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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.3.19
Chapter 8, Problem 8.3.19

9–61. Trigonometric integrals Evaluate the following integrals.
19. ∫[0 to π/3] sin⁵x cos⁻²x dx

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Step 1: Recognize that the integral involves powers of sine and cosine. To simplify, consider using trigonometric identities or substitution methods. For example, you can use the identity sin²x + cos²x = 1 if needed.
Step 2: Rewrite the integral to make it more manageable. Since the integrand is sin⁵x cos⁻²x, you can split sin⁵x into sin³x sin²x. Then, use the identity sin²x = 1 - cos²x to express it in terms of cosine.
Step 3: Perform substitution. Let u = cosx, which implies du = -sinx dx. Adjust the limits of integration accordingly: when x = 0, u = cos(0) = 1, and when x = π/3, u = cos(π/3) = 1/2.
Step 4: Substitute into the integral. Replace sinx dx with -du, and express the integrand in terms of u. The integral becomes ∫[1 to 1/2] (1 - u²)² u⁻² (-du).
Step 5: Expand and simplify the integrand. Multiply out (1 - u²)² to get 1 - 2u² + u⁴. Then divide each term by u² to obtain u⁻² - 2 + u². Integrate each term separately 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 Functions

Trigonometric functions, such as sine and cosine, are fundamental in calculus, particularly in integration. They describe relationships between angles and sides of triangles and are periodic functions. Understanding their properties, such as their ranges and symmetries, is essential for evaluating integrals involving these functions.
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Integration Techniques

Integration techniques, including substitution and integration by parts, are crucial for solving complex integrals. In the case of the integral ∫ sin⁵x cos⁻²x dx, recognizing patterns and applying appropriate techniques can simplify the process. Mastery of these techniques allows for the evaluation of integrals that may not be straightforward.
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Definite Integrals

Definite integrals calculate the area under a curve between two specified limits, in this case, from 0 to π/3. Understanding the properties of definite integrals, such as the Fundamental Theorem of Calculus, is vital for evaluating them. This theorem connects differentiation and integration, providing a method to compute the value of the integral using antiderivatives.
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