Study of the function $g(\theta)$ for relativistic aspects

Question

Given this function, where $\beta=v/c$ with $v$ the magnitude of the velocity of a charge $q$, $k_e$ without the dimension is $\approx 9\cdot 10^9$,

$$g(\theta)=f(\theta)\sin \theta=\left(k_e \,q\,\frac{1-\beta^{2}}{\left[1-\beta^{2}\sin^2\theta\right]^{\tfrac{3}{2}}}\right)\sin \theta$$ how I can see that if I study the function $g(\theta)$ for $v$ near $c$, after is strongly peaked near $θ_0=π/2$?

Surely I should do the study of the function $g(\theta)$, but there is a quick way to verify that $g(\theta)$ is it strongly peaked near $θ_0=π/2$?

Related question: https://physics.stackexchange.com/questions/521479/gauss-theorem-in-relativistic-conditions

Answer 1

I'll assume that $\Phi(\theta)$ is really supposed to be $\Phi(\theta_0)$. Taking a look at the integrand, we can approximate

$$(1-\beta^2\sin^2\theta)^{\frac{3}{2}} \approx \cos^3\theta$$

So the integral approximately evaluates to

$$\Phi(\theta_0) \approx \frac{1}{2\gamma^2 \epsilon_0}\int_0^{\theta_0} \frac{\sin\theta}{\cos^3\theta}d\theta = \frac{\sec^2(\theta_0) - 1}{4\gamma^2 \epsilon_0} = \frac{\tan^2(\theta_0)}{4\gamma^2 \epsilon_0}$$

which goes to infinity at $\theta_0 = \frac{\pi}{2}$.