Berkeley integration bee 2020 final #1: $\int_{-\infty}^\infty\frac{2020^{-|x|}}{1+5^{\arcsin(\sin^5(x))}}dx$ I was just recently digging through some integration bee problems, when I found this from Berkeyley 2020.
$$
\int_{-\infty}^\infty\frac{2020^{-|x|}}{1+5^{\arcsin(\sin^5(x))}}dx
$$
The guy in the video solved in about 2 minutes or less, but I don't seem to be able to understand how he did it. If anyone knows how to solve this or knows if someone else has already asked the question (I couldn't find one), it would be much appreciated if you could answer the question for me. It would be preferred if the answer is something up to about first year university, but if you have an alternative method that is still fine.
 A: As long as the integral exists as an extended real number (real or $-\infty$ or $+\infty$), we always have $$\int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^\infty f(-x)\,dx$$
which arguably is not terribly useful, but as an immediate consequence, the integral is also equal to the average of the two integrals:
$$\int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^\infty \frac{f(x)+f(-x)}2 \,dx$$
which often will simplify the integrand.

Turning to your problem, setting $$f(x) = \frac{2020^{-|x|}}{1+5^{\arcsin(\sin^5(x))}},$$
we note first that $|f(x)| \leq 2020^{-|x|}$ and $\int_{-\infty}^\infty 2020^{-|x|}\,dx = \frac{2}{\ln(2020)} < \infty,$ so the integral in the question not only exists as an extended real number, it's finite.
Now applying the property I outlined above to simplify the integral:
$$\begin{align*}\int_{-\infty}^\infty f(x)\,dx &= \int_{-\infty}^\infty \frac{1}{2}\left(\frac{2020^{-|x|}}{1+5^{\arcsin(\sin^5(x))}} + \frac{2020^{-|-x|}}{1+5^{\arcsin(\sin^5(-x))}}\right)\,dx \\ &= \int_{-\infty}^\infty\frac{1}{2}\left(\frac{2020^{-|x|}}{1+5^{\arcsin(\sin^5(x))}}+\frac{2020^{-|x|}}{1+5^{-\arcsin(\sin^5(x))}}\cdot\frac{5^{\arcsin(\sin^5(x))}}{5^{\arcsin(\sin^5(x))}}\right)\,dx \\ &= \int_{-\infty}^\infty\frac{1}{2}\left(\frac{\left(1+5^{\arcsin(\sin^5(x))}\right) \cdot 2020^{-|x|}}{1+5^{\arcsin(\sin^5(x))}}\right)\,dx \\ &= \frac{1}{2}\int_{-\infty}^\infty 2020^{-|x|}\,dx \\ &= \frac{1}{2}\cdot\frac{2}{\ln(2020)} = \frac{1}{\ln(2020)}\end{align*}$$
