note for jenna

Maybe there's a faster way to do this, but here's what I thought about the integral we discussed yesterday, which was basically $\int \root\of{1-\sin(x)}~dx$.

Notice that if the integral were instead $\int \root\of{1-\cos(x)}~dx$, then we could use the formula

(1)
\begin{align} \sin^2(x) = \frac{1}{2}(1-\cos(2x)) \end{align}

or its cousin (which is tailor made for this integral):

(2)
\begin{align} \sin^2\left(\frac{x}{2}\right) = \frac{1}{2}(1-\cos(x)) \end{align}

With this identity in hand, we can just do a substitution:

(3)
\begin{align} \int \root\of{1-\cos(x)} ~dx = \int \root\of{2\sin^2\left(\frac{x}{2}\right)}~dx = \root\of{2}\int |\sin\left(\frac{x}{2}\right)|~dx. \end{align}

This last integral might look bad because of the absolute value sign, but since you're working with a definite integral you can determine exactly which intervals the function is positive and negative, thus making computations easier.

Of course, you DON'T have the integral $\int \root\of{1-\cos(x)}~dx$. Fortunately, though, I think you should be able to convert your integral into this same form using the identity $\cos(x-\frac{\pi}{2}) = \sin(x)$. Now you're just a change of variables away from having the integral involving cosine, which should make life easier.