Integrating $\int_{0}^{2 \pi} \cos^{2020}(x)dx$, $\int_{0}^{\pi/2} \frac{1}{\tan^{\sqrt{2020}}(x)+1}dx$, $\int_{0}^{\infty} x^5 e^{-x^4}dx$ I've been working through the 2020 MIT Integration Bee qualifier questions (20 total) for fun, and there are three that I haven't been able to crack yet. (The complete list of problems and answers are all online (PDF link via mit.edu). However, there are no worked solutions with them.)
(9) $\quad\displaystyle\int_{0}^{2 \pi} \cos^{2020}(x) \, dx =  2^{-2019}\pi\binom{2020}{1010}$
(15) $\quad\displaystyle\int_{0}^{\pi/2} \frac{1}{\tan^{\sqrt{2020}}(x)+1} \, dx = \frac{\pi}{4}$
(20) $\quad\displaystyle\int_{0}^{\infty} x^5 e^{-x^4} \, dx = \frac{\sqrt\pi}{8}$
I think the binomial theorem might be needed for (9) since $2020 \choose 1010$ appears in the solution. I've tried substitution and integration by parts for (20) with no luck. Haven't made too much progress with (15), probably need a clever algebra trick. Any ideas would be much appreciated.
Is rather simple with $\cos x=\dfrac{e^{ix}+e^{-ix}}{2}$
$$\cos^n(x)=\frac{1}{2^n}\sum\limits_{k=0}^n
{n\choose k}e^{ikx}\cdot e^{-i(n-k)x}
=\frac{1}{2^n}\sum\limits_{k=0}^n
{n\choose k}e^{i(2k-n)x}$$
Then we group terms: first with last, second with second last, etc to get cosines back
$$2\cos^n(x)=\frac{1}{2^{n-1}}\sum\limits_{k=0}^n
{n\choose k}\frac{e^{i(2k-n)x}+e^{i(n-2k)x}}{2}=
\frac{1}{2^{n-1}}\sum\limits_{k=0}^n
{n\choose k}\cos((n-2k)x)$$
But with integration over $[0;2\pi]$ all the terms cancels except for $n-2k=0$ thus $$\int_0^{2\pi}\cos^n(x)\,\mathrm{d}x=\frac{1}{2^n}{n\choose k}\cdot 2\pi$$
where $k=\frac{n}{2}$ for an even $n$.
hint
For the first
$$\int_0^{2\pi}\cos^{2020}(x)dx=$$
by $t=x-\pi$
$$2\int_0^\pi\cos^{2020}(t)dt=$$
by $ u=t-\frac{\pi}{2}$
$$4\int_0^{\frac{\pi}{2}}\sin^{2020}(u)du$$
This is known as Wallis integral.