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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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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. $\pi$.

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    $\begingroup$ Oh boy now I feel foolish for not thinking about that haha. $\endgroup$ – Kai Nov 8 at 22:39
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We have that $$\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx =0$$ since the integrand is point symmetrix 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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