# The integral of sin^2(x).

I’m scheduled to teach Math 1320 this fall, so I’ve been writing some lecture notes.  The course is simply the second part of the ‘standard’ calculus sequence, which means we’re going to be talking about ways to integrate just about anything before me move on to sequences and series, then Taylor’s theorem, and possibly some polar coordinate exercises.  I’ll know more when I get the course memo. In any case, I’ve been going through the book and I’ve noticed that the following integrals are always done the same way:

$\displaystyle \int_0^\pi \sin^2(x) dx$ and $\displaystyle \int_0^\pi \cos^2(x) dx$.

We make the substitution $\sin^2(x) = \frac{1-\cos(2x)}{2}$ or $\cos^2(x) = \frac{1+\cos(2x)}{2}$, and then integrate that normally.  There’s nothing wrong with this approach and students need to know it, but mathematicians are lazy and writing all that is too much work for such a simple integral.  There’s a trick that you can do so that the integral becomes (almost) trivial.

Look at the graphs of $\sin^2(x)$ and $\cos^2(x)$. Let’s say that $\alpha = k \pi/2$ for some $k \in \mathbb{Z}$.  We know that

$\displaystyle \int_0^\alpha {\sin^2(x)}dx = \int_0^\alpha\cos^2(x) dx$.

Thus we get that each of the above integrals is equal to the average of the two, i.e.

$\displaystyle \int_0^\alpha \sin^2(x) dx = \frac{1}{2} \left( \int_0^\alpha \sin^2(x) dx + \int_0^\alpha \cos^2(x) dx \right)$,

$\displaystyle \int_0^\alpha \sin^2(x) dx = \frac{1}{2} \left( \int_0^\alpha (\sin^2(x) + \cos^2(x) )dx \right)$.

Since $\sin^2(x) + \cos^2(x) = 1$, we immediately see that the integral is equal to half of the length of the interval, namely

$\displaystyle \int_0^\alpha \sin^2(x) dx=\int_0^\alpha \cos^2(x) dx=\frac{\alpha}{2}$.

Sure, it only works for certain intervals, but it seems like whenever these problems come up $\alpha$ has a ‘nice’ value that lets you do this.