Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 36

Find the integrals $\displaystyle \int^{\pi/3}_0 \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta$

$\begin{equation} \begin{aligned} \int \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta &= \int \frac{\sin \theta (1 + \tan^2 \theta)}{\sec^2 \theta} d \theta && \text{Apply Pythagorean Identities } \left(1 + \tan^2 \theta = \sec^2 \theta\right) \\ \\ \int \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta &= \int \frac{\sin \theta \cancel{\sec^2 \theta} }{\cancel{\sec^2 \theta}} d \theta\\ \\ \int \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta &= \sin \theta d \theta\\ \\ \int \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta &= - \cos \theta + C\\ \\ \int^{\pi/3}_0 \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta &= - \cos \left( \frac{\pi}{3} \right) + C \left[ - \cos (0) + C \right]\\ \\ \int^{\pi/3}_0 \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta &= - \frac{1}{2} + C + 1 - C\\ \\ \int^{\pi/3}_0 \frac{\sin \theta + \sin \theta \tan^2\theta}{\sec^2 \theta} d \theta &= \frac{1}{2} \end{aligned} \end{equation}$