# What is the arc length of f(x)=sqrt(sinx)  in the interval [0,pi]?

Jan 8, 2018

The formula for arc length on interval $\left[a , b\right]$ is

$A = {\int}_{a}^{b} \sqrt{1 + {\left(f ' x\right)}^{2}} \mathrm{dx}$

The derivative of $f \left(x\right)$ will be obtained using the chain rule.

$f ' \left(x\right) = \cos x \cdot \frac{1}{2 \sqrt{\sin x}}$

f'(x) = cosx/(2sqrt(sinx)

Using the given formula:

$A = {\int}_{0}^{\pi} \sqrt{1 + {\left(\cos \frac{x}{2 \sqrt{\sin x}}\right)}^{2}} \mathrm{dx}$

$A = {\int}_{0}^{\pi} \sqrt{1 + {\cos}^{2} \frac{x}{4 \sin x}} \mathrm{dx}$

$A = {\int}_{0}^{\pi} \sqrt{1 + \frac{1}{4} \cot x \cos x} \mathrm{dx}$

Which according to the integral calculator has no solution through elementary antiderivatives. A numerical approximation for arc length gives

$A = 4.04$ units

to three significant figures.

Hopefully this helps!