# 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.

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Substitute what? – J. M. May 8 '11 at 2:23
@J. M. - I apologize for being unclear. I edited/clarified the question. – Sir Winford May 8 '11 at 2:35
It's separable; you will end up with $2\pi\left(\int_0^\pi\sin\,\phi\mathrm d\phi\right)\left(\int_0^2 \frac{p^2}{\sqrt{p^2+3}}\mathrm dp\right)$... – J. M. May 8 '11 at 2:54
@Sean: don't forget to transform the measure! In fact the substitution $p=\sinh \sqrt{3} x$ helps... – Fabian May 8 '11 at 7:36
@Sean Carruthers: no. If you substitute $u=p^2+3$ then you have to be careful to not forget the term $\frac{du}{dp}=2p$ (en.wikipedia.org/wiki/Integration_by_substitution). – Fabian May 9 '11 at 6:02

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

$$I=4\pi \left( \sqrt{7}-\frac{3}{2}\ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right) \right) .$$

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