Help with integral (pretty easy but my brain is stuck)

[homestar]

Homework Statement

This is to find the arc length of r= $$\theta$$ from 0 < $$\theta$$ < 2pi

I ended up with $$\int$$$$\sqrt{\theta^2 + 1}$$

Homework Equations

$$\int$$$$\sqrt{\theta^2 + 1}$$

The Attempt at a Solution

I couldn't see a relevant way to do integration by parts so I went with trig substitution, subbing tan(t) for theta. I got lost at the point where I'm to integrate sec^3(t)dt.

Can anyone help? Am I on the right track?

 

[rocomath]

Looks good to me.

$$\int\sec^{3}tdt$$

Let's start by breaking secant to the cube.

$$\int\sec t\sec^2 tdt$$

Which would you make u and dV?

 

[HallsofIvy]

Homework Statement

This is to find the arc length of r= $$\theta$$ from 0 < $$\theta$$ < 2pi

I ended up with $$\int$$$$\sqrt{\theta^2 + 1}$$

Homework Equations

$$\int$$$$\sqrt{\theta^2 + 1}$$

You actually mean $$\int\sqrt{\theta^2+ 1} d\theta$$ don't you?

The Attempt at a Solution

I couldn't see a relevant way to do integration by parts so I went with trig substitution, subbing tan(t) for theta. I got lost at the point where I'm to integrate sec^3(t)dt.

Can anyone help? Am I on the right track?

Homework Statement

Looks to me like a pretty standard integral. Let $\theta= tan(x)$ so that $d\theta= sec^2(x) dx$ and $\sqrt{\theta^2+ 1}= \sqrt{tan^2(x)+ 1}= sec(x)$. Then
$$\int\sqrt{\theta^2+ 1}d\theta= \int sec^3(x)dx= \int \frac{1}{cos^3(x)}dx$$
$$= \int\frac{cox(x)}{cos^3(x)}dx= \int\frac{(cos(x)dx)}{(1- sin^2(x))^2}$$

and the substitution u= sin(x) converts to a rational integral:
[tex]\int \frac{1}{(1-u^2)^2}du[/itex]<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2>[/QUOTE][/tex]

 

[homestar]

ahhhh...thank you. it's relatively early in the semester and i may have partied too hard over winter break ;)