$g_n(x)=\sin^2(x+1/n)$, $f_n(x)=\int_0^x g_n(t)\,\text dt$, where $x\in[0,\infty)$. Question is whether $f_n(x)$ is converges uniformly on $[0, \infty )$? Clearly this question has been answered here but I really can't get that hint properly.
What I only guess that $g_n(x)$ converges pointwise to $\sin^2(x)$, and also know that if $g_n(x)$ would convergent uniformly to $\sin^2(x)$, then $\int_a^b g_n(t)\,\text dt$ converges to $\int_a^b \sin^2(t)\,\text dt$ where $a, b\in\mathbb R$. But I'm confused whether $g_n(x)$ is uniformly convergent or not and here limit of the integration isn't finite.
Please help me to answer it in the answer section. Thanks in advance.
$|g_n(x)-g(x)| \leq \frac 2 n$ by MVT, so $g_n \to g$ uniformly. For $f_n$ the best thing to do is to evaluate the integral. $f_n(x)=\int_0^{x} \frac 1 2(1-\cos 2(t+\frac 1 n))dt=\frac x 2 -\frac 1 4 \sin (2(x+\frac 1 n)$. Similarly $f(x)=\frac x 2 -\frac 1 4 \sin (2x)$. it is now easy to see (by MVT again) that $f_n(x) \to f(x)$ uniformly on $[0\infty)$.