How could I show that $\int_{0}^{\frac{\pi}{2}} \frac{\cos (nx)}{\cos x}dx$ converges to zero as $n\rightarrow \infty$? Consider $$\int_{0}^{\frac{\pi}{2}} \frac{\cos (nx)}{\cos x}dx$$ 
From here, I can show that $$\int_{0}^{\frac{\pi}{2}} \frac{\cos (nx)}{\cos x}dx > \int_{0}^{\frac{\pi}{2}} \cos (nx)dx \rightarrow 0$$ as $n$ approaches infinity, so I've constructed a lower bound for the integral.
However, I cannot see how to construct an upper bound that also converges to zero, so then I can squeeze the two integrals between two things which approach zero.
 A: This integral does not converge to 0 as $n \to \infty$.
Case $n = 2m$. In this case, the integrand has simple pole at $x = \frac{\pi}{2}$
$$ \frac{\cos 2mx}{\cos x} \sim \frac{(-1)^m}{\frac{\pi}{2} - x} \quad \text{as } x \to \tfrac{\pi}{2},$$
and hence the integral diverges:
$$ \int_{0}^{\frac{\pi}{2}} \frac{\cos 2mx}{\cos x} \, dx = \begin{cases}
+\infty & \text{if } m \text{ is odd} \\
-\infty & \text{if } m \text{ is even}.
\end{cases} $$
Case $n = 2m+1$. In this case, the integral converges but the resulting sequence still does not converge. Indeed,
$$ \int_{0}^{\frac{\pi}{2}} \frac{\cos(2m+1)x}{\cos x} \, dx + \int_{0}^{\frac{\pi}{2}} \frac{\cos(2m-1)x}{\cos x} \, dx
= \int_{0}^{\frac{\pi}{2}} 2\cos(2mx) \, dx
= 0 $$
and iterating this relation $m$ times, we get
$$ \int_{0}^{\frac{\pi}{2}} \frac{\cos(2m+1)x}{\cos x} \, dx = (-1)^m \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x} \, dx = (-1)^m \frac{\pi}{2}. $$
Consequently, the integral does not converge even when we consider odd terms only.
