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Question Number 140296 by Satyendra last updated on 06/May/21

∫_2 ^5 (dx/( (√(1−x^4 ))))

$$\int_{\mathrm{2}} ^{\mathrm{5}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$

Answered by Dwaipayan Shikari last updated on 06/May/21

∫(dx/( (√(1−x^4 )))) =∫Σ_(n≥0) ((((1/2))_n )/(n!))x^(4n) =xΣ_(n≥0) ((((1/2))_n )/(n!(4n+1)))x^(4n) =(x/4)Σ_(n≥0) ((((1/2))_n Γ(n+(1/4)))/(n!Γ(n+(5/4))))x^(4n) =(x/4).((Γ((1/4)))/(Γ((5/4))))Σ_(n≥0) ((((1/2))_n ((1/4))_n )/(((5/4))_n n!))(x^4 )^n =x _2 F_1 ((1/2),(1/4);(5/4);x^4 )+C ∫_2 ^5 (dx/( (√(1−x^4 ))))=5_2 F_1 ((1/2),(1/4);(5/4);625)−2 _2 F_1 ((1/2),(1/4);(5/4);16)

$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$$$=\int\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{x}^{\mathrm{4}{n}} ={x}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{4}{n}+\mathrm{1}\right)}{x}^{\mathrm{4}{n}} =\frac{{x}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{{n}!\Gamma\left({n}+\frac{\mathrm{5}}{\mathrm{4}}\right)}{x}^{\mathrm{4}{n}} \\ $$$$=\frac{{x}}{\mathrm{4}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{{n}} }{\left(\frac{\mathrm{5}}{\mathrm{4}}\right)_{{n}} {n}!}\left({x}^{\mathrm{4}} \right)^{{n}} ={x}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}};{x}^{\mathrm{4}} \right)+{C} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{5}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}=\mathrm{5}_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}};\mathrm{625}\right)−\mathrm{2}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}};\mathrm{16}\right) \\ $$

Commented by Satyendra last updated on 06/May/21

thanks a lot.

$${thanks}\:{a}\:{lot}. \\ $$

Answered by Lordose last updated on 06/May/21

Ω = ∫_2 ^( 5) (1/( (√(1−x^4 ))))dx sin^(−1) (x) = Σ_(n=0) ^∞ (((x^(2n+1) ((1/2))_n )/(n!(2n+1)))) Putting x = x^2 sin^(−1) (x^2 ) = Σ_(n=0) ^∞ (((x^(2(2n+1)) ((1/2))_n )/(n!(2n+1)))) Differentiate both sides w.r.t x ((2x)/( (√(1−x^4 )))) = 2Σ_(n=0) ^∞ (((x^(4n+1) ((1/2))_n )/(n!))) Dividing by 2x, (1/( (√(1−x^4 )))) = Σ_(n=0) ^∞ (((x^(4n) ((1/2))_n )/(n!))) Ω = ∫_2 ^( 5) Σ_(n=0) ^∞ (((x^(4n) ((1/2))_n )/(n!)))dx = Σ_(n=0) ^∞ (((((1/2))_n )/(n!)))∫_2 ^( 5) x^(4n) dx Ω = ∣(x^(4n+1) /(4n+1))∣_2 ^5 = (5^(4n+1) /(4n+1))−(2^(4n+1) /(4n+1)) Ω = Σ_(n=0) ^∞ (((5^(4n+1) ((1/2))_n )/((4n+1)n!))) − Σ_(n=0) ^∞ (((2^(4n+1) ((1/2))_n )/((4n+1)n!))) Ω = (5/4)Σ_(n=0) ^∞ (((625^n ((1/2))_n )/((n+(1/4))n!))) − (1/2)Σ_(n=0) ^∞ (((16^n ((1/2))_n )/((n+(1/4))n!))) N.B :: ((𝚪(1+x))/(𝚪(x))) = x (a)_n = ((𝚪(a+n))/(𝚪(a))) (a)_n = Pochammer′s symbol 𝚪(x) = Gamma Function _2 F_1 (a,b,c;x) = Σ_(n=0) ^∞ ((((a)_n (b)_n )/((c)_n n!))x^n ) _2 F_1 (a,b,c;x) = Generalized Hypergeometric function Ω = (5/4)Σ_(n=0) ^∞ (((625^n 𝚪(n+(1/4))((1/2))_n )/(𝚪(n+(5/4))n!))) − (1/2)Σ_(n=0) ^∞ (((16^n 𝚪(n+(1/4))((1/2))_n )/(𝚪(n+(5/4))n!))) Ω = ((5𝚪((1/4)))/(4𝚪((5/4))))Σ_(n=0) ^∞ (((625^n ((1/4))_n ((1/2))_n )/(((5/4))_n n!))) − ((𝚪((1/4)))/(2𝚪((5/4))))Σ_(n=0) ^∞ (((16^n ((1/4))_n ((1/2))_n )/(((5/4))_n n!))) 𝛀 = 5∙_2 F_1 ((1/2),(1/4),(5/4);625) − 2∙_2 F_1 ((1/2),(1/4),(5/4);16)

