(b) Evaluate \( : \int_{0}^{\pi} \cos \theta \tan \theta d \theta \)

To evaluate the integral \( \int_{0}^{\pi} \cos \theta \tan \theta \, d\theta \), we can start by rewriting \(\tan \theta\) in terms of sine and cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] Thus, we can rewrite the integral: \[ \int_{0}^{\pi} \cos \theta \tan \theta \, d\theta = \int_{0}^{\pi} \cos \theta \cdot \frac{\sin \theta}{\cos \theta} \, d\theta = \int_{0}^{\pi} \sin \theta \, d\theta \] Now we compute the integral of \(\sin \theta\): \[ \int \sin \theta \, d\theta = -\cos \theta + C \] Evaluating the definite integral from \(0\) to \(\pi\): \[ \int_{0}^{\pi} \sin \theta \, d\theta = [-\cos \theta]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \] Therefore, the value of the integral is: \[ \int_{0}^{\pi} \cos \theta \tan \theta \, d\theta = 2 \]