## Calculus: Early Transcendentals 8th Edition

$$\int^{\pi/3}_0\frac{\sin\theta+\sin\theta\tan^2\theta}{\sec^2\theta}d\theta=\frac{1}{2}$$
$$A=\int^{\pi/3}_0\frac{\sin\theta+\sin\theta\tan^2\theta}{\sec^2\theta}d\theta$$ $$A=\int^{\pi/3}_0Bd\theta$$ *Consider B: $$B=\frac{\sin\theta+\sin\theta\tan^2\theta}{\sec^2\theta}$$ $$B=\frac{\sin\theta(1+\tan^2\theta)}{\sec^2\theta}$$ $$B=\frac{\sin\theta\sec^2\theta}{\sec^2\theta}$$ $$B=\sin\theta$$ So, $$A=\int^{\pi/3}_0(\sin\theta)d\theta$$ According to Table 1, we have $$\int(\sin x)dx=-\cos x+C$$ Therefore, $$A=-\cos\theta\Bigg]^{\pi/3}_0$$ $$A=-(\cos\frac{\pi}{3}-\cos0)$$ $$A=-(\frac{1}{2}-1)$$ $$A=\frac{1}{2}$$