# Calculus

Calc length of arc of y=ln(x) from x=1 to x=2
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So far:
Definite Integral over x=(1,2) of sqrt(1 + 1/x) dx

1/x = tan^2 t
x = 1/tan^2 t

sqrt(1+1/x) = sqrt(1+tan^2 t) = sec t
dx = -2 * tan^-3 t * sec^2 t dt

Integrate over x=(1,2): sec^3 t / tan^3 t dt
Integrate over x=(1,2): 1 / sin^3 t dt

How do I solve?

Thanks!

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1. I will assume that your integral for the path length is correct, and just worry about the integral.

The indefinite integral of sin^-3 t dt is, according to my Table of Integrals,
(-1/3)cos t/sin^2t +(1/2)ln[sin t/(1 + cos t)]
Since you have chosen to do a substitution to the variable t, the range of integration is the t range from where x = 1 to where x = 2
When x = 1, tan t = 1 and t = pi/4 =.7854
When x = 2, tan^2t = 1/2,
tan t = 1/sqrt 2 and t = 0.6155
Evaluate the indefinite integral at those two values of t, and take the difference.

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posted by drwls

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