How do I find the arc length of the curve $y = ln (sec x)$ $y=ln(sec x)$ from $(0, 0)$ $(0,0)$ to $(\frac{π}{4}, \frac{ln (2)}{2})$ $(pi/ 4, ln(2)/2)$ ?

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Answer 1 · 2018-05-31T10:30:35

$L = ln (1 + \sqrt{2})$ $L=ln(1+sqrt2)$ units.

$y = ln (sec x)$ $y=ln(secx)$

$y'=tanx$

Arc length is given by:

$L=int_0^(pi/4)sqrt(1+tan^2x)dx$

Simplify:

$L=int_0^(pi/4)secthetad theta$

Integrate directly:

$L=[ln|sectheta+tantheta|]_0^(pi/4)$

Hence

$L=ln(1+sqrt2)$

How do I find the arc length of the curve y=ln(sec x)y=ln(secx)y=ln(sec x) from (0,0)(0,0)(0,0) to (pi/ 4, ln(2)/2)(π4,ln(2)2)(pi/ 4, ln(2)/2)?