hint
$$\sin^2(t+\frac 1n)=\frac{1-\cos(2t+\frac 2n)}{2}$$
$$f_n(x)=\int_0^x\sin^2(t+\frac 1n)dt=\frac{1}{2}\Big[t-\frac{\sin(2t+\frac 2n)}{2}\Bigr]_0^x$$
$$=\frac x2-\frac 14\sin(2x+\frac 2n)+\frac 14\sin(\frac 2n)$$
The pointwise limit is
$$f(x)=\frac x2-\frac{\sin(2x)}{4}$$
For the uniform convergence, use MVT to get
$$|\sin(2x+\frac 2n)-\sin(2x)|\le \frac 2n$$
and
$$|f_n(x)-f(x)|\le \frac 1n.$$
The convergence is uniform at $[0,\infty)$.
Suppose $g_n: [0,1] \longrightarrow [0,1]$, $\,f_n:[0,1] \longrightarrow \mathbb R$ $(n=1,2,\ldots)$ are two sequences of continuous functions that converge uniformly to $g$ and $f$, respectively.
Let $\varepsilon>0$ be given.
By the uniform limit theorem and by the Heine-Cantor theorem, $f$ is uniformly continuous on $[0,1]$. Thus there is $\delta>0$ so that $|\,f(u)-f(v)|<\frac12 \varepsilon$ whenever $|u-v|<\delta$ and $u,v \in [0,1]$. Since $g_n \to g$ uniformly, there is a positive integer $N_1$ so that for all $x \in [0,1]$ we have
$$|g_n(x)-g(x)|<\delta \text{ whenever } n \geq N_1.$$
Since $f_n \to f$ uniformly, there is a positive integer $N_2$ so that for all $x \in [0,1]$ we have
$$|\,f_n(x)-f(x)|<\frac{\varepsilon}{2} \text{ whenever} n \geq N_2.$$
Set $N=\max\{N_1,N_2\}$. So if $n \geq N$ and $x \in [0,1]$, then we have
\begin{equation} \begin{split} | \,f_n(g_n(x)) - f(g(x))| &= |\,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\\
& \leq |\,f_n(g_n(x)) - f(g_n(x))|+|\,f(g_n(x)) - f(g(x))| \\
& <\frac{\varepsilon}{2} +\frac{\varepsilon}{2}\\
& = \varepsilon.
\end{split}\end{equation}
Therefore, $f_n \circ g_n \to f \circ g$ uniformly on $[0,1]$ (definition of uniform convergence).
Best Answer
Let $f_n(x)=nx$ for $0 \leq x \leq 1/n$, $2-nx$ for $ 1/n \leq x \leq 2/n$ and $0$ for $x >2/n$. $g_n(x)=\frac 1 n$. Let $f(x)=0$ for all $x$ and $g(x)=0$ for all $x$. Then $f_n(g_n(x))\to 1\neq f(g(x))$.