I came across a picture of a solution on calculating a limit with integral.
$$\begin{align} \lim_{n\to\infty}\int_{0}^{\pi/4}\cos^nx\,dx &= \lim_{n\to\infty}\int_{0}^{\pi/4}e^{n\ln\cos x}dx \tag1 \\[8pt] &= \lim_{n\to\infty}\int_{0}^\infty e^{-\frac12nx^2}dx \tag2\\[8pt] &= \lim_{n\to\infty}\sqrt{\frac{\pi}{2n}} \tag3\\[8pt] &= 0 \tag4 \end{align}$$
I understand that $(1)$ is obtained by using the property of exponentiation, while $(3)$ and $(4)$ can be done with a few different methods.
My question is how it gets $(2)$?
It seems like the Maclaurin Series is used here $$\ln \cos{x} = -\frac{x^2}{2} - \frac{x^4}{12} + O(x^6)$$ but how is upper bound of the integral changed from $\frac{\pi}{4}$ to $\infty$ ?
Like you, I do not understand why to write things like that since we can simply write $$\int_{0}^{\frac \pi 4} e^{-\frac12nx^2}\,dx =\sqrt{\frac{\pi }{2n}} \,\,\text{erf}\left(\frac{\pi}{4} \sqrt{\frac{n}{2}} \right)\tag 1$$ which is asymptotic to $\sqrt{\frac{\pi }{2n}}$.
As you know, the asymptotic of $$\int_{0}^{\pi/4}\cos^n(x)\,dx $$ is $$\frac{\sqrt{\pi }}{2}\,\frac{\Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)}\tag 2$$
Just for your curiosity, put on the same graph the two functions for, say, $10 \leq n \leq 100$.