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The arc length of some function $f(x)$ is given by $$L=\int{\sqrt{1+f'\left(x\right)^2}}dx$$ Plugging $\cos x$ in for $f$ and using basic algebra, this simplifies as follows $$L=\int{\sqrt{1+\cos'\left(x\right)^2}}dx$$ $$L=\int{\sqrt{1-\sin\left(x\right)^2}}dx$$ $$L=\int{\sqrt{\cos\left(x\right)^2}}dx$$ $$L=\int{\pm\cos\left(x\right)}dx$$ $$L=\pm\sin x$$ Obviously this is wrong. The real answer involves elliptic integrals, but my question is, why does this approach give such an incorrect result? What am I overlooking?

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$$(-\sin x)^2 \ne -(\sin x)^2$$

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  • $\begingroup$ For OP: Also note that $-\sin(x)^2$ is interpreted as $-(\sin(x)^2)$ because of order of operations. $\endgroup$ – M. Winter Sep 10 '17 at 7:59
  • $\begingroup$ I figured it was something simple like this. facepalm Thanks, Kenny. :) $\endgroup$ – Steven Fontaine Sep 10 '17 at 10:54

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