Integral for the End of the Year

Calculus Level 3

$\large \int_{0}^{\frac{2017\pi}{2}} \cos^{2017}(x) \ \mathrm{d}x = \frac{2^a(b!)^2}{c!}$

If the equation above holds true for some $a, b, c \in \mathbb N$ , evaluate $a+b+c$ .

The answer is 5041.

1 solution

Lorenzo Calogero
Dec 22, 2017

The integrand function $f(x)=\cos^{2017}(x)$ is periodical with $T=2\pi=4\frac{\pi}{2}$ . Therefore:

$\int_{0}^{\frac{2017\pi}{2}} \cos^{2017}(x)\mathrm{d}x = \int_{0}^{\frac{\pi}{2}} \cos^{2017}(x)\mathrm{d}x$

since $2017=4\cdot504 + 1$ .

By using the integration reduction formula of $\int\cos^n(x)\mathrm{d}x$ (which can be easily derived):

$\int_{0}^{\frac{\pi}{2}}\cos^n(x)\mathrm{d}x = \underbrace{\left. \frac{1}{n}\sin x \cos^{n-1}(x) \right|_{0}^{\pi/2}}_\text{=0} + \frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}\cos^{n-2}(x)\mathrm{d}x$

By applying the formula $\frac{n-1}{2}$ times, we obtain that:

$\int_{0}^{\frac{\pi}{2}}\cos^n(x)\mathrm{d}x = \frac{(n-1)(n-3)(n-5)\cdots}{n(n-2)(n-4)\cdots}\cdot\int_{0}^{\frac{\pi}{2}}\cos(x)\mathrm{d}x=\frac{(n-1)(n-3)(n-5)\cdots}{n(n-2)(n-4)\cdots}=I$

With $n=2017$ , $I=\frac{2016\cdot2014\cdot2012\cdots}{2017\cdot2015\cdot2013\cdots}=\frac{(2016\cdot2014\cdot2012\cdots)^2}{2017!}=\frac{(2^{1008}\cdot1008!)^2}{2017!}=\frac{2^{2016}\cdot(1008!)^2}{2017!}$

Thus, the solution is: $2016+1008+2017=\boxed{5041}$

Can you prove that there exists a unique set of three positive integers (a,b,c) corresponding to your solution? For example, if instead of 2017 we used 17, there would be two possible answers: (16,8,17) and (22,7,17).

D G - 3 years, 5 months ago

With $\cos^{17}(x)$ you can write the solution in two different ways since $b=2^k$ , with $k=3$ : $2^a\cdot(b!)^2=2^{16}\cdot(8!)^2=2^{16}\cdot(2^{3})^2\cdot(7!)^2=2^{22}\cdot(7!)^2$ . With $\cos^{2017}(x)$ , $1008\neq2^k\ \forall k \in \mathbb{N}$ , so the solution is unique.

Lorenzo Calogero - 3 years, 5 months ago

Integral for the End of the Year

The answer is 5041.

1 solution

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