Compute $$ \int_{0}^{2\pi} d\phi \ \delta(\sin^2\phi-a^2) \ \tan^2\phi, $$ where $\delta(x)$ is the Dirac delta function and $0<a^2<1$.
My attempt:
$$ \int_{0}^{2\pi} d\phi \ \delta(\sin^2\phi-a^2) \ \tan^2\phi = \frac{1}{2|a|}\int_{0}^{2\pi} d\phi \left( \delta(\sin\phi-a) + \delta(\sin\phi+a) \right) \tan^2\phi. $$
I guess I should use some change of variable, but I'm stuck here. Any help?
On
Hints: First notice that the integrand is even and $\pi$-periodic. Next use integration by substitution.
$$\begin{align}\int_{-\pi}^{\pi} \!\mathrm{d}\phi ~ \delta(\sin^2\phi-a^2) ~ \tan^2\phi ~=~~~~&4\int_0^{\pi/2} \!\mathrm{d}\phi ~ \delta(\sin^2\phi-a^2) ~ \tan^2\phi\cr ~\stackrel{s=\sin^2\phi}{=}~&2\int_0^1 \!\mathrm{d}s ~ \delta(s-a^2) ~ \frac{s^{1/2}}{(1-s)^{3/2}}\cr ~\stackrel{0<a^2<1}{=}~&\frac{2|a|}{(1-a^2)^{3/2}}. \end{align}$$
The formula that you used is a corollary of a more general form. $$\int_{-\infty}^\infty\delta(g(x))f(x)dx=\sum_i\frac{f(x_i)}{g'(x_i)}$$ Here $x_i$ are the roots of $g(x)$. In your case, the roots of $\sin^2\phi-a^2$ are at $$\phi\in\{\arcsin a, \pi-\arcsin a,\pi+\arcsin a,2\pi-\arcsin a\}$$ The derivative of $g$ is $2\sin\phi\cos\phi$.