Show that $\int_{-\pi}^\pi ~f(x) \cos (nx) \mathrm{d}\mu(x)$ converges to $0$ I need some help on the following problem.

Let $f\in L_1([-\pi,\pi])$. Then $\int_{-\pi}^\pi ~f(x) \cos (nx) \mathrm{d}\mu(x) \to 0$, where $\mu$ is the Lebesgue measure on $[\pi,\pi]$. 

Any hints and suggestions on how to begin is very much welcomed. Thanks.
 A: First, let's examine the functions $C_n(x)=\cos (nx)$.  For large  $n$, the graph of this function consists of many cosine waves of small common period. If you integrate this function over $[-\pi,\pi]$, you'll, of course, obtain 0. 
Now if $I$ is any interval in $[-\pi,\pi]$, for $n$ big, the graph of $C_n$ over $I$   will consist of  many cosine waves of small common period together with a portion  of a cosine wave near each endpoint  of $I$. Here, $\int_I C_n$ will  be the same as integrating $C_n$ just "near the endpoints" of $I$ (the middle portion will integrate to 0). But as $n$ grows large, the measure of those portions gets small, and as a result, 
$\lim\limits_{n\rightarrow\infty} \int_I C_n =0$.
Thus, for any interval $I$ and for any number $a$, we have 
$$\lim_{n\rightarrow\infty} \int_I a\cdot C_n =0.$$
Using the above result and linearity of integration, we can show  that if $g$ is a function of the form $$g(x)=\sum_{i=1}^n a_i \chi_{I_i},{\text{ where } }\ I_j\cap I_k\ \buildrel{j\ne k}\over=\ \emptyset{\text{ and }}\bigcup_{i=1}^n I_i=[-\pi,\pi],$$ 
then
$$\lim_{n\rightarrow\infty} \int_{[-\pi,\pi]} g C_n =0.$$
Thus, the theorem is true for any step function in $L_1$.
The general result follows from the fact that the step functions are dense in $L_1$ (that is, given $f\in L_1$ and $\epsilon>0$, there is a step function $g$ with $\Vert f-g\Vert_{L_1}<\epsilon$).  
I can provide more details here if you like, just let me know.
A: This statement is true for any $f\in C^1([-\pi,\pi])$. Note that $C^1([-\pi,\pi])$ is dense in $L_1([-\pi,\pi])$ hence equality holds for all $f\in L_1([-\pi,\pi])$
