# Math Help - Arclength Integration Help

1. ## Arclength Integration Help

Hey,

Ok I've been having a bit of trouble on a homework question and I was just wondering if I am even going about this the right way.

QUESTION:

Find the Arclength of y = 1/x from x=1 to x=2.

So I got the following integral:

$I = \int^2_1 \sqrt{1 + \big(\tfrac{-1}{x^2}\big)^2}dx$

Then I did:

$\tan\theta = \frac{1}{x^2} \implies \sec^2\theta d\theta = \frac{-2}{x^3} dx$

So I get:

$I = \int^{\arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} \frac{\sec^3\theta \cdot x^3 \cdot d\theta}{-2} = -\tfrac{1}{2} \int^{\arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} \frac{\sec^3\theta \cdot d\theta}{\tan\theta\sqrt{\tan\theta}}$

And after that I used integration by parts to get:

$I = -\bigg(\frac{\sqrt{\tan\theta}}{\sin\theta}\bigg|^{ \arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} + \int^{\arctan(\tfrac{1}{4})}_{\tfrac{\pi}{4}} \frac{\csc\theta \cdot d\theta}{\sqrt{\tan\theta}}\bigg)$

And this is where I hit a wall. I don't know how to solve that last integral. And I might be going about this the entirely wrong way.

Thanks,

Trevor

2. When you make a substitution, we require to get the new integration limits.

Since you made $\tan \theta = \frac{1}
{{x^2 }}$
plug $x=(1,2)$ to get integration limits for $\theta.$

3. Yeah you're right, I do normally write that in on my papers I hand in. I'll add that to my post though.

4. BUMP.

Anyone got any ideas?

5. Nevermind my prof just sent out a message saying that it should be y = ln(x). Which should make this question...well considerably easier.