2}^{{n}} \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}^{{n}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} \\ $$$$={x}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} .\mathrm{2}^{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}^{{n}} .\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)}\:.{x}^{\mathrm{4}{n}} +{c} \\ $$$$={x}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} .\left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{{n}} }{\left(\frac{\mathrm{5}}{\mathrm{4}}\right)_{{n}} }.{x}^{\mathrm{4}{n}} +{c} \\ $$$$={x}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{5}};{x}^{\mathrm{4}} \right)+{c} \\ $$$$ \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 25/Feb/20

$${great}\:{sir}\: \\ $$

Commented by jagoll last updated on 26/Feb/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by M±th+et£s last updated on 25/Feb/20

I=∫(dx/( (√(1+x^2 )) (√(1−x^2 )))) =∫((1/( (√(1−x^2 ))))/( (√(1−(−1)(sin(sin^(−1) (x))^2 )))) dx =F(sin^(−1) x∣−1)+c

$${I}=\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=\int\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{1}−\left(−\mathrm{1}\right)\left({sin}\left({sin}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} \right.}}\:{dx} \\ $$$$={F}\left({sin}^{−\mathrm{1}} {x}\mid−\mathrm{1}\right)+{c} \\ $$

Commented by M±th+et£s last updated on 25/Feb/20

notice (u∣a) is 1st kind eliptical intrgral

$${notice}\:\left({u}\mid{a}\right)\:{is}\:\mathrm{1}{st}\:{kind}\:{eliptical}\:{intrgral} \\ $$

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