# Trigonometric substitutions

1. Sep 3, 2015

http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx

In example one, the author drops the absolute value bars and makes the following statement:

"Without limits we won’t be able to determine if $\tan{\theta}$ is positive or negative, however, we will need to eliminate them in order to do the integral. Therefore, since we are doing an indefinite integral we will assume that $\tan{\theta}$ will be positive and so we can drop the absolute value bars."

Why should we assume that $\tan{\theta}$ will be positive?

2. Sep 3, 2015

### RUber

Assuming it to be positive allows you to go through the general steps. You could also have assumed it to be negative, or worked out both cases. In the end, you'll see that the answers are very similar.
In the case that you have a limits of integration, I feel like this page explains the steps well.

3. Sep 4, 2015

But isn't the indefinite integral the most general antiderivative? It would therefore make sense to assume that $x$ can take on any value. Can we avoid this problem by restricting $\theta$ in the first and fourth quadrant only? This would work for all $x$.
In the first example the author implicitly assumes that $\theta$ lies in $(0,\frac{\pi}{2})$ and that $x$ is positive. The right triangle wouldn't make sense otherwise. This is getting really confusing!