Given that $f\colon [-1,1] \to \mathbb{R}$ is a continuous function such that $ \int_{-1}^{1} f(t) \ dt =1$, how do I evaluate the limit of this integral: $$\lim_{n \to \infty} \int_{-1}^{1} f(t) \cos^{2}{nt} \,dt$$
What I did was to write $\cos^{2}{nt} = \frac{1+\cos{2nt}}{2}$ and substitute it in the integral so that I can make use of the given hypothesis of $\int_{-1}^{1} f(t) \ dt =1$. So the integral becomes,
\begin{align*} \int_{-1}^{1} f(t)\cos^{2}{nt} \ dt = \int_{-1}^{1} f(t) \biggl[\frac{1+\cos{2nt}}{2}\biggr] \ dt & \\ \hspace{3cm} = \frac{1}{2}\int_{-1}^{1}f(t) \ dt + \int_{-1}^{1} \frac{f(t)\cos{2nt}}{2} \ dt \end{align*}
But I don't really know how I can evaluate the second integral and also I can't realize as to why that integral condition on $f$ has been assumed. Moreover without assuming that condition on $f$ is it possible to evaluate this integral? If yes, then what would the answer be?