0-pi-2-dx-97-sin-2-x-Hypergeometric-form-

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

\int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{97 + {s i n}^{2} x}} H y p e r g e o m e t r i c f o r m

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

I h a v e f o u n d \frac{π}{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{π}{2}} \frac{d x}{\sqrt{1 - (- \frac{1}{97} {s i n}^{2} x)}} k^{2} = - \frac{1}{97}

= \frac{1}{\sqrt{97}} \int_{0}^{\frac{π}{2}} \frac{1}{\sqrt{1 - k^{2} {s i n}^{2} x}} d x

= \frac{1}{\sqrt{97}} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n!} \int_{0}^{\frac{π}{2}} k^{2 n} {s i n}^{2 n} x d x = \frac{1}{\sqrt{97}} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n!} k^{2 n} \int_{0}^{\frac{π}{2}} {s i n}^{2 n} x d x

= \frac{1}{2 \sqrt{97}} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n!} k^{2 n} . \frac{Γ (n + \frac{1}{2}) Γ (\frac{1}{2})}{Γ (n + 1)} = \frac{\sqrt{π}}{2 \sqrt{97}} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}^{2} Γ (\frac{1}{2})}{{(1)}_{n} n!} {(k^{2})}^{n}

= \frac{π}{2 \sqrt{97}}_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 1; k^{2}) = \frac{π}{2 \sqrt{97}}_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 1; - \frac{1}{97})

Γ (n + \frac{1}{2}) = Γ (\frac{1}{2}) {(\frac{1}{2})}_{n}