How to determine that this series is conditionally convergent?

Notice that $\sin(\pi x)$ doesn't change its sign on the interval $[n,n+1]$ since its zeros are $x=k\in\mathbb{Z}$. By the mean value theorem there's a $c_n\in[n,n+1]$ such that $$\int_n^{n+1}\frac{\sin(\pi x)}{x^p+1}\,dx=\frac{1}{(c_n)^p+1}\int_n^{n+1}\sin(\pi x)\,dx $$ For the last integral, we have $$\int_n^{n+1}\sin(\pi x)\,dx =-\frac{1}{\pi}\cos(\pi x)\Bigg\vert_n^{n+1}\\ =-\frac{1}{\pi}\left(\cos((n+1)\pi)-\cos(n\pi)\right)=\frac{2\cdot(-1)^n}{\pi}$$ so you have to establish the convergence of $$\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{(c_n)^p+1} $$ which follows from the alternating series test, and the divergence of $$\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{(c_n)^p+1} $$ which follows from a comparison with the divergent series $\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{(n+1)^p+1}=\frac{2}{\pi}\sum_{n=2}^{\infty}\frac{1}{n^p+1} $.