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

∫_0 ^(π/2) (dx/( (√(97+sin^2 x)))) Hypergeometric form

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\mathrm{97}+{sin}^{\mathrm{2}} {x}}}\:\:{Hypergeometric}\:{form} \\ $$

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

I have found (π/(2(√(97)))) _2 F_1 ((1/2),(1/2);1;−(1/(97)))

$${I}\:{have}\:{found}\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{97}}}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};−\frac{\mathrm{1}}{\mathrm{97}}\right) \\ $$

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

(1/( (√(97))))∫_0 ^(π/2) (dx/( (√(1−(−(1/(97))sin^2 x))))) k^2 =−(1/(97)) =(1/( (√(97))))∫_0 ^(π/2) (1/( (√(1−k^2 sin^2 x))))dx =(1/( (√(97))))Σ_(n=0) ^∞ ((((1/2))_n )/(n!))∫_0 ^(π/2) k^(2n) sin^(2n) x dx=(1/( (√(97))))Σ_(n=0) ^∞ ((((1/2))_n )/(n!))k^(2n) ∫_0 ^(π/2) sin^(2n) x dx =(1/( 2(√(97))))Σ_(n=0) ^∞ ((((1/2))_n )/(n!))k^(2n) .((Γ(n+(1/2))Γ((1/2)))/(Γ(n+1)))=((√π)/( 2(√(97))))Σ_(n=0) ^∞ ((((1/2))_n ^2 Γ((1/2)))/((1)_n n!))(k^2 )^n =(π/( 2(√(97)))) _2 F_1 ((1/2),(1/2);1;k^2 )=(π/(2 (√(97)))) _2 F_1 ((1/2),(1/2);1;−(1/(97))) Γ(n+(1/2))=Γ((1/2))((1/2))_n

$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{97}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{97}}{sin}^{\mathrm{2}} {x}\right)}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{97}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{97}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}}{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{97}}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {k}^{\mathrm{2}{n}} {sin}^{\mathrm{2}{n}} {x}\:{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{97}}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{k}^{\mathrm{2}{n}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} {x}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{97}}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{k}^{\mathrm{2}{n}} .\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}=\frac{\sqrt{\pi}}{\:\mathrm{2}\sqrt{\mathrm{97}}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{1}\right)_{{n}} {n}!}\left({k}^{\mathrm{2}} \right)^{{n}} \\ $$$$=\frac{\pi}{\:\mathrm{2}\sqrt{\mathrm{97}}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};{k}^{\mathrm{2}} \right)=\frac{\pi}{\mathrm{2}\:\sqrt{\mathrm{97}}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};−\frac{\mathrm{1}}{\mathrm{97}}\right) \\ $$$$\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \\ $$$$ \\ $$

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