Find the limit $$\lim_{n\to\infty} \int_0^{2\pi} (x+\frac{\pi}{2})^2 \frac{\sin((n+\frac{1}{2})x + x\cos nx}{\sin\frac{x}{2}}\ dx$$
So we may define $f = (x+\frac{\pi}{2})^2$ and then look at the convolution:
$$\int_0^{2\pi}(x+\frac{\pi}{2})^2 \frac{\sin((n+\frac{1}{2})x}{\sin\frac{x}{2}}\ dx = 2\pi f\star D_n$$ where $D_n$ is Dirichlet kernel
Now, we know that $f\star D_n = S_N(f, x)$, the partial sum of the Fourier series of $f(x)$.
So basically at this point I considered evaluating the Fourier coefficients of $f(x)$ but it got relatively complicated so I guess I should take another approach.
As for the other part of the integral, I think we may look at $$g(x)=\frac{(x+\frac{\pi}{2})^2}{\sin\frac{\pi}{2}}$$
And from Riemann–Lebesgue lemma we have that: $$\lim_{n\to\infty}\int_0^{2\pi} g(x)\cos nx = 0$$
That's not quite right. In your equation for the convolution, the expression on the left is a number and the expression on the right is a function. Let's fix this: Set $h(x) = (\pi/2-x)^2.$ Then your integral equals
$$\int_0^{2\pi}h(0-x) D_n(x)\,dx = 2\pi h * D_n (0).$$
So that makes sense, but I don't really see how to use this here. Note that as a $2\pi$-periodic function, $h$ has a jump discontinuity at $0 \sim 2\pi.$
Why not just do the following:
$$\int_0^{2\pi}(x+\pi/2)^2 D_n(x)\,dx = \int_0^{2\pi}[x^2 + x\pi + (\pi^2/4)] D_n(x)\,dx.$$
Now $D_n(x) = \sum_{k=-n}^{n}e^{ikx}.$ The integrals $\int_0^{2\pi}xe^{ikx}\,dx, \int_0^{2\pi}x^2e^{ikx}\,dx$ don't look that bad to me, and of course integrating $D_n$ against a constant is easy.