The [url=http://en.wikipedia.org/wiki/Lagrangian_mechanics]lagrange equation[/url] for the system is $L=T-V = \tfrac{1}{2}m (L\dot\theta)^2+\tfrac{1}{2}m (L\sin\theta \dot{\phi})^2- (-mgL\cos\theta)$

from $\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}})-\frac{\partial L}{\partial \phi}=0$
you will find $m L^2 \sin\theta^2 \dot{\phi}=const$

The reason is:
since there is no $\phi$ in lagrange equation so $\frac{\partial L}{\partial \phi}=0$
which mean $\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}})=0$ ,
so $(\frac{\partial L}{\partial \dot{\phi}})$ must be a constant. i.e. $m L^2 \sin\theta^2 \dot{\phi}=const$