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Trig substitution into integrals 
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#1
Feb1514, 08:32 PM

P: 5

I was testing for convergence of a series:
∑[itex]\frac{1}{n^2 1}[/itex] 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? 


#2
Feb1514, 09:10 PM

Mentor
P: 21,214

Not that you asked, but it's probably simpler and quicker to break up 1/(n^{2}  1) using partial fractions. 


#3
Feb1614, 08:29 PM

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? 


#4
Feb1614, 09:01 PM

P: 435

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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