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.


1 Answer 1


You may want to write the numerator as


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

  • 1
    $\begingroup$ Nice. I totally missed it. Thank you very much! $\endgroup$
    – lafinur
    Commented Aug 5, 2022 at 20:40

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