Integral and Beta Function

Note that $I \; = \; \int_0^\pi \frac{dx}{\sqrt{3 + \cos x}} \; = \; \int_0^\pi \frac{dx}{\sqrt{2(1 + \cos^2\frac12x)}} \; = \; \sqrt{2}\int_0^{\frac12\pi}\frac{dy}{\sqrt{1 + \cos^2y}}$ The substitution $u = \cos^2y$ now gives $I \; = \; \sqrt{2}\int_1^0 \frac{1}{\sqrt{1+u}} \times \frac{-du}{2\sqrt{u(1-u)}} \; = \; \frac{1}{\sqrt{2}}\int_0^1 \frac{du}{\sqrt{u(1-u^2)}}$ and, finally, the substitution $v = u^2$ gives us $I \; = \; \frac{1}{2\sqrt{2}}\int_0^1 \frac{dv}{(1-v)^{\frac12}v^{\frac34}} \; = \; \tfrac{1}{2\sqrt{2}}B\big(\tfrac14,\tfrac12\big)$ Thus $a^2 = bc = 8$ , and so the answer is $\boxed{1}$ .

Solving $\displaystyle \int ^ \pi _ 0 \frac { dx } { \sqrt { 3 + \cos x } } = \frac 1aB ( \frac 1b, \frac 1c )$ , then we get $\displaystyle a ^ 2 = bc$ , so $\displaystyle \frac { a ^ 2 } { bc } = \boxed1$