# How do you find the arc length of the curve y=2sinx over the interval [0,2pi]?

Jan 15, 2017

$\approx 5.27037$

#### Explanation:

The Arc Length of curve $y = f \left(x\right)$ is calculated using the formula:

$L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \setminus \mathrm{dx}$

So with $f \left(x\right) = 2 \sin x$, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cos x$

And so the required Arc Length is given by:

$L = {\int}_{0}^{\pi} \sqrt{1 + {\left(2 \cos x\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus = {\int}_{0}^{\pi} \sqrt{1 + 4 {\cos}^{2} x} \setminus \mathrm{dx}$

This integrand does not have an elementary solution

Using Wolfram Alpha this integral evaluates to:

$L \approx 5.27037$