Trig substitution for a triple integral This problem involves calculating the triple integral of the following fraction, first with respect to $p$:
$$
\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^{2}  \frac{p^2\sin(\phi)}{\sqrt{p^2 + 3}} dp d\phi d\theta
$$
This involves trig-substitution (I believe), and I am just hoping for an explanation of how to do it.

Update:  I don't use a trig-substitution because I don't see which one could work. To integrate $p^{2}/\sqrt{p^{2}+3}$ I applied a technique derived from this more general one. 
If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that
$$\int \frac{P(x)}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x=Q(x)\sqrt{ax^{2}+bx+c}+\int \frac{C}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x.$$
(Described in Cálculo Integral em $\mathbb{R}$ by M. Olga Baptista.)

EDIT: simplified the exposition. The integral is separable, as already noted by J. M. in a comment. Assuming the limits of integration are as follows, we have
$$\begin{eqnarray*}
I &:&=\int_{\theta =0}^{2\pi }\int_{\phi =0}^{\pi
}\int_{p=0}^{2}\frac{p^{2}\sin (\phi )}{\sqrt{p^{2}+3}}\;\mathrm{d}p\;%
\mathrm{d}\phi \;\mathrm{d}\theta \\&=&\left( \int_{0}^{2\pi }\mathrm{d}\theta \right) \left(
\int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{%
p^{2}+3}}\;\mathrm{d}p\right) \\
&=&2\pi \left(
\int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{%
p^{2}+3}}\;\mathrm{d}p\right).
\end{eqnarray*}$$
The antiderivative of $\sin (\phi )$ is $-\cos \left( \phi \right) $; hence 
$$I =4\pi \int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p.$$
For the evaluation of this last integral we are going to apply the following property:
If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that
$$\int \frac{P(x)}{\sqrt{x^{2}+c}}\;\mathrm{d}x=Q(x)\sqrt{x^{2}+c}+\int \frac{C}{\sqrt{%
x^{2}+c}}\;\mathrm{d}x.$$
In the present case, we have $P(x)=x^{2}$ and $Q(x)$ is of the form $Q(x)=Ax+B$ 
$$\int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\left( Ax+B\right) \sqrt{%
x^{2}+c}+\int \frac{C}{\sqrt{x^{2}+c}}\;\mathrm{d}x.$$
Differentiating both sides, we get
$$\begin{eqnarray*}
\frac{x^{2}}{\sqrt{x^{2}+c}} &=&A\sqrt{x^{2}+c}+\frac{x\left( Ax+B\right) }{%
\sqrt{x^{2}+c}}+\frac{C}{\sqrt{x^{2}+c}} \\
&=&\frac{1}{\sqrt{x^{2}+c}}\left[ A\left( x^{2}+c\right) +x\left(
Ax+B\right) +C\right], 
\end{eqnarray*}$$
and so
$$x^{2} =A\left( x^{2}+c\right) +x\left( Ax+B\right) +C=2Ax^{2}+Bx+Ac+C.$$
Comparing coefficients, we get $A=1/2,\;B=0,\;C=-c/2$. Hence
$$\begin{eqnarray*}
\int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x &=&\frac{1}{2}x\sqrt{x^{2}+c}-%
\frac{c}{2}\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x \\
&=&\frac{1}{2}x\sqrt{x^{2}+c}-\frac{c}{2}\ln \left( x+\sqrt{x^{2}+c}\right) 
\end{eqnarray*}$$
because
$$\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\ln \left( x+\sqrt{x^{2}+c}%
\right)+\text{ constant}.$$
Thus
$$\int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p=\sqrt{7}-\frac{3}{2}%
 \ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right).$$
And finally
$$I=4\pi \left( \sqrt{7}-\frac{3}{2}\ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right) \right) .$$
