So I've been going through some material in acoustics and came across the this integral $$ \int \left( \frac{\sin[\frac{1}{2} k L \sin\theta]}{\frac{1}{2} k L \sin\theta} \right)^2 d\Omega $$
I've gone through two sources both that assert that doing the substitution $v=\frac{1}{2} k L \sin(\theta)$ gives you
$$ \frac{8\pi}{kL}\int_0^{kL/2} \left( \frac{\sin v}{v} \right)^2 dv $$ I know that $d\Omega = \sin\theta \: d\theta d\phi$, giving me
$$ 2\pi\int_0^{\pi/2} \left( \frac{\sin[\frac{1}{2} k L \sin\theta]}{\frac{1}{2} k L \sin\theta} \right)^2 \sin\theta d\theta $$ From the substitution point, I have $dv = \frac{1}{2} k L \cos\theta d\theta$, therefore $d\theta = \frac{2 dv}{k L \cos \theta}$. The bounds I have are now $0$ and $kL/2$. After performing the substitution, I now have $$ \frac{4\pi}{kL}\int_0^{kL/2} \left( \frac{\sin v}{v} \right)^2 \tan\theta dv= \frac{4\pi}{kL}\int_0^{kL/2} \left( \frac{\sin v}{v} \right)^2 \tan\left( \arcsin\left(\frac{2v}{kL}\right) \right) dv\\ = \frac{4\pi}{kL}\int_0^{kL/2} \left( \frac{\sin v}{v} \right)^2 \left( \frac{2v}{\sqrt{(kL)^2-4v^2}} \right) dv $$
Clearly where I end up is not matching what is asserted to be true, unless I'm missing something that necessitates that $\frac{2v}{\sqrt{(kL)^2-4v^2}} = 1$.
EDIT: $d\Omega = \sin \theta d\theta d\phi$
EDIT #2: Fixed the upper bound on the second integration equation presented