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

$${\int }_2^5\frac{dx}{\sqrt{1-x^4}}$$

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

$$\int \frac{dx}{\sqrt{1-x^4}}$$$$=\int \sum _{n⩾0}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{x}^{4n}=x\sum _{n⩾0}\frac{{\left(\frac{1}{2}\right)}_n}{n!(4n+1)}{x}^{4n}=\frac{x}{4}\sum _{n⩾0}\frac{{\left(\frac{1}{2}\right)}_n\mathrm{\Gamma }(n+\frac{1}{4})}{n!\mathrm{\Gamma }(n+\frac{5}{4})}{x}^{4n}$$$$=\frac{x}{4}.\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{5}{4}\right)}\sum _{n⩾0}\frac{{\left(\frac{1}{2}\right)}_n{\left(\frac{1}{4}\right)}_n}{{\left(\frac{5}{4}\right)}_{n}n!}{\left(x^4\right)}^n=x{}_2{F}_1(\frac{1}{2},\frac{1}{4};\frac{5}{4};x^4)+C$$$${\int }_2^5\frac{dx}{\sqrt{1-x^4}}=5_2{F}_1(\frac{1}{2},\frac{1}{4};\frac{5}{4};625)-2{}_2{F}_1(\frac{1}{2},\frac{1}{4};\frac{5}{4};16)$$

Commented by Satyendra last updated on 06/May/21

$$thanksalot.$$

Answered by Lordose last updated on 06/May/21

$$$$$$\mathrm{\Omega }={\int }_2^5\frac{1}{\sqrt{1-{\mathrm{x}}^4}}\mathrm{dx}$$$${\sin }^{-1}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{\mathrm{x}}^{2\mathrm{n}+1}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{\mathrm{n}!(2\mathrm{n}+1)}\right)$$$$\mathrm{Putting}\mathrm{x}={\mathrm{x}}^2$$$${\sin }^{-1}\left({\mathrm{x}}^2\right)=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{\mathrm{x}}^{2(2\mathrm{n}+1)}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{\mathrm{n}!(2\mathrm{n}+1)}\right)$$$$\mathrm{Differentiate}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}$$$$\frac{2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^4}}=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{\mathrm{x}}^{4\mathrm{n}+1}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{\mathrm{n}!}\right)$$$$\mathrm{Dividing}\mathrm{by}2\mathrm{x},$$$$\frac{1}{\sqrt{1-{\mathbf{x}}^4}}=\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{\mathbf{x}}^{4\mathbf{n}}{\left(\frac{1}{2}\right)}_{\mathbf{n}}}{\mathbf{n}!}\right)$$$$\mathrm{\Omega }={\int }_2^5\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{\mathrm{x}}^{4\mathrm{n}}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{\mathrm{n}!}\right)\mathrm{dx}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{\mathrm{n}!}\right){\int }_2^5{\mathrm{x}}^{4\mathrm{n}}\mathrm{dx}$$$$\mathrm{\Omega }=\mid \frac{{\mathrm{x}}^{4\mathrm{n}+1}}{4\mathrm{n}+1}{\mid }_2^5=\frac{5^{4\mathrm{n}+1}}{4\mathrm{n}+1}-\frac{2^{4\mathrm{n}+1}}{4\mathrm{n}+1}$$$$\mathrm{\Omega }=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{5^{4\mathrm{n}+1}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{(4\mathrm{n}+1)\mathrm{n}!}\right)-\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{2^{4\mathrm{n}+1}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{(4\mathrm{n}+1)\mathrm{n}!}\right)$$$$\mathrm{\Omega }=\frac{5}{4}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{625}^{\mathrm{n}}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{(\mathrm{n}+\frac{1}{4})\mathrm{n}!}\right)-\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{16}^{\mathrm{n}}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{(\mathrm{n}+\frac{1}{4})\mathrm{n}!}\right)$$$$\mathbf{N}.\mathbf{B}::\frac{\mathbf{\Gamma }(1+\mathbf{x})}{\mathbf{\Gamma }\left(\mathbf{x}\right)}=\mathbf{x}$$$${\left(\mathbf{a}\right)}_{\mathbf{n}}=\frac{\mathbf{\Gamma }(\mathbf{a}+\mathbf{n})}{\mathbf{\Gamma }\left(\mathbf{a}\right)}$$$${\left(\mathbf{a}\right)}_{\mathbf{n}}={\mathbf{Pochammer}}^{\prime }\mathbf{s}\mathbf{symbol}$$$$\mathbf{\Gamma }\left(\mathbf{x}\right)=\mathbf{Gamma}\mathbf{Function}$$$${}_2{\mathbf{F}}_1(\mathbf{a},\mathbf{b},\mathbf{c};\mathbf{x})=\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{\left(\mathbf{a}\right)}_{\mathbf{n}}{\left(\mathbf{b}\right)}_{\mathbf{n}}}{{\left(\mathbf{c}\right)}_{\mathbf{n}}\mathbf{n}!}{\mathbf{x}}^{\mathbf{n}}\right)$$$${}_2{\mathbf{F}}_1(\mathbf{a},\mathbf{b},\mathbf{c};\mathbf{x})=\mathbf{Generalized}\mathbf{Hypergeometric}\mathbf{function}$$$$\mathrm{\Omega }=\frac{5}{4}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{625}^{\mathrm{n}}\mathbf{\Gamma }(\mathrm{n}+\frac{1}{4}){\left(\frac{1}{2}\right)}_{\mathrm{n}}}{\mathbf{\Gamma }(\mathrm{n}+\frac{5}{4})\mathrm{n}!}\right)-\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{16}^{\mathrm{n}}\mathbf{\Gamma }(\mathrm{n}+\frac{1}{4}){\left(\frac{1}{2}\right)}_{\mathrm{n}}}{\mathbf{\Gamma }(\mathrm{n}+\frac{5}{4})\mathrm{n}!}\right)$$$$\mathrm{\Omega }=\frac{5\mathbf{\Gamma }\left(\frac{1}{4}\right)}{4\mathbf{\Gamma }\left(\frac{5}{4}\right)}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{625}^{\mathrm{n}}{\left(\frac{1}{4}\right)}_{\mathrm{n}}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{{\left(\frac{5}{4}\right)}_{\mathrm{n}}\mathrm{n}!}\right)-\frac{\mathbf{\Gamma }\left(\frac{1}{4}\right)}{2\mathbf{\Gamma }\left(\frac{5}{4}\right)}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{16}^{\mathrm{n}}{\left(\frac{1}{4}\right)}_{\mathrm{n}}{\left(\frac{1}{2}\right)}_{\mathrm{n}}}{{\left(\frac{5}{4}\right)}_{\mathrm{n}}\mathrm{n}!}\right)$$$$\mathbf{\Omega }=5{\cdot }_2{\mathbf{F}}_1(\frac{1}{2},\frac{1}{4},\frac{5}{4};625)-2{\cdot }_2{\mathbf{F}}_1(\frac{1}{2},\frac{1}{4},\frac{5}{4};16)$$

Commented by Satyendra last updated on 06/May/21

$$thanksalot.$$

Commented by mohammad17 last updated on 06/May/21

$$sircanyouhelpmeinthisintegral$$$$\int \frac{dx}{\sqrt{1+x^4}}$$$$$$

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