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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 8.6.38
Chapter 8, Problem 8.6.38

7–84. Evaluate the following integrals.
38. ∫ from π/6 to π/2 [cos x · ln(sin x)] dx

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Step 1: Recognize that the integral involves a product of functions, specifically cos(x) and ln(sin(x)). This suggests that substitution might simplify the problem. Identify a substitution that simplifies the integral.
Step 2: Let u = sin(x). Then, compute the derivative of u with respect to x: du = cos(x) dx. This substitution transforms the integral into terms of u.
Step 3: Change the limits of integration to match the substitution. When x = π/6, u = sin(π/6) = 1/2. When x = π/2, u = sin(π/2) = 1. The new limits of integration are from u = 1/2 to u = 1.
Step 4: Rewrite the integral in terms of u using the substitution. The integral becomes ∫ from 1/2 to 1 [ln(u)] du.
Step 5: Evaluate the integral ∫ ln(u) du using integration by parts. Recall the formula for integration by parts: ∫ v dw = v*w - ∫ w dv. Let v = ln(u) and dw = du, then proceed to solve.

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Key Concepts

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

Definite Integrals

A definite integral calculates the accumulation of a function's values over a specific interval, represented as ∫ from a to b f(x) dx. It provides the net area under the curve of the function f(x) between the limits a and b. The Fundamental Theorem of Calculus connects differentiation and integration, allowing us to evaluate definite integrals using antiderivatives.
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Integration by Parts

Integration by parts is a technique used to integrate products of functions, based on the formula ∫ u dv = uv - ∫ v du. Here, u and dv are chosen parts of the integrand, where u is typically a function that simplifies upon differentiation, and dv is a function that can be easily integrated. This method is particularly useful when dealing with products of logarithmic and trigonometric functions, as seen in the given integral.
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Logarithmic and Trigonometric Functions

Logarithmic functions, such as ln(x), and trigonometric functions, like sin(x) and cos(x), are fundamental in calculus. The integral in question involves the product of cos(x) and ln(sin(x)), which requires understanding their properties and behaviors. Recognizing how these functions interact, especially within the limits of integration, is crucial for evaluating the integral accurately.
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