# Math Help - Arc Length Trig substitution

1. ## Arc Length Trig substitution

Find the arc length:
$
y = \frac{1}{2} x^2$
from point P to point Q: $P(-3,\frac{9}{2})$ $Q(3,-\frac{9}{2})$

Here is what I have so far:

Length = $2 \int\limits^{3}_{0} \sqrt{1 + x^2} dx$

$= 2 \int\limits^{\sqrt{10}}_{1} \sqrt{tan^2\theta + 1} \cdot sec^2\theta d\theta$

$
= 2 \int\limits^{\sqrt{10}}_{1} sec^3\theta d\theta$

Now what?

2. I=integration(sec^3 A dA)=INT.[secA*sec^2AdA]

now,apply integration by parts,
I=secA*INT[sec^2AdA]-int.[secA*tan^2AdA]

=secA*tanA-int.(sec^3AdA)+log(secA+tanA)

so,2I=secA*tanA+log(secA+tanA)

3. Originally Posted by sbcd90
I=integration(sec^3 A dA)=INT.[secA*sec^2AdA]

now,apply integration by parts,
I=secA*INT[sec^2AdA]-int.[secA*tan^2AdA]

=secA*tanA-int.(sec^3AdA)+log(secA+tanA)

so,2I=secA*tanA+log(secA+tanA)

I am not getting what you typed. Under the "now apply int. by parts" you did two separate int by parts there? I can't tell what that is. Can you tell me what you used for your substitutions u & dv?
Thanks!

4. u=secA

v=sec^2 A.