I've found on Wikipedia that energy $E$ and angular momentum $L$ of a particle are conserved quantities in Schwarzschild metric. It's written:
$$L=mr^2 \frac {d\phi} {d\tau},$$
$$E=mc^2\left(1-\frac{r_s}{r}\right)\frac{dt}{d\tau}.$$
And, from the metric, it finds these results:
$$\left(\frac{d\phi}{d\tau}\right)^2=\frac{L^2}{m^2r^4}$$
$$\left(\frac{dr}{d\tau}\right)^2=\frac{E^2}{m^2c^2}-\left(1-\frac{r_s}{r}\right)\left(c^2+\frac{L^2}{m^2r^2}\right).$$
$$\left(\frac{dt}{d\tau}\right)^2=\frac{E}{\left(1-\frac{r_s}{r}\right)mc^2}$$
That I need is to get the results including also the $\theta$ coordinate. So I tried this:
$$p_\phi=mr^2 \frac {d\phi} {d\tau}\qquad\qquad p_\theta=mr^2 \frac {d\theta} {d\tau}$$
$$L^2=p_\theta^2 + \sin^2\theta\ p_\phi^2.$$
$E$ is the same?
$$\left(\frac{d\phi}{d\tau}\right)^2=\frac{p_\phi^2}{m^2r^4}\qquad\qquad \left(\frac{d\theta}{d\tau}\right)^2=\frac{p_\theta^2}{m^2r^4}.$$
$\frac{dr}{d\tau}$ is the same?
$\frac{dt}{d\tau}$ is the same?
But I'm not sure. However, is it possible to get a similar result adding the $\theta$ coordinate?
PS: I am a beginner about GR, so I don't know many things about it.