$$ \\ $$$$\Omega\:=\:\int_{\mathrm{2}} ^{\:\mathrm{5}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\mathrm{dx} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\mathrm{n}!\left(\mathrm{2n}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{Putting}\:\mathrm{x}\:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} \right)\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{x}^{\mathrm{2}\left(\mathrm{2n}+\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\mathrm{n}!\left(\mathrm{2n}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{Differentiate}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{w}.\mathrm{r}.\mathrm{t}\:\mathrm{x} \\ $$$$\frac{\mathrm{2x}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\:=\:\mathrm{2}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{x}^{\mathrm{4n}+\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\mathrm{n}!}\right) \\ $$$$\mathrm{Dividing}\:\mathrm{by}\:\mathrm{2x}, \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{4}} }}\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\boldsymbol{\mathrm{x}}^{\mathrm{4}\boldsymbol{\mathrm{n}}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}!}\right) \\ $$$$\Omega\:=\:\int_{\mathrm{2}} ^{\:\mathrm{5}} \underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{x}^{\mathrm{4n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\mathrm{n}!}\right)\mathrm{dx}\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\mathrm{n}!}\right)\int_{\mathrm{2}} ^{\:\mathrm{5}} \mathrm{x}^{\mathrm{4n}} \mathrm{dx} \\ $$$$\Omega\:=\:\mid\frac{\mathrm{x}^{\mathrm{4n}+\mathrm{1}} }{\mathrm{4n}+\mathrm{1}}\mid_{\mathrm{2}} ^{\mathrm{5}} \:=\:\frac{\mathrm{5}^{\mathrm{4n}+\mathrm{1}} }{\mathrm{4n}+\mathrm{1}}−\frac{\mathrm{2}^{\mathrm{4n}+\mathrm{1}} }{\mathrm{4n}+\mathrm{1}} \\ $$$$\Omega\:=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{5}^{\mathrm{4n}+\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\left(\mathrm{4n}+\mathrm{1}\right)\mathrm{n}!}\right)\:−\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}^{\mathrm{4n}+\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\left(\mathrm{4n}+\mathrm{1}\right)\mathrm{n}!}\right) \\ $$$$\Omega\:=\:\frac{\mathrm{5}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{625}^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{n}!}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{16}^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{n}!}\right) \\ $$$$\boldsymbol{\mathrm{N}}.\boldsymbol{\mathrm{B}}\:::\:\frac{\boldsymbol{\Gamma}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)}\:=\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}\right)_{\boldsymbol{\mathrm{n}}} \:=\:\frac{\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{n}}\right)}{\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}\right)_{\boldsymbol{\mathrm{n}}} \:=\:\boldsymbol{\mathrm{Pochammer}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{symbol}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\boldsymbol{\mathrm{Gamma}}\:\boldsymbol{\mathrm{Function}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:_{\mathrm{2}} \boldsymbol{\mathrm{F}}_{\mathrm{1}} \left(\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}};\boldsymbol{\mathrm{x}}\right)\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(\boldsymbol{\mathrm{a}}\right)_{\boldsymbol{\mathrm{n}}} \left(\boldsymbol{\mathrm{b}}\right)_{\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{c}}\right)_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{n}}!}\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:_{\mathrm{2}} \boldsymbol{\mathrm{F}}_{\mathrm{1}} \left(\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}};\boldsymbol{\mathrm{x}}\right)\:=\:\boldsymbol{\mathrm{Generalized}}\:\boldsymbol{\mathrm{Hypergeometric}}\:\boldsymbol{\mathrm{function}} \\ $$$$\Omega\:=\:\frac{\mathrm{5}}{\mathrm{4}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{625}^{\mathrm{n}} \boldsymbol{\Gamma}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\boldsymbol{\Gamma}\left(\mathrm{n}+\frac{\mathrm{5}}{\mathrm{4}}\right)\mathrm{n}!}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{16}^{\mathrm{n}} \boldsymbol{\Gamma}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\boldsymbol{\Gamma}\left(\mathrm{n}+\frac{\mathrm{5}}{\mathrm{4}}\right)\mathrm{n}!}\right) \\ $$$$\Omega\:=\:\frac{\mathrm{5}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{625}^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\left(\frac{\mathrm{5}}{\mathrm{4}}\right)_{\mathrm{n}} \mathrm{n}!}\right)\:−\:\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{16}^{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)_{\mathrm{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{n}} }{\left(\frac{\mathrm{5}}{\mathrm{4}}\right)_{\mathrm{n}} \mathrm{n}!}\right) \\ $$$$\boldsymbol{\Omega}\:=\:\mathrm{5}\centerdot_{\mathrm{2}} \boldsymbol{\mathrm{F}}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{5}}{\mathrm{4}};\mathrm{625}\right)\:−\:\mathrm{2}\centerdot_{\mathrm{2}} \boldsymbol{\mathrm{F}}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{5}}{\mathrm{4}};\mathrm{16}\right) \\ $$

Commented by Satyendra last updated on 06/May/21

thanks a lot.

$${thanks}\:{a}\:{lot}. \\ $$

Commented by mohammad17 last updated on 06/May/21

sir can you help me in this integral ∫(dx/( (√(1+x^4 ))))

$${sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{this}\:{integral}\: \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }} \\ $$$$ \\ $$

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