I want to show uniform convergence for this question somehow. I was thinking that $f(x)sin^n(x)\rightarrow 0 \space \forall x \in [a,b]$, that $f_{n+1}(x) \leq f_n(x)$ and applying dini's theorem to show that $f(x)sin^n(x)$ is uniformly convergent, but I forgot to account for places in $[a,b]$ where $sin(x) = 1$. How do I get around this?
Using continuity, we know that $\sup_{a\leq x\leq b}|f(x)|\leq m$. So $$ \left|\int_a^bf(x)\sin^n(x)\mathrm dx\right|\leq m\int_a^b|\sin(x)|^n\mathrm dx $$ now taking the limit of both sides and using the dominated convergence theorem on the right ($|\sin(x)|^n\leq 1$, an integrable function on $[a,b]$), $$ \lim_{n\to \infty}\left|\int_a^bf(x)\sin^n(x)\mathrm dx\right|\leq m\lim_{n\to\infty}\int_a^b|\sin(x)|^n\mathrm dx\\ = m\int_a^b\lim_{n\to\infty}|\sin(x)|^n\mathrm dx $$ but $$ \lim_{n\to \infty}|\sin(x)|^n=\begin{cases}0,&\;x\ne \frac{(2n+1)\pi}{2}\\ 1,&\;x=\frac{(2n+1)\pi}{2}\end{cases}\\ =\mathbb{1}_{\left\{\frac{(2n+1)\pi}{2}\right\}} $$ so $0$ except at isolated points, sets of measure zero. So you may conclude that the limit on the right is $0$, $$ \lim_{n\to \infty}\left|\int_a^bf(x)\sin^n(x)\mathrm dx\right|\leq m\int \mathbb{1}_{\left\{\frac{(2n+1)\pi}{2}\right\}\cap [a,b]}\mathrm dx=0 $$