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The question is about a calculation in the first section of this page: https://openstax.org/books/university-physics-volume-1/pages/17-3-sound-intensity

Essentially, it states: "To find the time-averaged intensity over one period $T=2\pi/\omega$ for a position $x$, we integrate over the period" and they present the result:$$\int_0^{2\pi/\omega}[\beta k \omega s_\text{max}^2\sin^2(kx-\omega t+\phi)]dt=\frac{\beta k\omega s_\text{max}^2}{2}$$

I am having trouble coming to the same conclusion. This is my calculation so far:$$\begin{align*}\beta k \omega s_\text{max}^2\int_0^{2\pi/\omega}\sin^2(kx-\omega t+\phi)dt=\beta k \omega s_\text{max}^2\int_0^{2\pi/\omega}(\frac12-\frac12\cos[2(kx-\omega t+\phi)])dt\end{align*}$$

Now, my evaluation of the integral, using $u=2(kx-\omega t+\phi)$, results in$$\beta k \omega s_\text{max}^2\left[-\frac u{4\omega}+\frac{\sin(u)}{4\omega}\right]|_0^{2\pi/\omega}=\beta k \omega s_\text{max}^2\left[-\frac {2kx-2\omega t+2\phi}{4\omega}+\frac{\sin(2kx-2\omega t+2\phi)}{4\omega}\right]|_0^{2\pi/\omega}=\beta k\omega s_\text{max}^2\left[\left(\frac{-2kx+4\pi-2\phi+\sin\left(2kx-4\pi+2\phi\right)}{4\omega}\right)+\left(\frac{2kx+2\phi-\sin\left(2kx+2\phi\right)}{4\omega}\right)\right]=\beta ks_\text{max}^2\pi$$

This does not look like the correct solution so where did I take the wrong turn? Also, I had to assume that $\sin(2kx-4\pi+2\phi)$ and $\sin(2kx+2\phi)$ subtract to zero (since it represents the same function but phase shifted?)

The answer to this question (Relation between Displacement Amplitude and Intensity of Sound Waves) looks very similar to my calculation but the book I mentioned, first does not explain how it got there and second, results in a different final expression of intensity at the end

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The time-averaged intensity is the integral of intensity over a period, divided by the period. You have done the integral correctly, but have not divided by the period $2\pi/\omega$.

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