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:
\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]

 

[homestar]

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