Question

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

• A. \( \pi + \sqrt{3} \)
• B. \( 3\sqrt{3} - \pi \)
• C. \( \sqrt{3} - \pi \)
• D. \( \pi - \sqrt{3} \)

Hint:

For integrals involving trigonometric identities, simplify the integrand using known identities such as \( \tan^2 \theta = \sec^2 \theta - 1 \). Also, remember to perform substitution for integrals with functions of \( \theta \).

Answer 1

The Correct Option is B

We are asked to evaluate the integral \( \int_{0}^{\pi} \tan^2 \left( \frac{\theta}{3} \right) d\theta \). First, use the identity \( \tan^2 \theta = \sec^2 \theta - 1 \), so the integral becomes: \[ \int_{0}^{\pi} \left( \sec^2 \left( \frac{\theta}{3} \right) - 1 \right) d\theta \] This can be split into two integrals: \[ \int_{0}^{\pi} \sec^2 \left( \frac{\theta}{3} \right) d\theta - \int_{0}^{\pi} 1 \, d\theta \] For the first integral, apply substitution \( u = \frac{\theta}{3} \), which gives \( du = \frac{1}{3} d\theta \), so the limits change as follows: when \( \theta = 0 \), \( u = 0 \); and when \( \theta = \pi \), \( u = \frac{\pi}{3} \). The first integral becomes: \[ \int_{0}^{\pi/3} 3 \sec^2 u \, du = 3 \left[ \tan u \right]_{0}^{\pi/3} = 3 \left( \tan \left( \frac{\pi}{3} \right) - 0 \right) = 3 \times \sqrt{3} \] The second integral is straightforward: \[ \int_{0}^{\pi} 1 \, d\theta = \pi \] Now, combining both parts: \[ 3\sqrt{3} - \pi \] Thus, the value of the integral is \( 3\sqrt{3} - \pi \), matching option (B).