comparison theorem for integral of of 1 / sqrt(4x^2 - 1)

[Clifford]

using the comparison theorem, determine whether integral from 1 to infinity of 1 / sqrt(4x^2 - 1) converges or diverges.

I am only familiar with doing the comparison theorem with bounded functions like sin and cos. I don't know if I need to transfer this integral into a trig one and then apply the comparison theorem or if there is an easier way.

I tried to let x = secu, dx = secutanudu

sqrt(4x^2-1) = sqrt(4secu^2-1)
From here usually you can factor out something like a root 4 and then get something under the root which will be secu^2-1 which is tanu, but I am stuck here since 4 isn't a common in both terms.

 

[tkhunny]

Re: comparison theorem

Why are you trying to find the antiderivative?

As x increases without bound in the positive direction,

\(\displaystyle \frac{1}{\sqrt{4x^{2}-1}}\;>\;\frac{1}{\sqrt{4x^{2}}}\;=\;\frac{1}{2x}\)

This is quite sufficiently helpful. Why?

 

[Deleted member 4993]

Re: comparison theorem

using the comparison theorem, determine whether integral from 1 to infinity of 1 / sqrt(4x^2 - 1) converges or diverges.

I....From here usually you can factor out something like a root 4 and then get something under the root which will be secu^2-1 which is tanu, but I am stuck here since 4 isn't a common in both terms.

Assume

\(\displaystyle 4x^2 =\, sec^2(u)\)

\(\displaystyle x =\frac{1}{2}\cdot sec(u)\)

But as tkhunny explained above, you really don't need to go this route.