complex analysis - Show $int_{0}^infty frac{x^q}{x^p + 1} dx = frac{pi}{p}csc frac{pi(q+1)}{p}$, where $p>q+1$.

Suppose that $p,q$ are positive integers with $p>q+1$. By integrating around the boundary of the region $\{z\in\mathbb{C}:|z|$$\int_{0}^\infty \frac{x^q}{x^p + 1} dx = \frac{\pi}{p}\csc \frac{\pi(q+1)}{p}.$$

My attempt

I feel like I know the general procedure for this... but I feel a bit lost.

Let $C$ be the join of the contours $C_1,C_2,C_3$ where $C_1$ is the interval $[0,R]$, $C_2$ is the arc of the circle of radius $R$, with inclination angle $\frac{2\pi}{p}$, and $C_3$ is the line connecting the origin to $Re^{\frac{2\pi}{p}}$.

Define $f(z) = \frac{z^q}{z^p+1}$.
We would like to calculate $$\int_{C}f(z)dz.$$
The roots to $z^p + 1$ are $z= e^{i\mathrm{Arg(z)}}$, where
$\mathrm{Arg(z)} = \frac{\pi}{p},\frac{-\pi}{p},\frac{3\pi}{p},\frac{-\pi}{p},\ldots$

Suppose $R>1$. Then the only singularity of $f$ inside the region is $z=e^{\frac{i\pi}{p}}$.

Thus, the residue of $f$ at this singularity is
$$\mathrm{Res}(f(z),e^{\frac{i\pi}{p}}) = \lim_{z\rightarrow e^{\frac{i\pi}{p}}} \frac{z^q}{z^p + 1} \\ = \frac{(e^{\frac{i\pi}{p}})^q}{p(e^{\frac{i\pi}{p}})^{p-1}} \\ = \frac{1}{p} e^{i\pi\left(\frac{1-(p-q)}{p}\right)}.$$

Now looking at my answer, I'm not sure how to involve a $\csc$ there, or even go ahead with the problem.