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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 8.2.42

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ sin(2x) cos(4x) dx

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1
Recognize that the integral involves the product of sine and cosine functions with different arguments: \(\int \sin(2x) \cos(4x) \, dx\).
Recall the product-to-sum identities, which help simplify products of sine and cosine into sums of trigonometric functions. Specifically, use the identity: \(\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]\).
Apply the identity to rewrite the integral as: \(\int \sin(2x) \cos(4x) \, dx = \int \frac{1}{2} [\sin(2x + 4x) + \sin(2x - 4x)] \, dx = \frac{1}{2} \int [\sin(6x) + \sin(-2x)] \, dx\).
Simplify the expression inside the integral, noting that \(\sin(-2x) = -\sin(2x)\), so the integral becomes \(\frac{1}{2} \int [\sin(6x) - \sin(2x)] \, dx\).
Now, integrate each sine term separately using the basic integral formula \(\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C\). Write the integral as the sum of these two integrals and proceed accordingly.

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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, such as product-to-sum formulas, allow the transformation of products of sine and cosine functions into sums or differences of trigonometric functions. For example, sin(A)cos(B) = ½[sin(A+B) + sin(A−B)], which simplifies integration by converting complex products into easier integrals.
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