Question

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

Accepted Answer

Start by rewriting the integral $$ \int \frac{1}{\cos 	heta + \sin 2	heta} \, d	heta $$ and recall the double-angle identity for sine: $$ \sin 2	heta = 2 \sin 	heta \cos 	heta . $$Substitute this into the integral to get $$ \int \frac{1}{\cos 	heta + 2 \sin 	heta \cos 	heta} \, d	heta . $$Factor $$ \cos 	heta $$ from the denominator: $$ \cos 	heta + 2 \sin 	heta \cos 	heta = \cos 	heta (1 + 2 \sin 	heta) . So $$the integral becomes $$ \int \frac{1}{\cos 	heta (1 + 2 \sin 	heta)} \, d	heta . $$Rewrite the integral as $$ \int \frac{1}{\cos 	heta (1 + 2 \sin 	heta)} \, d	heta = \int \frac{\sec 	heta}{1 + 2 \sin 	heta} \, d	heta $$, since $$ \sec 	heta = \frac{1}{\cos 	heta} . $$Use the substitution $$ u = 1 + 2 \sin 	heta . $$Then, compute $$ du = 2 \cos 	heta \, d	heta $$, which implies $$ d	heta = \frac{du}{2 \cos 	heta} . $$Rewrite the integral in terms of $$ u $$ and $$ 	heta $$, and express $$ \sec 	heta \, d	heta $$ using the substitution. This will allow you to simplify the integral and proceed with integration techniques such as partial fractions or further substitutions.

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

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Key Concepts

Trigonometric Identities

Integration of Rational Trigonometric Functions

Substitution Method in Integration