# 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.

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 $$u=t$$: $$=\frac{1}{2} \left(-\frac{1}{2}te^{-t^2} \bigg \rvert_0^{\infty}+ \frac{1}{2} \int_0^{\infty} e^{-t^2} \; dt \right)$$ $$=\frac{1}{4} \int_0^{\infty} e^{-t^2} \; dt$$ $$=\boxed{\frac{\sqrt{\pi}}{8}}$$

1. 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 $$n$$.

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.

9 - You can show $$\cos^{2k}(x) = C_k+\sum_{j\in J} \cos(jx)$$ for some finite set of positive integers $$J$$; I'm being crass with $$J$$ because they integrate to zero. So once you determine $$C_k$$, that's basically the problem.

15 - Try differentiating under the integral sign or using symmetry with cotangent.

20 - There is a clear substitution $$y=x^4$$; you can then rewrite the integral in terms of gamma functions.

• To 15: The answer here is really cool 🙂 – Maximilian Janisch Jun 27 at 22:09

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.

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.