Find the ARC LENGTH of the curve y= (picture of problem included), please help!?

Question

I didn't know how to type out an integral symbol so here goes

http://i3.tinypic.com/6z3unid.jpg

Dubya, doesn't it become the integral of sec^2 t -2 and not just sec^2 t ? I thought there were two 1's being subtracted and not +1 -1

Yeah, I definitely didn't copy the problem down wrong... which makes me SO sad

Answer 1

formula for the length of arc...

L = ∫ √(f'(x)² + 1) dx

f(x) = ∫ √(tan²t - 1) dt ... with the limits...
f'(x) = √(tan²x - 1) {use fundamental theorem of calculus}

L = ∫ [0,π/4] tanx dx
= ln|sec x| ... from 0 to π/4
= (1/2) ln2


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Answer 2

If you're just having troubles integrating that then:

sin^2(x)+cos^2(x)=1; divide both sides by cos^2(x) and you get:

tan^2(x)+1=sec^2(x)
tan^2(x)=sec^2(x)-1

So replace that in the integral:
∫(sec^2(x)+1-1)^(1/2)=∫secxdx

Multiply that by (secx+tanx)/(secx+tanx) and you get:
∫(sec^2x+secxtanx)/(secx+tanx)dx

Now let u=secx+tanx then du=(sec^2x+secxtanx)dx and the integral becomes:
∫(1/u)du = ln(u) = ln(secx+tanx)

Now, I'm not too sure why you're integrating from pi/4 to x . . . I'm pretty sure you should integrating from 0 to pi/4. But either way, you just plug in the numbers.

Just as a note, arc length is L is

a
∫ds
b

where ds = ((dx/dt)^2 + (dy/dt)^2)^(1/2)
Or ds = (1+(dx/dt)^2)^(1/2)

But I don't see that in your work anywhere. Are you sure you're doing it correctly?

edit:
Whoops. The problem is tan^(x)-1, not +1. Since that's the case, the integral is a lot more complicated. Not too sure how you would integrate that.

Make sure you didn't copy the problem down wrong?

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So I'll concede that DC is significantly smarter than I am, and you should probably listen to him, but I still don't see how it comes out to be ∫tanxdx.