Advanced integration problem The solution to Schrodinger's equation are wave functions $\Psi (r,\theta ,\phi )$ of the form, $\Psi (r,\theta ,\phi )= R(r)\Theta(\theta)\Phi(\phi)$
Where, the probability of finding an electron in a region, V:$0<r<\infty     ,    0<\theta <\pi     ,    0<\phi<2\pi $,   is found using the volume integral;
$$N^2\int_{0}^{\infty }r^2R^2(r)dr\int_{0}^{2\pi }\Phi^2(\phi)d\phi\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta =1$$
Now, consider the $2p_{y}$ orbital;
                                                         $\Psi _{2p_{y}}=\frac{1}{4\sqrt{2\pi a_{0}^{5}}}re^{-\frac{r}{2a_{0}}}sin \theta sin \phi$
And now the question itself;
Evaluate the three integrals,
One. $\int_{0}^{\infty }r^2R^2(r)dr$ 
Two. $\int_{0}^{2\pi }\Phi^2(\phi)d\phi$
Three. $\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta$
and show $\int_{V}^{}\Psi ^2dV=1$
I understand this may be alot but any help would be greatly appreciated. Integration is something I really need to work on. Thanks guys.
 A: From your expression for $\Psi$ one can deduce
\begin{align}
R(r)           &= re^{-r/(2a_0)} \\
\Phi(\phi)     &= \sin\phi       \\
\Theta(\theta) &= \sin\theta     \\
N              &= \frac{1}{4\sqrt{2\pi a_0^5}}
\end{align}
so that
\begin{align}
&\int_0^{\infty}r^2R^2(r)dr = \int_0^{\infty}r^4e^{-r/a_0}dr = \\
&\quad= -a_0^5\left.\left[
      \left(\frac{r}{a_0}\right)^4
      +4\left(\frac{r}{a_0}\right)^3
      +12\left(\frac{r}{a_0}\right)^2
      +24\left(\frac{r}{a_0}\right)
      +24
    \right]e^{-r/a_0}\right|_0^{\infty}=24a_0^5\\
&\int_0^{2\pi}\Phi^2(\phi)d\phi=\int_0^{2\pi}\sin^2\phi d\phi=\\
&\quad=\left.\frac{1}{2}(\phi-\sin\phi\cos\phi)\right|_0^{2\pi}=\pi\\
&\int_0^{\pi}\Theta^2(\theta)\sin\theta d\theta=\int_0^{\pi}\sin^3\theta d\theta=\\
&\quad=\left.\left(-\cos\theta+\frac{1}{3}\cos^3\theta\right)\right|_0^{\pi}=\frac{4}{3}
\end{align}
Putting all together:
$$
N^2\int_0^{\infty}r^2R^2(r)dr\int_0^{2\pi}\Phi^2(\phi)d\phi\int_0^{\pi}\Theta^2(\theta)\sin\theta d\theta=\frac{1}{16(2\pi a_0^5)}\cdot24a_0^5\cdot\pi\cdot\frac{4}{3}=1
$$
