Answer
$$\int^{\pi}_{0}\tan^2\frac{\theta}{3} d\theta=3\sqrt3-\pi$$
Work Step by Step
$$A=\int^{\pi}_{0}\tan^2\frac{\theta}{3} d\theta$$
Recall the identity: $$\tan^2a=\sec^2a-1$$
Apply the identity here, we have
$$A=\int^{\pi}_{0}(\sec^2\frac{\theta}{3}-1)d\theta$$ $$A=\Big(3\tan\frac{\theta}{3}-\theta\Big)\Big]^{\pi}_{0}$$ $$A=\Big(3\tan\frac{\pi}{3}-\pi\Big)-\Big(3\tan0-0\Big)$$ $$A=3\sqrt3-\pi-(0-0)$$ $$A=3\sqrt3-\pi$$
