Remember when we are doing a trig substitution, for instance for an integral with: $\sqrt{a^2-x^2}$

We said that we should choose $x=a\sin\theta$, which seemed reasonable, but we also said that $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. Where did this last bit come from?

Well, we want a couple of things to hold true. The first is that any substitution that we make, we have to be able to undo. That is, we will substitute $x$ for a function of $\theta$ but in the end we need to convert back to $x$ and so to do that we have to be able to write the inverse function of, in this case $x=a\sin\theta$. The $\sin$ function is itself not invertible because it’s not one to one, so we have to choose a range over which it is one to one. We could choose $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ or we could choose $\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$ (amongst an infinite set of possibilities). That would also be invertible. However, remember that we are going to end up with a term of the form: $\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}$

So if we want this to simplify, we had better choose our range of $\theta$ such that $\cos\theta$ is positive, so that we can write $\sqrt{\cos^2\theta}=\cos\theta$. A simple choice for this would be $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. Let’s look at the graph of $\sin\theta$ (in red) and $\cos\theta$ (in blue) to convince ourselves of that fact. In the darker regions $\sin$ is one to one and $\cos$ is positive.

We won’t go through the full argument for the other trig substitutions, but we will show the plots so you can convince yourselves that it is true for them to.

For $x=a\tan\theta$ we have $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$: And for $x=a\sec\theta$ we have $0\le\theta<\frac{\pi}{2}$ or $\pi\le\theta<\frac{3\pi}{2}$: How clear is this post?