find the length of the arc y lnsecx between 0 x pi4Soluti

Solution

the derivative of y = ln (sec x) is (1/sec(x))*sec(x)*tan(x) = tan(x). Arc Length = ? (v(1 + [f \'(x)]^2)) dx {from a to b} ? (v(1 + [tan(x)]^2)) dx {from 0 to pi/4} Use the trig identity sec(x)^2 = 1 + tan(x)^2 = ? v(sec(x)^2) dx {from 0 to pi/4} = ? sec(x) dx {from 0 to pi/4} = ? 1/cos(x) dx {from 0 to pi/4} Multiply the integrand by cos(x)/cos(x). = ? cos(x) / cos(x)^2 dx {from 0 to pi/4} = ? cos(x) / (1 - sin(x)^2) dx {from 0 to pi/4} Now do a substitution. Let u=sin(x), then du=cos(x) dx. ? cos(x) / (1 - sin(x)^2) dx {from 0 to pi/4}