$\int_{0}^{2\Pi }\sqrt{4\cos^{2}t +sin^{2}t}dt$
$=\int_{0}^{2\Pi }\sqrt{3\cos^{2}t +1}dt$
$=\sqrt{3}\int_{0}^{2\Pi }\sqrt{\cos^{2}t +\left ( \frac{1}{\sqrt{3}} \right )^{2}}dt$
Now put in direct formula
$=\frac{cost\sqrt{cos^{2}t+\frac{1}{3}}}{2}+\frac{1/3}{2}log\begin{vmatrix} cos^{2}t & +&\sqrt{cos^{2}t+\frac{1}{3}} \end{vmatrix}$
As we know, $\int \sqrt{x^{2}+a^{2}}dx=\frac{x\sqrt{x^{2}+a^{2}}}{2}+\frac{a^{2}}{2}log\begin{vmatrix} x &+ & \sqrt{x^{2}+a^{2}} \end{vmatrix}$
Now put the range on it