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

##### 1 Answer
Aug 29, 2016

$S = 4.28154$

#### Explanation:

Arc length $S = {\int}_{0}^{\pi} \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$

$y ' = - \sin x - 2 \sin x \cos x$

$S = {\int}_{0}^{\pi} \sqrt{1 + {\left(- \sin x - 2 \sin x \cos x\right)}^{2}} \mathrm{dx}$

$= {\int}_{0}^{\pi} \sqrt{1 + {\sin}^{2} x + 4 {\sin}^{2} x {\cos}^{2} x + 4 {\sin}^{2} x \cos x} \mathrm{dx}$

$= {\int}_{0}^{\pi} \sqrt{1 + {\sin}^{2} x + {\sin}^{2} 2 x + 2 \sin 2 x \sin x} \mathrm{dx}$

That doesn't simplify easily so computer solution is:

$S = 4.28154$