# Arc Length intregration

1. Sep 16, 2008

### jplandreneau

I am trying to figure out the following arc length problem, and it's really coming down to a question over intregration.

Compute the length of the curve r(t)=(4t)i +(4t)j+(t^2+6k) over the interval 0 to 6.

I have dr/dt = (4, 4, 2t) , and then used the arc length equation:

L= integral 0,6 ( sqrt((4^2)+(4^2) +(2t)^2)

and have reduced this to 2 int((sqrt(t^2+8))

I'm doing this to prepare for a test, and have "cheated" by using my calculator to get an answer of 2(6*sqrt(11) + 4*arcsinh(3/sqrt(2))) . I have no idea how to do the intregration of the sqrt(t^2 +8) and would greatly appreciate some advice. Thank you

2. Sep 16, 2008

### sutupidmath

$$\int \sqrt{x^2+8}dx$$ well, as i can see probbably a trig substitution would clean things up, let me have a crack at it.

$$x=\sqrt{8}tan(u)=>dx=\frac{\sqrt{8}}{cos^2u}$$ now,

$$\int \sqrt{8tan^2u+8}\frac{\sqrt{8}}{cos^2u}du$$

3. Sep 16, 2008

### jplandreneau

OK thanks a lot - its a little messy but definately helps

4. Sep 17, 2008

### sutupidmath

Actually i was going to type the whole thing, but i was havin' latex problems in this comp. so thats what i eventually posted.

5. Sep 17, 2008

### HallsofIvy

Staff Emeritus
The point is that $sec^2(\theta)= 1+ tan^2(\theta)$ so that the substitution $x= \sqrt{8}tan u$ gives $\sqrt{x^2+ 8}= \sqrt{8tan^2u+ 8}= \sqrt{8}sec u$. Since sec u= 1/cos u, the integral becomes
$$8\int \frac{du}{cos^3 u}$$
Since that involves an odd factor of cos u, you can multiply numerator and denominator by cos u to get
$$8\int \frac{cos u du}{cos^4 u}= 8\int \frac{cos u du}{(1- sin^2u)^2}$$
and use the substitution y= sin u to reduce to a rational function which can then be done by partial fractions.

Last edited: Sep 17, 2008