When doing trigonometric substitution what is the advantage of using $\tan^2(\theta)+1=\sec^2(\theta)$ over $\cot^2(\theta)+1=\csc^2(\theta)$? I remember my professor said something related to the domain and how using $\tan^2(\theta)+1=\sec^2(\theta)$ over $\cot^2(\theta)+1=\csc^2(\theta)$ will make it so that we can get rid of the absolute value of our final result. Can someone elaborate why is this?
 A: If you work with an indefinite integral with a radical ($y^2 = x$ implies $y = | \sqrt {x}|$ because $y = \pm \sqrt {x}$) and know that the integral you're working with is positive, absolute value bars are not required.  If you're working with limits on the integral, then you would have to keep the absolute value bars to determine the limits on the new integral.
Example: For $\int \sqrt {x^2 + 25} \ \ dx$, when we make the substitution $x = \tan \theta$, when squaring $x$ to get rid of the radical, $\tan^2 \theta$ will be positive, so the absolute value bars are not necessary.  On the other hand $\int_{-5}^{5} \sqrt {x^2 + 25} \ \ dx$, we have to be careful of the limits, so the absolute value is required.  (However, once we find the correct integral, we can substitute the values right in.)
I don't think I've seen a subsitution using $x = \cot \theta$ - mainly the ones used are $x = \sin \theta$, $x = \sec \theta$, and $x = \tan \theta$.  EDIT: In certain circumstances, $x = \cot \theta$ can be used, but very rarely.
