$\lim\int_{-1}^{1} f(t) \cos^2(nt) dt$ = ? when $f$ is a continuous function on $[-1,1]$.
My try:
Let $\left| \int_{-1}^{1} f(t) \cos^2(nt) dt\right|\leq \int_{-1}^{1} | f(t) \cos^2(nt)| dt\leq k\int_{-1}^{1}\cos^2(nt) dt$ as $f$ is a continuous function on $[-1,1]$. So $\lim |a_n| = 0$ as $\lim \int_{-1}^{1}\cos^2(nt) dt = 0$. So $\lim a_n = 0$.
[$\int_{-1}^{1}\cos^2(nt) dt = \int_{-1}^{1}\cos(2nt)/2 dt = \frac{\sin(2n)}{2n}$, and $\lim \frac{\sin(2n)}{2n} = 0$]
Can anyone please let me know where have I gone wrong?
The mean value of $\cos^2(t)$ or $\cos^2(nt)$ is not zero, it is $\frac{1}{2}$. So the value of the wanted limit is half the integral of $f$ over $(-1,1)$, since by the Riemann-Lebesgue lemma $\lim_{n\to +\infty}\int_{-1}^{1}f(t)\cos(2nt)\,dt = 0$.