Problem 36 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(13-48\) . $$ \int \frac{6 \cos t}{(2+\sin t)^{3}} d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( -\frac{3}{(2 + \sin t)^2} + C \).

Step 1: 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 \( du = \cos t \cdot dt \).

Step 2: 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 \( 6 \int \frac{1}{u^3} du \).

Step 3: 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 \( -\frac{3}{u^2} \).

Step 4: Substitute back to the Original Variable

We substitute back \( u = 2 + \sin t \) to express the answer in terms of \( t \). This gives \( -\frac{3}{(2 + \sin t)^2} \).

Step 5: 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 \( -\frac{3}{(2 + \sin t)^2} + C \).

Key Concepts

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.

• 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.

• 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.

In this exercise, the trigonometric function \( \sin t \) was chosen as part of our substitution to simplify the denominator.
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".

• 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.

In this particular case, we identified \( \int \frac{1}{u^3} \, du \) by substituting \( u \) back into terms of \( t \).
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.

• Mathematically, it's added to the final solution as "\(+ C\)".
• The constant ensures the inclusion of every possible antiderivative.

This constant is vital when expressing the final answer post-integration.
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.