Solve $\int_0^{\frac{\pi}{3}} \frac{\sin\theta + \sin\theta\tan^2\theta}{\sec^2\theta}d\theta$ without using integration by parts or substitution

Question

I was required to solve $$\int_0^{\frac{\pi}{3}} \frac{\sin\theta + \sin\theta\tan^2\theta}{\sec^2\theta}d\theta$$

without using integration by parts or substitution. This is, I must use trigonometric identities to boil the expression down to identifiable antiderivatives and apply the fundamental theorem of calculus directly. Clearly

$$\frac{\sin\theta + \sin\theta\tan^2\theta}{\sec^2\theta} = \frac{\sin\theta}{\sec^2 \theta} + \sin\theta$$

The second term is easily integrable because $\int \sin x \space dx = -\cos x + C$. But I can't seem to transform $\frac{\sin\theta}{\sec^2 \theta}$ into an identifiable antiderivative. I'm sure there's some trigonometric identity or trick I'm missing, but after consulting Stewart's Calculus and the web I still did not find a possible solution.

Thanks in advance.

Ref: Problem 38, Section 5.4 in Stewart's Calculus: Early transcendentals, 7 ed.

Answer 1

You may want to write the numerator as

$\sin\theta\color{blue}{(1+\tan^2\theta)}$

and compare the blue factor with $\sec^2\theta$ in the denominator. The integral is really a softball in disguise.