I saw this today, I checked in Mathematica and the integral comes out to $\pi$, but I have no idea how to solve it.

FREE Wi-Fi: The Wi-Fi password is the first $10$ digits of the answer. $$\int_{-2}^2\left(x^3\cos\frac x2+\frac12\right)\sqrt{4-x^2}\ dx$$

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  • 1
    $\begingroup$ Motivation: think about even and odd functions whenever you see the integral from some $-a$ to $a$. $\endgroup$
    – kushal
    Sep 17, 2020 at 8:30
  • $\begingroup$ So not only do you have to do the integral to get $\pi$, you have to know $10$ digits of it. Or at least be able to find those $10$ digits on your laptop/phone without using Wifi. $\endgroup$
    – GEdgar
    May 4, 2021 at 0:02

2 Answers 2


The integrand is the sum of an odd and even function, and only the latter contributes, so it's $\int_0^2\sqrt{4-x^2}dx$. This is a quarter of the area of a radius-$2$ circle, i.e. $\pi$.


We have that $$\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx =0$$ since the integrand is point symmetric in the origin. Since $\sqrt{4-x^2}$ on $[-2;2]$ is the formula for the upper part of a circle we find that $$\int_{-2}^2 \sqrt{4-x^2} dx=\frac 12 \pi r^2=\frac 12 \pi 2^2=2\pi$$ So the whole integral is: $$\int_{-2}^2(x^3\cos \frac x2 +\frac 12 )\sqrt{4-x^2} dx=\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx+\frac 12\int_{-2}^2 \sqrt{4-x^2} dx=0+\frac 12 2\pi=\pi$$


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