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Question Number 197525 by mokys last updated on 20/Sep/23

$$\int \sqrt[n]{tanx}dx$$

Answered by Frix last updated on 20/Sep/23

$$t=\sqrt[n]{\tan x}\Rightarrow$$$$n\int \frac{t^n}{t^{2n}+1}dt$$$$\mathrm{I}\mathrm{think}\mathrm{this}\mathrm{should}\mathrm{be}\mathrm{possible}\mathrm{using}\mathrm{the}$$$$\mathrm{Hypergeometric}\mathrm{Function}{}_2{F}_1(a,b;c,x)$$

Commented by Frix last updated on 20/Sep/23

$$\mathrm{Woframalpha}\mathrm{gives}$$$$n\int \frac{t^n}{t^{2n}+1}dt=\frac{{nt}^{n+1}}{n+1}{}_2{F}_1(1,\frac{n+1}{2n};\frac{3n+1}{2n},-t^{2n})+C$$

Commented by mokys last updated on 20/Sep/23

$$yessirbathowcantotakethissolution?$$

Answered by witcher3 last updated on 21/Sep/23

$$\mathrm{y}=\tan \left(\mathrm{x}\right)$$$${\int }_0^{\mathrm{x}}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}}}{1+{\mathrm{t}}^2}\mathrm{dt}=\mathrm{f}\left(\mathrm{x}\right)$$$$=\sum _{\mathrm{k}⩾0}{(-1)}^{\mathrm{k}}{\int }_0^{\mathrm{x}}{\mathrm{t}}^{2\mathrm{k}+\frac{1}{\mathrm{n}}}\mathrm{dt}$$$$=\frac{1}{2}{\mathrm{x}}^{\frac{1}{\mathrm{n}}+1}(\frac{2}{1+\frac{1}{\mathrm{n}}}+\sum _{\mathrm{k}⩾1}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+\frac{1}{2}+\frac{1}{2\mathrm{n}}}{\mathrm{x}}^{2\mathrm{k}})$$$$=\frac{1}{2}{\mathrm{x}}^{\frac{1}{\mathrm{n}}+1}(\frac{2}{1+\frac{1}{\mathrm{n}}}+\sum _{\mathrm{k}⩾1}\frac{{(-{\mathrm{x}}^2)}^{\mathrm{n}}}{\mathrm{k}!}.\frac{\mathrm{k}!.\mathrm{\Gamma }(\mathrm{k}+\frac{1}{2}+\frac{1}{2\mathrm{n}})}{\mathrm{\Gamma }(\mathrm{k}+\frac{3}{2}+\frac{1}{2\mathrm{n}})}))$$$$=\frac{\mathrm{\Gamma }(\mathrm{k}+\frac{1}{2}+\frac{1}{2\mathrm{n}})}{\mathrm{\Gamma }(\frac{1}{2}+\frac{1}{2\mathrm{n}})}={(\frac{1}{2}+\frac{1}{2\mathrm{n}})}_{\left(\mathrm{k}\right)}$$$$\frac{\mathrm{\Gamma }(\mathrm{k}+\frac{3}{2}+\frac{1}{2\mathrm{n}})}{\mathrm{\Gamma }(\frac{1}{2}+\frac{1}{2\mathrm{n}})}=(\frac{1}{2}+\frac{1}{2\mathrm{n}}){(\frac{3}{2}+\frac{1}{2\mathrm{n}})}_{\left(\mathrm{k}\right)}$$$$=\frac{{\mathrm{x}}^{\frac{1}{\mathrm{n}}+1}}{2}(\frac{2}{1+\frac{1}{\mathrm{n}}}+\mathrm{\Sigma }\frac{{(-{\mathrm{x}}^2)}^{\mathrm{k}}}{\mathrm{k}!}.\frac{{\left(1\right)}_k{(\frac{1}{2}+\frac{1}{2n})}_k\mathrm{\Gamma }(\frac{1}{2n}+\frac{1}{2})}{{(\frac{3}{2}+\frac{1}{2\mathrm{n}})}_{\mathrm{k}}\mathrm{\Gamma }(\frac{1}{2}+\frac{1}{2\mathrm{n}})(\frac{1}{2\mathrm{n}}+\frac{1}{2})})$$$$=\frac{{\mathrm{x}}^{\frac{1}{\mathrm{n}}+1}}{2}(\frac{2}{1+\frac{1}{\mathrm{n}}}+\mathrm{\Sigma }\frac{{(-{\mathrm{x}}^2)}^{\mathrm{k}}}{\mathrm{k}!}.\frac{{(\frac{1}{2\mathrm{n}}+\frac{1}{2})}_{\mathrm{k}}{\left(1\right)}_{\mathrm{k}}}{{(\frac{3}{2}+\frac{1}{2\mathrm{n}})}_{\mathrm{k}}(\frac{1}{2\mathrm{n}}+\frac{1}{2})})+\mathrm{c}$$$$=\frac{{\mathrm{x}}^{\frac{1}{\mathrm{n}}+1}}{2}(\frac{1}{\frac{1}{2\mathrm{n}}+\frac{1}{2}}+\frac{1}{\frac{1}{2\mathrm{n}}+\frac{1}{2}}\sum _{\mathrm{k}⩾0}\frac{{\left(1\right)}_{\mathrm{k}}{(\frac{1}{2\mathrm{n}}+\frac{1}{2})}_{\mathrm{k}}}{{(\frac{3}{2}+\frac{1}{2\mathrm{n}})}_{\mathrm{k}}}.\frac{{(-{\mathrm{x}}^2)}^{\mathrm{k}}}{\mathrm{k}!})+\mathrm{c}$$$$=\frac{\mathrm{n}}{\mathrm{n}+1}{\mathrm{x}}^{\frac{1}{\mathrm{n}}+1}(1+\sum _{\mathrm{k}⩾0}\frac{{\left(1\right)}_{\mathrm{k}}{(\frac{1}{2\mathrm{n}}+\frac{1}{2})}_{\mathrm{k}}}{{(\frac{3}{2}+\frac{1}{2\mathrm{n}})}_{\mathrm{k}}}.\frac{{(-{\mathrm{x}}^2)}^{\mathrm{k}}}{\mathrm{k}!})+\mathrm{c}$$$$=\frac{{\mathrm{nx}}^{\frac{1}{\mathrm{n}}+1}}{\mathrm{n}+1}{.}_2{\mathrm{F}}_1(1,\frac{\mathrm{n}+1}{2\mathrm{n}};\frac{3\mathrm{n}+1}{2\mathrm{n}};-{\mathrm{x}}^2)+\mathrm{c},\mathrm{x}=\tan \left(\mathrm{t}\right)$$$$$$

Commented by mokys last updated on 21/Sep/23

$$thankyousir$$

Commented by witcher3 last updated on 21/Sep/23

$$\mathrm{you}\mathrm{are}\mathrm{welcom}$$$$\mathrm{have}\mathrm{You}\mathrm{the}\mathrm{name}\mathrm{if}\mathrm{the}\mathrm{files}$$$$\mathrm{withe}[\mathrm{Qustion}{\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{x}}^2+{\mathrm{a}}^2}.\frac{\mathrm{dx}}{{\ln }^2\left(\mathrm{x}\right)+{\pi }^2}$$

Commented by mokys last updated on 22/Sep/23

$$thenameoffilesis\left[nodalarena\right]andisiraqibook$$

Commented by witcher3 last updated on 23/Sep/23

$${\mathrm{can}}^{\prime }\mathrm{t}\mathrm{find}\mathrm{this}\mathrm{one}$$

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