The last integrand I absolutely have no idea how to solve. I tried to use the euler formula but to no avail.
$$\frac { 1 }{ (2\pi ^{ 2 }a^{ 3 }\hbar^{ 3 })^{ 1/2 } } \int _{ 0 }^{ \infty }{ r^{ 2 }dr \;e^{-r/a} \int _{ -1 }^{ 1 }{ d(\cos\theta)\;e^{-ipr \cos\theta/\hbar}} } $$ Screenshot of the integrand The next step in the integration process is
If I understand the question correctly, the question is about the integral $$ \int_{-1}^1 d(\cos \theta) e^{ipr (\cos \theta)/\hbar} $$
This is taken a little out of context (there may be text or formulas elsewhere in the same source that explain what this integral means), but a reasonable interpretation seems to be that it means
$$ \int_{-1}^1 e^{ipr u/\hbar} du $$
where $u = \cos\theta$. The reason for using the first form of the integral might be that whoever wrote this wanted the effect of a "$u$-substitution" but wanted to avoid having to break up the flow of equations in order to define the "$u$" they intended to use. So rather than define $u = \cos\theta$ and write the integral over the variable $u$, they wrote the integral as if it were an integral over $u$ where $u = \cos\theta$, but simply wrote $\cos\theta$ in each place where the $u$-substituted integral would have had $u$.
An examination of the lines before and after the integral (to see what other formulas it comes from, and how it was solved) could confirm or deny whether this interpretation was meant.