CalculusChallengeNon-Elementary Integration ShareNon-Elementary Integration · Challenge problemLet A=∫0π/2sinx dxA=\displaystyle\int_0^{\pi/2}\sqrt{\sin x}\,dxA=∫0π/2sinxdx and B=∫0π/2dxsinxB=\displaystyle\int_0^{\pi/2}\dfrac{dx}{\sqrt{\sin x}}B=∫0π/2sinxdx; neither has an elementary value. Using ∫0π/2sin2p−1xcos2q−1x dx=Γ(p)Γ(q)2 Γ(p+q)\displaystyle\int_0^{\pi/2}\sin^{2p-1}x\cos^{2q-1}x\,dx=\dfrac{\Gamma(p)\Gamma(q)}{2\,\Gamma(p+q)}∫0π/2sin2p−1xcos2q−1xdx=2Γ(p+q)Γ(p)Γ(q), Γ (12)=π\Gamma\!\left(\tfrac12\right)=\sqrt\piΓ(21)=π and Γ(s+1)=s Γ(s)\Gamma(s+1)=s\,\Gamma(s)Γ(s+1)=sΓ(s), find ABABAB.Check Show answer Walk through it step by step Practice more like this Review this concept