Answer
$$\frac{1}{2}\ln2.$$
Work Step by Step
We have
$$\int_0^{\pi/6}\tan 2\theta d\theta=\int_0^{\pi/6}\frac{\sin2\theta}{\cos2\theta} d\theta\\
=-\frac{1}{2}\int_0^{\pi/6}\frac{(\cos2\theta)'}{\cos2\theta}=-\frac{1}{2}\ln\cos2\theta|_0^{\pi/6}\\
=-\frac{1}{2}(\ln\cos(\pi/3)-\ln \cos 0)=-\frac{1}{2}\ln(1/2)=\frac{1}{2}\ln2.$$
