Limit of a weird Integral

When $x > 1$ and $x \neq k\pi$ for integer $k$ , $0 \leq \cos^2x < 1$ , and when $n \rightarrow \infty$ , $x^n \rightarrow \infty$ . Thus, $(\cos^2x)^{ \normalsize x^{\normalsize n}}\rightarrow 0$ .

When $0 \leq x < 1$ , $0 < \cos^2x \leq 1$ , and and when $n \rightarrow \infty$ , $x^n \rightarrow 0$ . Thus, $(\cos^2x)^{ \normalsize x^{\normalsize n}}\rightarrow 1$ .

When $x = k\pi$ for $k \in \mathbb{Z}^{+}$ , $\cos^2x = 1$ . Thus, $(\cos^2x)^{ \normalsize x^{\normalsize n}}\rightarrow 1$

Finally, when $x = 1$ , $\cos^2x=\cos^21$ , and $x^{n} = 1$ for every $n$ , so $(\cos^2x)^{ \normalsize x^{\normalsize n}}\rightarrow \cos^21$

The set $S = \{1\}\cup\{k\pi \, | \, k \in \mathbb{Z}^{+}\}$ is countably infinite, so it has measure $0$ , and when $x \in S$ , $(\cos^2x)^{ \normalsize x^{n}}$ is finite, so $S$ can be ignored in the integration.

Our limit now becomes $\displaystyle \int_{0}^{1} 1 dx + \int_{1}^{\infty} 0 dx\ = \boxed{1}$