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

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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:

$S = 4.28154$

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