# Evaluating the integral $\mathop{\lim}\limits_{n \to \infty} \int_{-1}^{1} f(t)\cos^{2}(nt) \ dt$

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?

Define $g(t):[-\pi,\pi]\to \mathbb{R}$ such that $g(t)=f(t)$ for $t\in [-1,1]$ and $g(t)=0$ elsewhere. Then $g$ is integrable and by Lebesgue-Riemann lemma $$\lim_{n\to\infty}\int_{-\pi}^\pi g(t)\cos 2nt dt=0.$$ But this is just the same as $$\lim_{n\to\infty}\int_{-1}^1 f(t)\cos 2nt dt=0.$$
You can use $\sin^2 nt = \cos^2(nt + \pi/2)$ and some manipulation to show that