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

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

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\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) \\
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\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}
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