At a time $t$, let: :the rod be at an angle $\theta$ to the vertical :the bob be travelling at a speed $v$ :the displacement of the bob from where it is when the rod is vertical, along its line of travel, be $s$. :350px At its maximum displacement, the speed of the bob is $0$, so its kinetic energy is $0$. By the Principle of Conservation of Energy: :$\dfrac 1 2 m v^2 = m g \paren {a \cos \theta - a \cos \alpha}$ We have that: :$s = a \theta$ :$v = \dfrac {\d s} {\d t} = a \dfrac {\d \theta} {\d t}$ The rate of change of $s$ at time $t$ is the speed of the bob. So: {{begin-eqn}} {{eqn | l = v       | r = \frac {\d s} {\d t} = a \frac {\d \theta} {\d t}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {a^2} 2 \paren {\frac {\d \theta} {\d t} }^2       | r = g a \paren {a \cos \theta - a \cos \alpha}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d \theta} {\d t}       | r = -\sqrt {\frac {2 g \paren {\cos \theta - \cos \alpha} } a}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d t} {\d \theta}       | r = -\sqrt {\frac a {2 g \paren {\cos \theta - \cos \alpha} } }       | c =  }} {{eqn | ll= \leadsto       | l = \frac T 4       | r = -\sqrt {\frac a {2 g} } \int_\alpha^0 \frac {\d \theta} {\sqrt {\cos \theta - \cos \alpha} }       | c =  }} {{eqn | r = \sqrt {\frac a {2 g} } \int_0^\alpha \frac {\d \theta} {\sqrt {\cos \theta - \cos \alpha} }       | c =  }} {{end-eqn}} Substituting: :$\cos \theta = 1 - 2 \sin^2 \dfrac \theta 2$ :$\cos \alpha = 1 - 2 \sin^2 \dfrac \alpha 2$ we get: :$\displaystyle T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {1 - 2 \sin^2 \frac \theta 2 - 1 + 2 \sin^2 \frac \alpha 2} } } = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {\sin^2 \frac \alpha 2 - \sin^2 \frac \theta 2} } }$ We now put $k = \sin \dfrac \alpha 2$: :$\displaystyle T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {k^2 - \sin^2 \frac \theta 2} } }$ Next, let us introduce the variable $\phi$, such that: :$\sin \dfrac \theta 2 = k \sin \phi$ and where $\phi$ goes from $0 \to \pi / 2$ as $\theta$ goes from $0 \to \alpha$. Differentiating with respect to $\phi$ we have: :$\dfrac 1 2 \cos \dfrac \theta 2 \dfrac {\d \theta} {\d {\phi}} = k \cos \phi$ Thus: {{begin-eqn}} {{eqn | l = \d \theta       | r = \frac {2 k \cos \phi} {\cos \frac \theta 2} \rd \phi       | c =  }} {{eqn | r = \frac {2 k \sqrt {1 - \sin^2 \phi} } {\sqrt {1 - \sin^2 \frac \theta 2} } \rd \phi       | c =  }} {{eqn | r = \frac {2 \sqrt {k^2 - \sin^2 \frac \theta 2} } {\sqrt {1 - k^2 \sin^2 \phi} } \rd \phi       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = T       | r = 2 \sqrt {\frac a g} \int_0^{\pi / 2} \frac {2 \sqrt {k^2 - \sin^2 \frac \theta 2} } {\sqrt {1 - k^2 \sin^2 \phi} } \frac {\d \phi} {\sqrt {k^2 - \sin^2 \frac \theta 2} }       | c = }} {{eqn | r = 4 \sqrt {\frac a g} \int_0^{\pi / 2} \frac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} }       | c =  }} {{end-eqn}} The integral: :$\displaystyle \int_0^{\pi / 2} {\frac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} } }$ is the complete elliptic integral of the first kind and is a function of $k$, defined on the interval $0 < k < 1$. Hence the result. {{qed}}