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Question Number 161830 by EnterUsername last updated on 22/Dec/21

∫(dx/( (√(1−x^(14) ))))

$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{14}} }} \\ $$

Answered by Ar Brandon last updated on 22/Dec/21

arcsin(x)=Σ_(n=0) ^∞ ((x^(2n+1) ((1/2))_n )/(n!(2n+1)))⇒arcsin(x^7 )=Σ_(n=0) ^∞ ((x^(14n+7) ((1/2))_n )/(n!(2n+1))) ⇒((7x^6 )/( (√(1−x^(14) ))))=7Σ_(n=0) ^∞ ((x^(14n+6) ((1/2))_n )/(n!))⇒(1/( (√(1−x^(14) ))))=Σ_(n=0) ^∞ ((((1/2))_n )/(n!))x^(14n) I=∫(dx/( (√(1−x^(14) ))))=∫Σ_(n=0) ^∞ ((((1/2))_n )/(n!))x^(14n) dx=Σ_(n=0) ^∞ ((((1/2))_n )/(n!(14n+1)))x^(14n+1) =(x/(14))Σ_(n=0) ^∞ ((((1/2))_n )/(n!(n+(1/(14)))))x^(14n) =(x/(14))Σ_(n=0) ^∞ ((((1/2))_n Γ(n+(1/(14))))/(n!Γ(n+((15)/(14)))))x^(14n) =(x/(14))∙((Γ((1/(14))))/(Γ(((15)/(14)))))Σ_(n=0) ^∞ ((((1/2))_n ((1/(14)))_n )/(n!(((15)/(14)))_n ))x^(14n) =x _2 F_1 ((1/2),(1/(14));((15)/(14));x^(14) )

$$\mathrm{arcsin}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}\Rightarrow\mathrm{arcsin}\left({x}^{\mathrm{7}} \right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{14}{n}+\mathrm{7}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{7}{x}^{\mathrm{6}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{14}} }}=\mathrm{7}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{14}{n}+\mathrm{6}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{14}} }}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{x}^{\mathrm{14}{n}} \\ $$$${I}=\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{14}} }}=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{x}^{\mathrm{14}{n}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{14}{n}+\mathrm{1}\right)}{x}^{\mathrm{14}{n}+\mathrm{1}} \\ $$$$\:\:\:=\frac{{x}}{\mathrm{14}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left({n}+\frac{\mathrm{1}}{\mathrm{14}}\right)}{x}^{\mathrm{14}{n}} =\frac{{x}}{\mathrm{14}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{14}}\right)}{{n}!\Gamma\left({n}+\frac{\mathrm{15}}{\mathrm{14}}\right)}{x}^{\mathrm{14}{n}} \\ $$$$\:\:\:=\frac{{x}}{\mathrm{14}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{14}}\right)}{\Gamma\left(\frac{\mathrm{15}}{\mathrm{14}}\right)}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{14}}\right)_{{n}} }{{n}!\left(\frac{\mathrm{15}}{\mathrm{14}}\right)_{{n}} }{x}^{\mathrm{14}{n}} ={x}\underset{\mathrm{2}} {\:}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{14}};\frac{\mathrm{15}}{\mathrm{14}};{x}^{\mathrm{14}} \right) \\ $$

Commented by Lordose last updated on 22/Dec/21

fast

$$\mathrm{fast} \\ $$

Commented by Ar Brandon last updated on 22/Dec/21

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