How to calculate the intensity of the interference of two waves in a given point? There are two different point sources which produce spherical waves with the same power, amplitude, ω, wavenumber and phase.
I can calculate the intensity of each wave in a point:
$$
    I_1 = P / (4 \pi r_1^2)
$$
$$
    I_2 = P / (4 \pi r_2^2)
$$
But how can I calculate the intensity of the resulting wave in that point?

I found the solution, but forgot to add it here. But since a considerable number of people has seen the question, the answer could be useful.
Notation


*

*$r_1$: distance between the given point and the focus 1.

*$r_2$: distance between the given point and the focus 2.

*$A_1$: amplitude in the given point of the wave generated by focus 1.

*$A_2$: amplitude in the given point of the wave generated by focus 2.

*$A_r$: amplitude in the given point of the resulting wave.

*$I_1$: intensity in the given point of the wave generated by focus 1.

*$I_2$: intensity in the given point of the wave generated by focus 2.

*$I_r$: intensity in the given point of the resulting wave.

*$P$: power of the initial waves.

*$\varphi$: phase difference between the initial waves in the given point.

*$k$: wavenumber of the initial and resulting waves.

*$\omega$: angular frequency of the initial and resulting waves.

*$v$ : velocity of the initial and resulting waves.


Preface
I will use the formula
$$
    I = \frac{1}{2}\rho v \omega^2 A^2
$$
The resulting wave will be a wave of amplitude $A_r$ with the same $k$ and $\omega$ than the initial waves. Moreover, $\rho$ and $v$ will also be the same because they only depend on the environment.
Then,
$$
\begin{cases}
I_r = \frac{1}{2}\rho v \omega^2 A_r^2 \\
I_2 = \frac{1}{2}\rho v \omega^2 A_2^2 \\
\end{cases}
\implies
I_r
= I_2 \left(\frac{A_r}{A_2}\right)^2
= \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2
$$
In order to express $I_r$ in terms of $r_1$ and $r_2$ instead of $A_1$ and $A_2$, I will use that the amplitude of an spherical wave is inversely proportional to the distance to the focus. That is:
$$
\frac{A_1}{A_2} = \frac{r_2}{r_1}
$$ 
Answer
In a point with destructive interference ($\varphi = \pi$)
The resulting amplitude will be the difference of amplitudes:
$$
    A_r = |A_1 - A_2|
$$
Then, the resulting intensity is
$$
I_r
= \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2
= \frac{P}{4\pi r_2^2} \left(\frac{A_1 - A_2}{A_2}\right)^2
= \frac{P}{4\pi r_2^2} \left(\frac{A_1}{A_2}-1\right)^2
= \frac{P}{4\pi r_2^2} \left(\frac{r_2}{r_1}-1\right)^2
= \frac{P}{4\pi} \left(\frac{r_2-r_1}{r_1 r_2}\right)^2
$$
In a point with constructive interference ($\varphi = 0$)
The resulting amplitude will be the sum of amplitudes:
$$
    A_r = A_1 + A_2
$$
Then, the resulting intensity is
$$
I_r
= \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2
= \frac{P}{4\pi r_2^2} \left(\frac{A_1 + A_2}{A_2}\right)^2
= \frac{P}{4\pi r_2^2} \left(\frac{A_1}{A_2}+1\right)^2
= \frac{P}{4\pi r_2^2} \left(\frac{r_2}{r_1}+1\right)^2
= \frac{P}{4\pi} \left(\frac{r_1+r_2}{r_1 r_2}\right)^2
$$
In general
The resulting amplitude will be ($\varphi$ is the phase difference):
$$
A_r
= \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos{\varphi}}
$$
Then, the resulting intensity is
$$
I_r
= \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2
= \frac{P}{4\pi r_2^2} \frac{A_1^2 + A_2^2 + 2 A_1 A_2 \cos{\varphi}}{A_2^2}
= \frac{P}{4\pi r_2^2} \left(\left(\frac{A_1}{A_2}\right)^2 + 1 + 2 \frac{A_1}{A_2} \cos{\varphi}\right)
= \frac{P}{4\pi r_2^2} \left(\left(\frac{r_2}{r_1}\right)^2 + 1 + 2 \frac{r_2}{r_1} \cos{\varphi}\right)
= \frac{P}{4\pi} \frac{r_1^2 + r_2^2 + 2 r_1 r_2 \cos{\varphi}}{(r_1 r_2)^2}
$$
