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Question Number 197525 by mokys last updated on 20/Sep/23
∫ ((tanx))^(1/n) dx
$$\int\:\sqrt[{{n}}]{{tanx}}\:{dx} \\ $$
Answered by Frix last updated on 20/Sep/23
t=((tan x))^(1/n) ⇒ n∫(t^n /(t^(2n) +1))dt I think this should be possible using the Hypergeometric Function _2 F_1 (a, b; c, x)
$${t}=\sqrt[{{n}}]{\mathrm{tan}\:{x}}\:\Rightarrow\: \\ $$$${n}\int\frac{{t}^{{n}} }{{t}^{\mathrm{2}{n}} +\mathrm{1}}{dt} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{possible}\:\mathrm{using}\:\mathrm{the} \\ $$$$\mathrm{Hypergeometric}\:\mathrm{Function}\:_{\mathrm{2}} {F}_{\mathrm{1}} \:\left({a},\:{b};\:{c},\:{x}\right) \\ $$
Commented by Frix last updated on 20/Sep/23
Woframalpha gives n∫(t^n /(t^(2n) +1))dt=((nt^(n+1) )/(n+1)) _2 F_1 (1, ((n+1)/(2n)); ((3n+1)/(2n)), −t^(2n) ) +C
$$\mathrm{Woframalpha}\:\mathrm{gives} \\ $$$${n}\int\frac{{t}^{{n}} }{{t}^{\mathrm{2}{n}} +\mathrm{1}}{dt}=\frac{{nt}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \:\left(\mathrm{1},\:\frac{{n}+\mathrm{1}}{\mathrm{2}{n}};\:\frac{\mathrm{3}{n}+\mathrm{1}}{\mathrm{2}{n}},\:−{t}^{\mathrm{2}{n}} \right)\:+{C} \\ $$
Commented by mokys last updated on 20/Sep/23
yes sir bat how can to take this solution ?
$${yes}\:{sir}\:{bat}\:{how}\:{can}\:{to}\:{take}\:{this}\:{solution}\:? \\ $$
Answered by witcher3 last updated on 21/Sep/23
y=tan(x) ∫_0 ^x (t^(1/n) /(1+t^2 ))dt=f(x) =Σ_(k≥0) (−1)^k ∫_0 ^x t^(2k+(1/n)) dt =(1/2)x^((1/n)+1) ((2/(1+(1/n)))+Σ_(k≥1) (((−1)^k )/(k+(1/2)+(1/(2n))))x^(2k) ) =(1/2)x^((1/n)+1) ((2/(1+(1/n)))+Σ_(k≥1) (((−x^2 )^n )/(k!)).((k!.Γ(k+(1/2)+(1/(2n))))/(Γ(k+(3/2)+(1/(2n))))))) =((Γ(k+(1/2)+(1/(2n))))/(Γ((1/2)+(1/(2n)))))=((1/2)+(1/(2n)))_((k)) ((Γ(k+(3/2)+(1/(2n))))/(Γ((1/2)+(1/(2n)))))=((1/2)+(1/(2n)))((3/2)+(1/(2n)))_((k)) =(x^((1/n)+1) /2)((2/(1+(1/n)))+Σ(((−x^2 )^k )/(k!)).(((1)_k ((1/2)+(1/(2n)))_k Γ((1/(2n))+(1/2)))/(((3/2)+(1/(2n)))_k Γ((1/2)+(1/(2n)))((1/(2n))+(1/2))))) =(x^((1/n)+1) /2)((2/(1+(1/n)))+Σ(((−x^2 )^k )/(k!)).((((1/(2n))+(1/2))_k (1)_k )/(((3/2)+(1/(2n)))_k ((1/(2n))+(1/2)))))+c =(x^((1/n)+1) /2)((1/((1/(2n))+(1/2)))+(1/((1/(2n))+(1/2)))Σ_(k≥0) (((1)_k ((1/(2n))+(1/2))_k )/(((3/2)+(1/(2n)))_k )).(((−x^2 )^k )/(k!)))+c =(n/(n+1))x^((1/n)+1) (1+Σ_(k≥0) (((1)_k ((1/(2n))+(1/2))_k )/(((3/2)+(1/(2n)))_k )).(((−x^2 )^k )/(k!)))+c =((nx^((1/n)+1) )/(n+1))._2 F_1 (1,((n+1)/(2n));((3n+1)/(2n));−x^2 )+c,x=tan(t)
$$\mathrm{y}=\mathrm{tan}\left(\mathrm{x}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{x}} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{k}} \int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{t}^{\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{n}}} \mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}+\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}}\mathrm{x}^{\mathrm{2k}} \right) \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}+\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} }{\mathrm{k}!}.\frac{\mathrm{k}!.\Gamma\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)}{\Gamma\left(\mathrm{k}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)}\right)\right) \\ $$$$=\frac{\Gamma\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)_{\left(\mathrm{k}\right)} \\ $$$$\frac{\Gamma\left(\mathrm{k}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)_{\left(\mathrm{k}\right)} \\ $$$$=\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}} }{\mathrm{2}}\left(\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}+\Sigma\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{k}} }{\mathrm{k}!}.\frac{\left(\mathrm{1}\right)_{{k}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)_{{k}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)_{\mathrm{k}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)\left(\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\right) \\ $$$$=\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}} }{\mathrm{2}}\left(\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}+\Sigma\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{k}} }{\mathrm{k}!}.\frac{\left(\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{k}} \left(\mathrm{1}\right)_{\mathrm{k}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)_{\mathrm{k}} \left(\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\right)+\mathrm{c} \\ $$$$=\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{1}\right)_{\mathrm{k}} \left(\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{k}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)_{\mathrm{k}} }.\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{k}} }{\mathrm{k}!}\right)+\mathrm{c} \\ $$$$=\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}} \left(\mathrm{1}+\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{1}\right)_{\mathrm{k}} \left(\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mathrm{k}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2n}}\right)_{\mathrm{k}} }.\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{k}} }{\mathrm{k}!}\right)+\mathrm{c} \\ $$$$=\frac{\mathrm{nx}^{\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}._{\mathrm{2}} \mathrm{F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2n}};\frac{\mathrm{3n}+\mathrm{1}}{\mathrm{2n}};−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{c},\mathrm{x}=\mathrm{tan}\left(\mathrm{t}\right) \\ $$$$ \\ $$
Commented by mokys last updated on 21/Sep/23
thank you sir
$${thank}\:{you}\:{sir}\: \\ $$
Commented by witcher3 last updated on 21/Sep/23
you are welcom have You the name if the files withe[Qustion ∫_0 ^∞ (1/(x^2 +a^2 )).(dx/(ln^2 (x)+π^2 ))
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcom}\: \\ $$$$\mathrm{have}\:\mathrm{You}\:\mathrm{the}\:\mathrm{name}\:\mathrm{if}\:\mathrm{the}\:\mathrm{files}\: \\ $$$$\mathrm{withe}\left[\mathrm{Qustion}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \:\:\:}.\frac{\mathrm{dx}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)+\pi^{\mathrm{2}} }\right. \\ $$
Commented by mokys last updated on 22/Sep/23
the name of files is [ nodal arena ] and is iraqi book
$${the}\:{name}\:{of}\:{files}\:{is}\:\left[\:{nodal}\:{arena}\:\right]\:{and}\:{is}\:{iraqi}\:{book}\: \\ $$
Commented by witcher3 last updated on 23/Sep/23
can′t find this one
$$\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{this}\:\mathrm{one}\: \\ $$

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