# Arc length of $x\sin x$

Can anyone tell me if it is possible to find a formula for the arc length of the function $f(x)= x\sin x$? I've been working on this on and off for a few days and I'm starting to think it's impossible, yet it seems like such a simple expression.

Wolfram Alpha can give me a determinate answer, but this won't work for me.

This is the integral that should solve for the arc length:

$$\int \sqrt{ (x \cos(x)+\sin(x))^2 + 1\,}\,\, \mathrm{d}x$$

I can't seem to find a solution for it. Can anyone help me, or explain why it can't be solved?

In principle, you can use the Risch algorithm to prove that $\sqrt{(x\cos(x)+\sin(x))^2+1}$ has no elementary antiderivative.
But, let us consider the case where you need to get a number for the arc lenght between $x=0$ and $x=a$. What you can do is to expand the integrand as a Taylor series around $x=0$ and use as many terms as you can handle. For example $$\sqrt{ (x \cos(x)+\sin(x))^2 + 1\,}=1+2 x^2-\frac{10 x^4}{3}+\frac{629 x^6}{90}-\frac{12329 x^8}{630}+\frac{1770127 x^{10}}{28350}+O\left(x^{11}\right)$$ Integrating between $0$ and $a$ gives for the arc length $$L=a+\frac{2 a^3}{3}-\frac{2 a^5}{3}+\frac{629 a^7}{630}-\frac{12329 a^9}{5670}+\frac{1770127 a^{11}}{311850}+\cdots$$ For sure, this will only work for small values of $a$. Just for your curiosity, I give below a few results $$\left( \begin{array}{ccc} a & approx & "exact" \\ 0.1 & 0.10066 & 0.10066 \\ 0.2 & 0.205132 & 0.205132 \\ 0.3 & 0.316566 & 0.316564 \\ 0.4 & 0.437144 & 0.437070 \\ 0.5 & 0.568825 & 0.567681 \\ 0.6 & 0.718789 & 0.708479 \\ 0.7 & 0.923335 & 0.858767 \end{array} \right)$$
If you need to work with larger values of $a$, you should need to consider numerical integration.