$\lim_{n\to\infty}\int_0^{\pi/2} \frac{\sin^n(x)}{1+x^2} \, dx$ $$\lim_{n\to\infty}\int_0^{\pi/2} \frac{\sin^n(x)}{1+x^2} \, dx$$
Is it right answer ?
$$
 \begin{cases}
  0, & x \ne \pi/2\\[8pt]
  \dfrac{2\pi}{4 + \pi^2}, & x = \pi/2
 \end{cases}
$$

Consider the sequence of functions $f_n (x) = {\sin^n (x)}/({1+x^2})$ on $[0,2\pi]$
If $a \neq \pi/2, 3\pi/2$ for some $k\in\mathbb{Z}$, then $|\sin(a)| < 1$, so as $n \to \infty$ we have $f_n (a) \to 0$. Thus, pointwise almost everywhere, as $n\to\infty$ we have $f_n \to 0$. 
Let $g(x) = 1/(1+x^2)$. Note that $|f_n (x)| \le |g(x)|$ and $g(x)$ is integrable on $[0,2\pi]$. Hence, by the dominated convergence theorem, it follows $$\lim_{n\to\infty} \int_{0}^{\pi/2} \frac{\sin^n (x)}{1+x^2} \, dx = \lim_{n\to\infty} \int_{0}^{\pi/2} f_n (x) \, dx = \int_{0}^{\pi/2} \lim_{n\to\infty} f_n (x) \, dx = \int_{0}^{\pi/2} 0 \, dx = 0.$$
