Uniform convergence of ${f_n(x)=\int_{0}^{x}g_n(t)}dt$ where $g_n(x)=\sin^2(x+\frac{1}{n})$.

Question

I need to prove that $${f_n(x)=\int_{0}^{x}g_n(t)}dt$$ where $g_n(x)=\sin^2(x+\frac{1}{n})$ is uniformly convergent on

1. $[0,\infty)$
2. $[0,1]$

How can I do this? Is it correct to argue that if $g_n(x)$ is uniformly convergent then $f_n$ is also uniformly convergent? How can I find some $N$ s.t $\forall n>N$ it is true? Any help please.

Edit So I compute $f_n(x)$ and got

$$\lim_{n\to \infty}f_n(x)=\frac{1}{2}\left(x-\frac{\sin2x}{2}\right)$$