# Theta ranges for trig substitution

My professor, when doing trig substitution in lecture, always defines theta between certain intervals and when he takes the square root, he adds an absolute value bar to the trig function and then makes sure its positive through the interval. For practical purposes, is it necessary to go through all that hassle or might I trip up on a problem If I ignore theta ranges and take square roots normally, will I always get the right answer?

## Answers and Replies

arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
"If I ignore theta ranges and take square roots normally, will I always get the right answer? "

No, you won't.

Do as your professor does.

can you give me an example? All the indefinite integrals I encountered, through his exercises as well as in high school, all worked without labeling the theta ranges.

mathman
Science Advisor
The indefinite integral of trig functions do not require absolute value. However, when there is a square root involved and you want to keep things real, you need absolute value.

CAF123
Gold Member
can you give me an example?
There was a thread the other day where we wanted to find the length of the curve r=9+9cosθ.
Following through the standard recipe of finding this, we arrive at the integral $$c \int_0^{2\pi} \sqrt{\cos^2\left(\frac{\theta}{2}\right)}\,d\theta,$$ where ##c## is a constant.

Doing as you say, writing ##\sqrt{\cos^2\left(\frac{\theta}{2}\right)} = \cos\left(\frac{\theta}{2}\right)## will give a zero length, which is clearly false.
The right way is of course ##\sqrt{\cos^2\left(\frac{\theta}{2}\right)} = |\cos\left(\frac{\theta}{2}\right)|## and this can be integrated by considering the sign of cos(θ/2) over the domain [0,2π].