$$\int_0^{\frac\pi{2}}\frac {\cos\left((1-2n)\arcsin\left(\frac{\sin(\theta)}{\sqrt 2}\right)\right)}{\sqrt{(1-\frac{\sin^2 \theta}{2})}}d\theta$$
I tried using some sort of substitutions but I think this must have some other way to solve and gave me another different integral and gamma functions and all, which now I'm uncertain if it's my cup of tea!
$$\begin{align*} \int_0^{\frac\pi{2}}\frac {\cos\left((1-2n)\arcsin\left(\frac{\sin(\theta)}{\sqrt 2}\right)\right)}{\sqrt{(1-\frac{\sin^2 \theta}{2})}}d\theta & = \frac{\pi}{2}{_2F_1}{(1-n, n;1;\frac 1{2})}\\ & = \frac {\pi}{2}\frac {\sqrt \pi}{\Gamma(1-\frac n{2})\Gamma(\frac 1{2} + \frac n{2})}\\ & = \frac {\pi}{2}P_{-n}(1-2x)|_{x = \frac 1{2}}\\ \end{align*}$$
In reply to the comment:
$$\left|\int_0^{\sin^{-1}\sqrt x} \frac {\cos\left((1-2n)\phi\right)}{\sqrt{(x-\sin^2\phi)}}d\phi = \frac {\pi}{2}{_2F_1}(1-n,n;1;x)\right|_{x = \frac 1{2}}$$
Another way to look at the solution
$$(1-y^2)^{-1/2}\cos(2n\sin^{-1} y) ={_2F_1}(\frac 1{2}+n, \frac 1{2}-n;\frac 1{2};y^2)$$$(y, 2n) ≡ (\sqrt x \sin \theta, 1-2n)$ & integrating w.r.t $\theta$ over $(0, \pi/2)$