# Math Help - Trig Integral - Direct Comparison Theorem

1. ## Trig Integral - Direct Comparison Theorem

Use the Comparison Theorem to determine whether the integral is convergent or divergent.

I can easily see that:

$\frac{1}{x \cos{x}} \leq \frac{1}{x}$ on $0 to \frac{\pi}{2}$

But I also know that $\int_0^{\frac{\pi}{2}}\frac{1}{x}dx$ is divergent because $\ln{0} = -\infty$

So, that substitution doesn't tell me anything about my original integral. Can anyone suggest another substitution that might help me?

Thanks.

2. ## Re: Trig Integral - Direct Comparison Theorem

Originally Posted by joatmon
Use the Comparison Theorem to determine whether the integral is convergent or divergent.

I can easily see that:

$\frac{1}{x \cos{x}} \leq \frac{1}{x}$ on $0 to \frac{\pi}{2}$ is that so?
...

3. ## Re: Trig Integral - Direct Comparison Theorem

Ack! I was thinking of x>=1. Thanks for the graph.