Evaluate the integrals in Exercises 39–54.
∫ 1 / (cos θ + sin 2θ) dθ

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Start by rewriting the integral \( \int \frac{1}{\cos \theta + \sin 2\theta} \, d\theta \) and recall the double-angle identity for sine: \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substitute this into the integral to get \( \int \frac{1}{\cos \theta + 2 \sin \theta \cos \theta} \, d\theta \).

Factor \( \cos \theta \) from the denominator: \( \cos \theta + 2 \sin \theta \cos \theta = \cos \theta (1 + 2 \sin \theta) \). So the integral becomes \( \int \frac{1}{\cos \theta (1 + 2 \sin \theta)} \, d\theta \).

Rewrite the integral as \( \int \frac{1}{\cos \theta (1 + 2 \sin \theta)} \, d\theta = \int \frac{\sec \theta}{1 + 2 \sin \theta} \, d\theta \), since \( \sec \theta = \frac{1}{\cos \theta} \).

Use the substitution \( u = 1 + 2 \sin \theta \). Then, compute \( du = 2 \cos \theta \, d\theta \), which implies \( d\theta = \frac{du}{2 \cos \theta} \).

Rewrite the integral in terms of \( u \) and \( \theta \), and express \( \sec \theta \, d\theta \) using the substitution. This will allow you to simplify the integral and proceed with integration techniques such as partial fractions or further substitutions.

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