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Question Number 129185 by Dwaipayan Shikari last updated on 13/Jan/21

$${\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{97+{sin}^{2}x}}Hypergeometricform$$

Commented by Dwaipayan Shikari last updated on 13/Jan/21

$$Ihavefound\frac{\pi }{2\sqrt{97}}{}_2{F}_1(\frac{1}{2},\frac{1}{2};1;-\frac{1}{97})$$

Commented by Dwaipayan Shikari last updated on 13/Jan/21

$$\frac{1}{\sqrt{97}}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{1-(-\frac{1}{97}{sin}^{2}x)}}{k}^2=-\frac{1}{97}$$$$=\frac{1}{\sqrt{97}}{\int }_0^{\frac{\pi }{2}}\frac{1}{\sqrt{1-k^2{sin}^{2}x}}dx$$$$=\frac{1}{\sqrt{97}}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{\int }_0^{\frac{\pi }{2}}{k}^{2n}{sin}^{2n}xdx=\frac{1}{\sqrt{97}}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{k}^{2n}{\int }_0^{\frac{\pi }{2}}{sin}^{2n}xdx$$$$=\frac{1}{2\sqrt{97}}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{k}^{2n}.\frac{\mathrm{\Gamma }(n+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(n+1)}=\frac{\sqrt{\pi }}{2\sqrt{97}}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n^2\mathrm{\Gamma }\left(\frac{1}{2}\right)}{{\left(1\right)}_{n}n!}{\left(k^2\right)}^n$$$$=\frac{\pi }{2\sqrt{97}}{}_2{F}_1(\frac{1}{2},\frac{1}{2};1;k^2)=\frac{\pi }{2\sqrt{97}}{}_2{F}_1(\frac{1}{2},\frac{1}{2};1;-\frac{1}{97})$$$$\mathrm{\Gamma }(n+\frac{1}{2})=\mathrm{\Gamma }\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}_n$$$$$$

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