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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.3.12

Evaluate the integrals in Exercises 1–22.
∫ cos³(2x) sin⁵(2x) dx

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1
Recognize that the integral involves powers of sine and cosine functions with the same argument, which suggests using trigonometric identities or substitution to simplify the expression.
Express one of the trigonometric functions in terms of the other using the Pythagorean identity. For example, rewrite \(\cos^3(2x)\) as \(\cos(2x) \cdot \cos^2(2x)\) and then use \(\cos^2(2x) = 1 - \sin^2(2x)\).
Rewrite the integral as \(\int \cos(2x) (1 - \sin^2(2x)) \sin^5(2x) \, dx\), which separates the powers of sine and cosine and prepares for substitution.
Use the substitution \(u = \sin(2x)\), then compute \(du = 2 \cos(2x) \, dx\), or equivalently, \(\cos(2x) \, dx = \frac{du}{2}\), to rewrite the integral entirely in terms of \(u\).
Substitute into the integral to get an integral in terms of \(u\): \(\int (1 - u^2) u^5 \frac{du}{2}\), then simplify and integrate the resulting polynomial expression with respect to \(u\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Trigonometric Identities

Trigonometric identities like power-reduction and product-to-sum formulas help simplify integrals involving powers of sine and cosine. For example, expressing cos³(2x) as cos(2x)·cos²(2x) and then using cos²(θ) = 1 - sin²(θ) can make the integral more manageable.
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Substitution Method

The substitution method involves changing variables to simplify the integral. In this problem, substituting u = sin(2x) or u = cos(2x) can transform the integral into a polynomial form, making it easier to integrate.
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Integration of Powers of Sine and Cosine

Integrals involving powers of sine and cosine often require reducing the powers step-by-step or using identities to rewrite the integrand. Recognizing patterns and applying reduction formulas is key to solving such integrals efficiently.
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