Find the arc length of
y = ln(cos x)
over x = [0, pi/4]
We have to integrate sqrt[1+y'2] from 0 to pi/4
1+y'^2 = 1+tan^2(x) = 1/cos^2(x)
So, the arc length is:
Integral from 0 to pi/4 of 1/cos(x)
= Integral from pi/4 to p1/2 of 1/sin(x)
We can write:
1/sin(x) =
1/[2sin(1/2 x) cos(1/2 x)] =
[sin^2(1/2x) + cos^2(1/2x)]
/[2sin(1/2 x) cos(1/2 x)] =
1/2 [tan(1/2 x) + cot(1/2 x)]
Which is trivial to integrate.