Solving for range with Arc Length Integral

Question

I encountered an interesting problem beyond my skill level, which is this

SOLVE FOR b

L = ∫_a^b √(1+(2wx)²) dx

In which I need to solve for b, but I don't know how to do that yet.

If anyone can help I'd greatly appreciate it :D!

Answer 1

You want to do a trigonometric substitution. If you have sqrt(1-x²) use sin, sqrt(x²-1) use sec, and sqrt(1-x²) use tan.

Let 2wx = tanθ It's a u-substitution, but it is convenient to use an angle variable in order to be able to derive all the other trig functions using some basic trig that you are used to.

2wdx = sec²θ dθ and dx = (sec²θ/2w)dθ

The integral become Int (from θ_a to θ_b) of (1/sqrt(1+tan²θ)( sec²θ)/(2w)dθ)θ

Using 1+tan² = sec², int (from θ_a to θ_b) of 1/(2w) secθ dθ

1/(2w) ln|secθ + tanθ| evaluated at θ_b - eval at θ_a

The usual way to proceed is to translate the trig functions of θ by making the right triangle with vertical side 2wx and horizontal side 1, and, therefore, hypotenuse sqrt(1+(2wx)²) which may look familiar.

Solution becomes (1/(2w)) ln |sqrt(1+(2wx)²) + 2wx| evaluated from a to b

I certainly might have made an error, but that's the logic of it.

Answer 2

Unless there is a trick I do not see, what needs to be done is to integrate which will give you

L=F(b)-F(a) where F is the integral. Then solve that (if you can) for b.

To integrate the substitution 2wx=tan y will give you an integrand which can be easily looked up in a table of integrals. Then convert the result back to a function of x. Messy...but it can be done.