What is the arc length of #f(x)=cosx-sin^2x# on #x in [0,pi]#?

Question

1784 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T19:44:30

#S = 4.28154#

Arc length #S = int_0^pi sqrt(1+(y')^2) dx#

#y' = -sin x - 2 sin x cos x#

#S = int_0^pi sqrt(1+(-sin x - 2 sin x cos x)^2) dx#

#= int_0^pi sqrt(1+sin^2 x + 4 sin^2 x cos^2 x + 4 sin^2 x cos x) dx#

#= int_0^pi sqrt(1+sin^2 x + sin^2 2x + 2 sin2x sin x ) dx#

That doesn't simplify easily so computer solution is:

#S = 4.28154#