How is taking the average of an integral over an interval justified?

Question

I have been studying classical mechanics. Often when going through a worked problem, I see a step where there is an integral from 0 to 2$\pi$ of $\sin^{2} \theta \ d\theta$. Instead of using the integral value, $\pi$, they will take the average value which is $\frac{1}{2}$.

An example is in this link, which shows that when going from equation 100 to equation 101, three averages are used.

My questions are as follows:

• How is this justified?
• Is this an approximation?

Answer 1

The integral evaluates to

$$ \int_0^{2\pi}\cos^2(\theta)\mathrm{d}\theta = \pi$$

and they use it there exactly that way (note the sneaky $\pi$ in $(101)$). No approximation involved. The talk of average values there is due to the fact that the average value is defined by

$$ \frac{1}{2\pi - 0}\int_0^{2\pi}\cos^2(\theta)\mathrm{d}\theta = \frac{1}{2}$$

which is equivalent to the above, but really rather tangential to the calculation.