Integral $\int_{\pi/2+\delta}^{3\pi/2-\delta} x^{R \cos \varphi} d \varphi$ bounded

Question

This is probably a silly question, or maybe I am missing a very simple slick trick, but I am trying to see how the following integral is bounded in terms of $\delta$: \begin{equation} \int_{\pi/2+\delta}^{3\pi/2-\delta} x^{R \cos \varphi} d \varphi. \end{equation} $R >0$ is just an arbitrary real number. (This is all to show that a contour integration vanishes). I have tried substitution to no avail as well as absolute value considerations. Any ideas?

EDIT: The only info about $x$ is that $x > 1$. (I think this is supposed to be a straightforward exercise in complex integration but I am a bit rusty, my apologies. The integral is \begin{equation} \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{x^s}{s} ds.) \end{equation}

Answer 1

Let $a=|\cos(\pi/2+\delta)|>0$. Then $$ \frac\pi2+\delta\le \varphi\le \frac{3\,\pi}2-\delta\implies-1\le\cos\varphi\le-a $$ and $$ 0<\int_{\pi/2+\delta}^{3\pi/2-\delta}x^{R\cos\varphi}\,d\varphi\le(\pi-2\,\delta)x^{-aR}. $$ Since $x>1$, the last expression converges to $0$ as $R\to\infty$.

Answer 2

If it is the real case...

Because zeros of $\cos x$ are cut off, the function under integral is bounded, the interval is bounded, so...