Find length of curve. Integral of sec?

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SUMMARY

The discussion centers on finding the length of the curve defined by the function y=ln(cos(x)) from 0 to π/3 using calculus. The derivative y' is calculated as -tan(x), leading to the integral of √(1 + (-tan(x))²). The transformation of this integral simplifies to √(sec(x)), confirming the approach taken by the user. The discussion highlights the complexity of the integral and seeks validation of the method employed.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with derivatives, particularly trigonometric functions.
  • Knowledge of the Pythagorean identity: 1 + tan²(x) = sec²(x).
  • Experience with integral calculus, including the evaluation of definite integrals.
NEXT STEPS
  • Study the evaluation of integrals involving trigonometric identities, specifically secant functions.
  • Learn about arc length formulas in calculus, particularly for parametric curves.
  • Explore advanced integration techniques, including substitution and integration by parts.
  • Investigate numerical methods for approximating integrals when analytical solutions are complex.
USEFUL FOR

Students studying calculus, particularly those focusing on integration and arc length problems, as well as educators seeking to clarify concepts related to trigonometric integrals.

phantomcow2
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Homework Statement


Find the length of the curve y=ln(cosx) from 0 to pi/3

Homework Equations


integral of (1+(y')^2)^1/2

The Attempt at a Solution


First, find y'. y' is equal to -sin/cos, or simply -tan(x).

[tex]\int[/tex][tex]\sqrt{1+(-tan(x)^{2}}[/tex]
= [tex]\int[/tex][tex]\sqrt{sec(x}[/tex]

Homework Statement



Can this possibly be right so far? This is a horrendous integral. I've expended so much energy on this problem, but if someone can at least validate that I am on the right track, I'll post my work from here out.
 
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phantomcow2 said:
[tex]\int[/tex][tex]\sqrt{1+(-tan(x)^{2}}[/tex]
= [tex]\int[/tex][tex]\sqrt{sec(x}[/tex]

[tex]1+tan^2x=sec^2x \Rightarrow \sqrt{1+tan^2x}= \sqrt{sec^2x}=secx[/tex]
 
:redface::cry:

Thank you so much.
 

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