The value of $\int\limits_{0}^{\pi} \tan^2\left(\frac{\theta}{3}\right) d\theta$ is:

$3\sqrt{3} - \pi$

The correct answer is Option (2) → $3\sqrt{3} - \pi$

$\int\limits_{0}^{\pi} \tan^2\left(\frac{\theta}{3}\right) d\theta = \int\limits_{0}^{\pi} \left[ \sec^2\left(\frac{\theta}{3}\right) - 1 \right] d\theta$

$= \int\limits_{0}^{\pi} \sec^2\left(\frac{\theta}{3}\right) d\theta - \int\limits_{0}^{\pi} d\theta$

$= \left[ \frac{\tan\left(\frac{\theta}{3}\right)}{\frac{1}{3}} \right]_{0}^{\pi} - \left[ \theta \right]_{0}^{\pi}$

$= \left( 3\tan\frac{\pi}{3} - 3\tan\frac{0}{3} \right) - (\pi - 0)$

$= 3 \times \sqrt{3} - \pi$