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Magazine Chapter 5: Problem 36
Evaluate the integrals in Exercises \(13-48\) . $$ \int \frac{6 \cos t}{(2+\sin t)^{3}} d t $$
Short Answer
Expert verified
The integral evaluates to \( -\frac{3}{(2 + \sin t)^2} + C \).
Step by step solution
01
Identify the Substitution
To simplify the integral, we look for a substitution that will make the denominator easier to manage. Observing the inner function, we can use the substitution \( u = 2 + \sin t \). Differentiating both sides with respect to \( t \), we find \( du = \cos t \cdot dt \).
02
Substitute and Simplify the Integral
By substituting \( u = 2 + \sin t \) and \( du = \cos t \cdot dt \), the integral becomes \( \int \frac{6 \cdot (\cos t \cdot dt)}{u^3} \). This simplifies to \( 6 \int \frac{1}{u^3} du \).
03
Integrate with Respect to the New Variable
The integral \( \int \frac{1}{u^3} du \) is a standard integral that evaluates to \( -\frac{1}{2u^2} \). Therefore, the integral \( 6 \int \frac{1}{u^3} du \) evaluates to \( -\frac{6}{2u^2} \), simplifying further to \( -\frac{3}{u^2} \).
04
Substitute back to the Original Variable
We substitute back \( u = 2 + \sin t \) to express the answer in terms of \( t \). This gives \( -\frac{3}{(2 + \sin t)^2} \).
05
Introduce the Constant of Integration
Because we have evaluated an indefinite integral, we add a constant of integration \( + C \) to our result. Thus, the final answer is \( -\frac{3}{(2 + \sin t)^2} + C \).
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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 powerful technique for evaluating integrals, especially when handling complex expressions.
It involves identifying a part of the integral that can be replaced with a single variable, simplifying the integration process.
In our exercise, we used the substitution \( u = 2 + \sin t \). This is a strategic choice, as it targets the inner function of the integral's denominator.
The key to successful substitution is choosing a \( u \) that simplifies the integral to a more recognizable form.
It involves identifying a part of the integral that can be replaced with a single variable, simplifying the integration process.
In our exercise, we used the substitution \( u = 2 + \sin t \). This is a strategic choice, as it targets the inner function of the integral's denominator.
The key to successful substitution is choosing a \( u \) that simplifies the integral to a more recognizable form.
- Deriving \( du \) requires differentiating \( u \) with respect to \( t \). For our function, \( u = 2 + \sin t \), differentiating results in \( du = \cos t \cdot dt \).
- Replacing all instances of the original variable with \( u \) and \( du \) transforms the integral, allowing for simpler computation.
Trigonometric Functions
Trigonometric functions like \( \sin t \) and \( \cos t \) are integral to the operation of finding derivatives and integrals.
In many calculus problems, they appear either in their basic form or embedded within more complex expressions.
Its derivative, \( \cos t \), conveniently matches the numerator, \( 6 \cos t \), allowing for straightforward integration post-substitution.
Understanding such correspondences is key to recognizing and selecting effective substitutions.
In many calculus problems, they appear either in their basic form or embedded within more complex expressions.
- Recognizing these functions as part of the differentiation or integration process is critical to setting the correct substitutions.
- Trigonometric identities and derivatives often play a role, requiring a solid understanding of their properties.
Its derivative, \( \cos t \), conveniently matches the numerator, \( 6 \cos t \), allowing for straightforward integration post-substitution.
Understanding such correspondences is key to recognizing and selecting effective substitutions.
Indefinite Integrals
Indefinite integrals represent the antiderivative of a function, expressing a family of functions rather than a fixed value.
Unlike definite integrals, they do not evaluate to a number but to a function, signified by adding a "constant of integration".
The integral evaluates to \( -\frac{1}{2u^2} \), reflecting a standard form and revealing steps towards the expression of the function.
Unlike definite integrals, they do not evaluate to a number but to a function, signified by adding a "constant of integration".
- Notation for indefinite integrals is expressed as \( \int f(x) \, dx \), aiming to find a function \( F(x) \) where \( F'(x) = f(x) \).
- Finding the antiderivative often involves reversing differentiation rules or employing techniques like u-substitution.
The integral evaluates to \( -\frac{1}{2u^2} \), reflecting a standard form and revealing steps towards the expression of the function.
Constants of Integration
When dealing with indefinite integrals, a "constant of integration" is essential. It arises because the differentiation of a constant is zero, making the original function ambiguous.
Every indefinite integral solution is therefore a family of functions differing by a constant.
It represents that despite solving specifically, the solution's derivative equally matches the original integrated function.
Keeping this in mind helps understand the integral's role in quantitatively describing the behavior of a function beyond fixed endpoints.
Every indefinite integral solution is therefore a family of functions differing by a constant.
- Mathematically, it's added to the final solution as "\(+ C\)".
- The constant ensures the inclusion of every possible antiderivative.
It represents that despite solving specifically, the solution's derivative equally matches the original integrated function.
Keeping this in mind helps understand the integral's role in quantitatively describing the behavior of a function beyond fixed endpoints.
