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Question Number 87409 by redmiiuser last updated on 04/Apr/20

$$\int {\left(sinx\right)}^{\frac{1}{5}}dx$$

Commented by Prithwish Sen 1 last updated on 04/Apr/20

$$\int \frac{{\sin }^{\frac{1}{5}}\mathrm{x}\mathrm{cosx}}{\mathrm{cosx}}\mathrm{dx}\mathrm{put}\mathrm{sinx}={\mathrm{u}}^5\mathrm{cosxdx}={5\mathrm{u}}^4\mathrm{du}$$$$=\int \frac{{5\mathrm{u}}^5\mathrm{du}}{\sqrt{1-{\mathrm{u}}^{10}}}=5\int {\mathrm{u}}^5{(1-{\mathrm{u}}^{10})}^{-\frac{1}{2}}\mathrm{dx}\mid \mathrm{x}\mid ⩽1$$$$\mathrm{now}\mathrm{apply}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{the}\mathrm{series}...$$

Commented by redmiiuser last updated on 04/Apr/20

$$Yesmisteryouare$$$$correct.Godblessyou!$$

Answered by mind is power last updated on 04/Apr/20

$$=\int \frac{t^{\frac{1}{5}}}{\sqrt{1-t^2}}dx,t=sin\left(x\right)$$$$\frac{1}{\sqrt{1-t^2}}=\sum _{n⩾0}\frac{\left(2n\right)!}{2^{2n}.{(n!)}^2}{t}^{2n}$$$$=\int \sum _{n⩾0}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}{t}^{2n+\frac{1}{5}}dt=\sum _{n⩾0}\frac{\left(2n\right)!5t^{2n+\frac{6}{5}}}{2^{2n}{(n!)}^2(10n+6)}$$$$=\frac{1}{2}{t}^{\frac{6}{5}}\sum _{n⩾0}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2(n+\frac{3}{5})}=\frac{t^{\frac{6}{5}}}{2}(\frac{5}{3}+\sum _{n⩾1}\frac{\left(2n\right)!x^{2n}}{2^{2n}{(n!)}^2(n+\frac{3}{5})})$$$$=\frac{5t^{\frac{6}{5}}}{6}(1+\sum _{n⩾1}\frac{2^{n}n!.\underset{k=0}{\prod ^{n-1}}(2k+1).\frac{3}{5}}{2^{2n}{(n!)}^2.(n+\frac{3}{5})}.t^{2n})$$$$=\frac{5t^{\frac{6}{5}}}{6}(1+\sum _{n⩾1}\frac{\underset{k=0}{\prod ^{n-1}}(\frac{1}{2}+k).\underset{k=0}{\prod ^{n-1}}(\frac{3}{5}+k)}{\underset{k=0}{\prod ^{n-1}}(\frac{8}{5}+k)}.\frac{{\left({\mathrm{t}}^2\right)}^n}{n!})$$$$=\frac{5}{6}{t}^{\frac{6}{5}}{}_2{F}_1(\frac{1}{2},\frac{3}{5};\frac{8}{5};{\mathrm{t}}^2)+\mathrm{c}$$$$=\frac{5}{6}\sin {\left(\mathrm{x}\right)}^{\frac{6}{5}}{}_2{F}_1(\frac{1}{2},\frac{3}{5};\frac{8}{5};{sin}^2\left(x\right))+c$$$$$$$$$$

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