Assistance with solving the integral

Question

Can you give me an idea how to handle this integral?

$\int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\,dx$

Answer 1

Since $1 - \cos(x) = 2 \sin^2\left(\frac{x}{2} \right) $ then \begin{align} \int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\, \mathrm{d}x & \overset{u= x -\pi}{=} 2\int_{0}^{\pi}2 \sin^2\left(\frac{u}{2} \right) \sqrt{2+2\sin^2\left(\frac{u}{2} \right)}\, \mathrm{d}u \\ & \overset{t = \frac{u}{2}}{=} 8 \sqrt{2} \int_0^{\frac{\pi}{2}} \sin^2(t) \sqrt{1 + \sin^2(t)}\, \mathrm{d}t\\ \end{align} And from this answer we know $$ \int_0^{\frac{\pi}{2}} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}\, \mathrm{d}\theta = \frac{(1-k^2)K(k)-(1-2k^2)E(k)}{3k^2} $$ So with $k=i$ you get $$ \int_{0}^{2\pi} (1+\cos{x})\sqrt{3+\cos{x}}\, \mathrm{d}x =8 \sqrt{2} E(i) - \frac{16 \sqrt{2}}{3} K(i) $$

Answer 2

Using the tangent half-angle substitution $$I=\int_{0}^{2\pi} (1+\cos(x))\sqrt{3+\cos(x)}\,dx=2\int_{0}^{\pi} (1+\cos(x))\sqrt{3+\cos(x)}\,dx$$ $$I=8 \sqrt{2} \int_0^\infty \sqrt{\frac{t^2+2}{\left(t^2+1\right)^5}}\,dt=4 \sqrt{2}\int_1^\infty \frac 1 {x^2}\sqrt{\frac{x+1}{(x-1) x}}$$ The antiderivative is nasty but its evaluation at $\infty$ is simply $$I=\frac{2 }{3 \sqrt{\pi }}\left(\Gamma \left(\frac{1}{4}\right)^2+12\, \Gamma \left(\frac{3}{4}\right)^2\right)$$ which is the same as $$I=16 \left(E\left(\frac{1}{2}\right)-\frac{1}{3}K\left(\frac{1}{2}\right)\right)$$ and $$I=8 \sqrt{2} \left(E(-1)-\frac{2 }{3}K(-1)\right)$$ already given by @Robert Lee.