Remember when we are doing a trig substitution, for instance for an integral with:




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:




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?