Hint: Showing $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$ converges in $L^{p}$ to $x^{-\frac{3}{2}}\sin{(x)}$ Let $f_{n}:]0, \infty[\to \mathbb R$ and $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$
Show that $f_{n}$ converges in $L^{p}$ to $f$ where $f(x):=x^{-\frac{3}{2}}\sin{(x)}$  and $p \in [1,2[$
My idea:
\begin{align}
\lVert f_{n}-f\rVert_{p}^{p}&=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}-x^{-\frac{3}{2}}\sin{(x)}\right\rvert^{p}dx\leq\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}}{1+nx^{2}}-x^{-\frac{3}{2}}\right\rvert^{p}dx \\
&=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}-x^{-\frac{3}{2}}(1+nx^{2})}{1+nx^{2}}\right\rvert^{p}dx=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}-x^{-\frac{3}{2}}+nx^{-\frac{1}{2}}}{1+nx^{2}}\right\rvert^{p}dx
\end{align} 
Am I on the right track? How do I continue from here?

First notice that
$$\frac{n\sqrt{x}\sin x}{1+nx^2} = \frac{\sqrt{x}\sin x}{\frac1n+x^2} \xrightarrow{n\to\infty} x^{-3/2}\sin x$$
pointwise.
Then notice that
$$\left|\frac{\sqrt{x}\sin x}{\frac1n+x^2}\right|^p \le \frac1{x^{p/2}}\chi_{(0,1]}(x) + \frac1{x^{3p/2}}\chi_{[1,\infty)}(x) =: g(x) \in L^1(0,\infty)$$
and also $|x^{-3/2}\sin x|^p$ is dominated by the same integrable function $g$.
Hence
$$\left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\le 2^p g(x)$$
so Lebesgue dominated convergence theorem yields
$$\lim_{n\to\infty}\int_0^\infty \left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\,dx = \int_0^\infty\lim_{n\to\infty} \left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\,dx = \int_0^\infty 0\,dx = 0$$
