# $s(x)$ is a arc length function, find $s'(x)$

Here is the problem is my textbook:

Suppose $s(x)$ is the arc length function for the curve $y=\sin x$ taking $(0,1)$ as the starting point. Find $s’(x)$.

According to arc length formula, I have :

$$L = \int_0^1 \sqrt{1+\left((\sin x)'\right)^2}dx = \int_0^1 \sqrt{1+\cos^2x}dx$$

But this integral is hard to solve (and When I calculate in Maple, I cannot have the exactly result.) So, I cannot find $s(x)$ to calculate $s’(x)$

Thanks :)

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Have you seen differentiation under the integral sign? en.wikipedia.org/wiki/Differentiation_under_the_integral_sign – Argon Jun 14 '12 at 17:39
You are supposed to look at $L(x)= \int_0^x f(y) dy$ – user20266 Jun 14 '12 at 17:42

You're given $$s(x):=\int_0^x\sqrt{1+\cos^2t}\,dt$$and since this Riemann integral exists (the function in the integral is continuous and thus integrable in the inverval $\,[0,1]\,$), the primitive $\,G(x)\,$ of the function is a derivable function of the upper limit, so if $$s(x)=\int_0^x\sqrt{1+\cos^2t}\,dt=G(x)-G(0)\Longrightarrow s'(x)=G'(x)=\sqrt{1+\cos^2x}$$By the way, the point$\,(0,1)\,$ is NOT on the curve $\,y=\sin x\,$ , so the function might be $\,\displaystyle{y=\sin\left(\frac{\pi}{2}+ x\right)}\,$ or else you meant something else.
The point $(x=0, y=1)$ does not lie on $y = \sin\pi x$ either. :) – Rahul Jun 14 '12 at 18:30
He...true, @Rahul. I thought of $\,\displaystyle{y=\sin\left(\frac{\pi}{2}+x\right)}\,$, but wrongly wrote the above. I'll edit it now. Thanks. – DonAntonio Jun 14 '12 at 18:38
The arc length function $s(X)$ is supposed to be the length of the curve $y = f(x)$ from $x=0$ to $x=X$. So the integral should go from $0$ to $X$, not $0$ to $1$. All you need to differentiate the integral with respect to $X$ is the Fundamental Theorem of Calculus: you don't need to actually evaluate the integral.
By the way, if it's $y = \sin x$ I don't understand "taking $(0,1)$ as the starting point". $\sin(0) = 0$, not $1$.