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Magazine Chapter 7: Problem 32
In Problems 1-54, perform the indicated integrations. \(\int \frac{(6 t-1) \sin \sqrt{3 t^{2}-t-1}}{\sqrt{3 t^{2}-t-1}} d t\)
Short Answer
Expert verified
The integral evaluates to \(-\cos(\sqrt{3t^2 - t - 1}) + C\)."
Step by step solution
01
Identify the Integration Technique
The integral given is \( \int \frac{(6t-1) \sin \sqrt{3t^2-t-1}}{\sqrt{3t^2-t-1}} \, dt \). The presence of the square root \( \sqrt{3t^2-t-1} \) and its derivative \( 6t-1 \) in the numerator suggests a substitution method. Specifically, we can use a substitution where the variable inside the square root is set as a new variable.
02
Perform the Substitution
Let \( u = \sqrt{3t^2-t-1} \). Then, squaring both sides gives \( u^2 = 3t^2 - t - 1 \). Taking derivatives yields \( 2u \frac{du}{dt} = 6t - 1 \). Therefore, \( dt = \frac{2u}{6t-1} \, du \). However, notice that the expression \( 6t-1 \) precisely matches the term needed for substitution.
03
Substitute and Simplify the Integral
Substitute \( u = \sqrt{3t^2 - t - 1} \) into the integral. The original integral becomes \( \int \frac{(6t-1) \sin u}{u} \cdot \frac{du}{6t-1} \). The terms \( 6t-1 \) cancel out, simplifying the integral to \( \int \sin u \, du \).
04
Integrate the Simplified Expression
The integral \( \int \sin u \, du \) is a basic integral. Its solution is \(-\cos u + C\), where \( C \) is the constant of integration.
05
Back-Substitute the Original Variable
Replace \( u \) back with the original variable substitution \( u = \sqrt{3t^2 - t - 1} \). Thus, the antiderivative becomes \(-\cos(\sqrt{3t^2 - t - 1}) + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a valuable tool in calculus for simplifying complex integrals. It's particularly useful when dealing with integrals involving composite functions. The idea is to substitute part of the integral with a new variable. This turns a daunting integral into a simpler one.
- Identify the part of the integral: Look for a function within another, such as a square root, that also appears in the derivative form. For our example, we have the expression \( \sqrt{3t^2 - t - 1} \) whose derivative, \( 6t-1 \), is conveniently positioned in the numerator.
- Define a new variable: Set \( u = \sqrt{3t^2 - t - 1} \). This change of variables aims to reduce the complexity.
- Rework the integral: Substitute \( u \) into the integral, changing all expressions of \( t \) to expressions of \( u \). This usually simplifies the integral considerably.
- Revert the substitution: After integration, don't forget to convert back to the original variable to get the final antiderivative.
Indefinite Integral
An indefinite integral is the reverse process of differentiation and represents a family of functions whose derivatives yield the integrand. These integrals are expressed generally with a constant of integration, \( C \), because differentiation does not indicate the original constant that might have existed.
- Symbolizing the process: The notation \( \int f(x) \, dx \) denotes finding the antiderivative of \( f(x) \).
- Including the constant of integration: Since derivatives of constants vanish, recovered functions from integration are added with \( C \).
- Example: The integral \( \int \sin u \, du = -\cos u + C \) demonstrates how a basic pattern leads to another function plus \( C \).
Trigonometric Integration
Trigonometric integration frequently appears when dealing with sine, cosine, and other trigonometric functions within integrals. Recognizing and utilizing trigonometric identities can simplify these processes.
- Basic integrals: Familiarity with integrals like \( \int \sin x \, dx = -\cos x + C \) and \( \int \cos x \, dx = \sin x + C \) is essential.
- Using identities: Trigonometric identities, such as \( \sin^2 x + \cos^2 x = 1 \), help replace or reduce expressions to simplify integration.
- Simplifying complex expressions: For complex trigonometric integrals, substitution might involve setting a trigonometric function as the new variable, as shown in our specifc example where integration reduces to \( \int \sin u \, du \).
