How to prove this definite integral does not depend on the parameter?
Put \begin{equation*} I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1+\sin^2\theta}{(\cos^4\theta +(\gamma\cos^2\theta-\sin \theta)^2)^{\frac{3}{4}}}\, d\theta \end{equation*} If $x = \dfrac{\sin\theta}{\cos^2\theta}$, $\, y = \gamma-x$ and $y = \sqrt{z}$ then \begin{equation*} dx = \dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta}\, d\theta \end{equation*} and \begin{gather*} I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta\left(1 +\left(\gamma-\frac{\sin \theta}{\cos^2\theta}\right)^2\right)^{\frac{3}{4}}}\, d\theta = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +\left(\gamma- x\right)^2\right)^{\frac{3}{4}}}\, dx = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +y^2\right)^{\frac{3}{4}}}\, dy = \\[2ex] \int_{0}^{\infty}\dfrac{z^{\frac{1}{2}-1}}{(1+z)^{\frac{1}{2}+\frac{1}{4}}}\, dz = {\rm B}\left(\frac{1}{4},\frac{1}{2}\right) \approx 5.244115109 \end{gather*} where ${\rm B}$ is the Beta function.
This is not an answer, but too long for a comment.
I still think it may be helpful to consider the derivative (I use $x$ as integration variable).
$$J=\frac{d}{d \gamma} I(\gamma)=\int_{-\pi/2}^{\pi/2} \frac{(1+\sin^2 x)(\gamma ~\cos^2 x-\sin x) \cos^2 x~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}$$
We need to prove $J \equiv 0$.
However, there's a stronger statement $^*$, which seems to be true numerically:
$$J_1=\int_{-\pi/2}^{\pi/2} \frac{\gamma (1+\sin^2 x) \cos^4 x~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}$$
$$J_2=\int_{-\pi/2}^{\pi/2} \frac{(1+\sin^2 x) \cos^2 x \sin x ~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}$$
$$J_1 \equiv J_2$$
Mathematica confirms it numerically for $\gamma \in (-1,1)$. For $|\gamma|>1$ there's some trouble with computing the integrals numerically.
Moreover, as we see, both integrals follow linear dependence on $\gamma$ with very good accuracy.
Using linear regression in Mathematica, I obtained, with amazing accuracy the following fit:
$$J_1(\gamma)= \frac{2}{3} L \gamma $$
Where $L$ is a lemniscate constant:
$$\frac{2}{3} L = \frac{1}{3 \sqrt{2 \pi}} \left( \Gamma \left( \frac{1}{4} \right) \right)^2 =\frac{4}{3} \int_0^1 \frac{dx}{\sqrt{1-x^4}}$$
Here are the plots with the proposed closed form (green) for comparison:
This gives us another task, to prove the proposed closed form for both integrals.
Note that for $J_1$ we can divide by $\gamma$ and obtain the following proposition:
$$\int_{-\pi/2}^{\pi/2} \frac{ (1+\sin^2 x) \cos^4 x~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}=\frac{4}{3} \int_0^1 \frac{dx}{\sqrt{1-x^4}}$$
Which makes $J_1/ \gamma$ yet another integral which doesn't depend on $\gamma$.
Edit, remark:
$^*$ this statement is not really stronger, because as we can see, the rest of the integrated function is non-negative for all $x$, which means that $J_1 \equiv J_2$ follows from $J \equiv 0$.