Finding $\int_0^{2\pi} \sqrt{x + \frac{1}{4x}}\; dx $

Question

I am to determine the length of the curve given by polar equation $r = \sqrt{\theta}$ for $0\leq \theta\leq 2\pi$. I know that this is the integral $$ \int_0^{2\pi} \sqrt{\theta + \frac{1}{4\theta}}\; d\theta. $$ But I am not sure how to find this integral. I know that I should write $$ \theta + \frac{1}{4\theta} $$ as the square of something to do away with the radical.

Answer 1

Letting $\theta=u^2$ then $d\theta=2u\,du$ so we get:

$$\int \sqrt{\theta +\frac{1}{4\theta}}\,d\theta=\int \sqrt{u^2+\frac{1}{4u^2}}\cdot 2u\,du=\int\sqrt{4u^4+1}\,du$$

Wolfram Alpha gave a monstrosity for this indefinite integral in terms of an elliptic integral.

Answer 2

Assuming we are considering the length of the graph of $\sqrt{\theta}$ over $[0,2\pi]$, this is the length of an arc of a parabola (the graph of $f(x)=x^2$ over $[0,\sqrt{2\pi}]$), which is also $$ \int_{0}^{\sqrt{2\pi}}\sqrt{1+4x^2}\,dx =\left[\frac{1}{2}x\sqrt{1+4x^2}+\frac{1}{4}\text{arcsinh}(2x)\right]_{0}^{\sqrt{2\pi}}=\sqrt{\frac{\pi}{2}(8\pi +1)}+\frac{1}{4}\text{arcsinh}(2\sqrt{2\pi}).$$

If $r=\sqrt{\theta}$ means that we are considering the equation of a curve in polar coordinates, we cannot avoid elliptic integrals, as shown by Thomas Andrews.