Prove that $\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0$. 
Let $f$ be a continuous function on the interval $[a,b]$. Prove that $$\displaystyle\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0.$$

Attempt:
Want to show that for a given $\epsilon > 0$, there exists $N \in \mathbb N$ s.t. if $n \geq N$, then $$\left|\int_a^b f(x) \sin^n (x) dx\right| < \epsilon.$$
Since $f$ is continuous on a closed interval, it is bounded, i.e., $|f(x)| \leq M, \forall x \in [a,b]$.
For a given $\epsilon > 0$, choose $N$ s.t. $|\sin^N(x)| < \dfrac{\epsilon}{M(b-a)}$. [Is this allowed? I am thinking of the case where $\sin(x) = 1$, then this doesn't hold.]
Hence, $$\left|\int_a^b f(x) \sin^n (x) dx\right| \leq \int_a^b |f(x)|| \sin^n(x) | dx < \epsilon$$ for $n \geq N$.
 A: Some points have $\sin x=1$, so you cannot take $|\sin^Nx|<\epsilon/M(b-a)$. The solution is to remove small neighborhoods around those points (bounding the integral by the length of those small neighborhoods, times a constant). For the remaining parts you have $|\sin x|<1-\delta<1$, so you can bound the integral by $(1-\delta)^N$ times a constant.
A: What you did is correct... except that you noticed that it doesn't work when $\vert \sin x\vert = 1$, i.e. when $x \in A= \pi \mathbb Z + \pi/2$.
To take care of this notice that $B =[a,b] \cap A$ is finite. Let says that $B$ has $m$ elements $\{b_1, \dots, b_m\}$. You can write $$[a,b] = C \cup ([a,b] \setminus C)$$
where $C=\displaystyle\bigcup_{1 \le i \le m}\left(b_i- \frac{\epsilon}{2M m}, b_i+\frac{\epsilon}{2M m}\right)$.
Then on $[a,b] \setminus C$, you can proceed as you did. And on $C$ you have
$$\left\vert \int_C f(x) \ dx \right\vert \le M m  \frac{\epsilon}{M m} = \epsilon.$$
You then get the desired conclusion.
A: Hint:
Divide $[a,b]=A\cup B$, where $A=\bigcup_\limits{(2k+1)\pi/2\in[a,b]}[(2k+1)\pi/2-\delta, (2k+1)\pi/2+\delta]$.
It follows by
$$
\left| \int_A f(x) \sin^n (x) dx\right|\leqslant (1-\epsilon)^n\left| \int_A f(x)dx\right|\to 0\quad\text{as}\:\:n\to\infty
$$
And $$
\left| \int_B f(x) \sin^n (x) dx\right|\leqslant \left| \int_B f(x)dx\right|\leqslant M m(B)<MN\delta
$$
A: The bounded convergence theorem will quickly yield the desired result.  Note that since $f$ is continuous, it is bounded on any interval $[a, b]$.
