Length of curve: cannot solve integral

Question

I was trying to calculate the length of the cardioid, the curve $\phi$ in $\mathbb{R^3}$ parametrized by $\rho = 1+ \cos \theta$ for $\theta \in (0, 2\pi)$.

Switching to Cartesian coordinates and simplifying a whole bunch offsides and cosines I managed to reduce the integral as such: $$ \int _0 ^{2\pi}||\phi||\text d \theta = 2\sqrt2\int_0^\pi\sqrt{1+\cos\theta}$$ (Taking advantage of the symmetry of $\cos\theta$).

Now, I am told it should come up with 8 as the answer, and Wolframalpha confirms it but I'm stuck.

As an extra: is there a way to do this integration by series?

Answer 1

Use the fact that $$ 1+\cos\theta=2\cos^2(\theta/2). $$

Answer 2

Just out of curiosity, using Taylor series of $\cos(\theta)$ and the generalized binomial theorem, you should arrive to $$\sqrt{1+\cos(\theta)}=\sqrt{2}-\frac{\theta ^2}{4 \sqrt{2}}+\frac{\theta ^4}{192 \sqrt{2}}-\frac{\theta ^6}{23040 \sqrt{2}}+O\left(\theta ^8\right)$$ Integrating $$\int \sqrt{1+\cos(\theta)}\, d\theta=\sqrt{2} \theta -\frac{\theta ^3}{12 \sqrt{2}}+\frac{\theta ^5}{960 \sqrt{2}}-\frac{\theta ^7}{161280 \sqrt{2}}+O\left(\theta ^9\right)$$ Using the bounds, $$\int_ 0^\pi \sqrt{1+\cos(\theta)}\, d\theta=\frac{\pi \left(322560-13440 \pi ^2+168 \pi ^4-\pi ^6\right)}{161280 \sqrt{2}}\approx 2.82798$$ Using one more term in the expansion, you should get $\approx 2.82844$.