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.
 A: 

*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 $n$.

A: An approach to #20 without use of the gamma function as other comments/answers show but with knowledge of the Gaussian integral is to let $t=x^2$:
$$\frac{1}{2} \int_0^{\infty} t^2 e^{-t^2} \; dt$$
Now, use integration by parts with $dv=te^{-t^2} \; dt$ and $u=t$:
$$=\frac{1}{2} \left(-\frac{1}{2}te^{-t^2} \bigg \rvert_0^{\infty}+ \frac{1}{2} \int_0^{\infty} e^{-t^2} \; dt \right)$$
$$=\frac{1}{4} \int_0^{\infty} e^{-t^2} \; dt$$
$$=\boxed{\frac{\sqrt{\pi}}{8}}$$
A: Let $ n $ be a positive integer, we have the following :
\begin{aligned}\int_{0}^{2\pi}{\cos^{2n}{x}\,\mathrm{d}x}&=\oint_{\left|z\right|=1}{\frac{1}{\mathrm{i}z}\left(\frac{z+\frac{1}{z}}{2}\right)^{2n}\,\mathrm{d}z}\\ &=-\frac{\mathrm{i}}{4^{n}}\oint_{\left|z\right|=1}{\frac{\left(z^{2}+1\right)^{2n}}{z^{2n+1}}\,\mathrm{d}z}\end{aligned}
Since $ f_{n} : z\mapsto\frac{\left(z^{2}+1\right)^{2n}}{z^{2n+1}} $ can be expanded as follows : $$ \frac{\left(z^{2}+1\right)^{2n}}{z^{2n+1}}=\sum_{k=0}^{2n}{\binom{2n}{k}z^{2k-2n-1}} $$ We get that : $$ \mathrm{Res}\left(f_{n},0\right)=\binom{2n}{n} $$
And thus : $$ \oint_{\left|z\right|=1}{\frac{\left(z^{2}+1\right)^{2n}}{z^{2n+1}}\,\mathrm{d}z}=2\pi\mathrm{i}\,\mathrm{Res}\left(f_{n},0\right)=2\pi\mathrm{i}\,\binom{2n}{n} $$ Which means $$ \int_{0}^{2\pi}{\cos^{2n}{x}\,\mathrm{d}x}=2^{1-2n}\pi\binom{2n}{n} $$
Taking $ n=1010 $, we get the final result.
A: Define $ I_{\alpha} $, $ \left(\forall \alpha\in\mathbb{R}_{+}\right) $, as follows : $$I_{\alpha}=\int_{0}^{\frac{\pi}{2}}{\frac{\cos^{\alpha}{x}}{\sin^{\alpha}{x}+\cos^{\alpha}{x}}\,\mathrm{d}x}$$
Define $ J_{\alpha} $, $ \left(\forall \alpha\in\mathbb{R}_{+}\right) $, as follows : $$I_{\alpha}=\int_{0}^{\frac{\pi}{2}}{\frac{\sin^{\alpha}{x}}{\sin^{\alpha}{x}+\cos^{\alpha}{x}}\,\mathrm{d}x}$$
Let $ \alpha\in\mathbb{R}_{+} $, adding them both together, we end with $ I_{\alpha}+J_{\alpha}=\frac{\pi}{2} \cdot $
Substituting $ u=\frac{\pi}{2}-x $, we can prove that $ I_{\alpha}=J_{\alpha} \cdot $
Hence : $$ I_{\alpha}=J_{\alpha}=\frac{\pi}{4} $$
Thus : $$ \int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{\tan^{\alpha}{x}+1}}=I_{\alpha}=\frac{\pi}{4} $$
Taking $ \alpha=\sqrt{2020} $, we get the final result.
A: 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.
