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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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    You are supposed to look at $L(x)= \int_0^x f(y) dy$ – 2012-06-14

2 Answers 2

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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.

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    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. – 2012-06-14
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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 $1$.