# Trig substitution into integrals

Tags: integral test, integrals, substitution, trig, trig substitution
 P: 5 I was testing for convergence of a series: ∑$\frac{1}{n^2 -1}$ from n=3 to infinity I used the integral test, substituting n as 2sin(u) so here's the question: when using the trig substitution, I realized the upperbound, infinity, would fit inside the sine. Is it still possible to make the substitution? Or is there a restriction when this happens?
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P: 21,081
 Quote by tadf2 I was testing for convergence of a series: ∑$\frac{1}{n^2 -1}$ from n=3 to infinity I used the integral test, substituting n as 2sin(u) so here's the question: when using the trig substitution, I realized the upperbound, infinity, would fit inside the sine.
What does "fit inside the sine" mean?
 Quote by tadf2 Is it still possible to make the substitution? Or is there a restriction when this happens?
Sure, you can make the substitution. The integral will be from 3 to, say b, and you take the limit as b → ∞.

Not that you asked, but it's probably simpler and quicker to break up 1/(n2 - 1) using partial fractions.
 P: 5 Inside the sine meaning, the argument of the 'arcsine' would only range from -1 to 1. So I'm guessing you can't make the substitution because arcsin(infinity) = error?
P: 338

## Trig substitution into integrals

If you're looking for an appropriate trig substitution for the definite integral (and not just one that gets you a correct antidierivative), then ##\sec u## is the way to go. But like Mark44 said, partial fractions is really the "right" technique of integration for this particular integral.

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