Problem 7

Question

Evaluate the indefinite integrals in Exercises $1-16$ by using the given substitutions to reduce the integrals to standard form. $$ \int \sin 3 x d x, \quad u=3 x $$

Step-by-Step Solution

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Answer

The indefinite integral is $-\frac{1}{3} \cos(3x) + C$.

1Step 1: Substitution of Variables

The problem suggests using the substitution $ u = 3x $. First, compute the differential $ du $. Since $ u = 3x $, differentiating both sides with respect to $ x $ gives $ du = 3 dx $.

2Step 2: Change in Differential

Solve for $ dx $ in terms of $ du $. We have $ du = 3 dx $, therefore, $ dx = \frac{du}{3} $. Substitute this into the integral.

3Step 3: Substitute in Integral

Substitute $ u = 3x $ and $ dx = \frac{du}{3} $ into the integral: $ \int \sin(3x) \, dx = \frac{1}{3} \int \sin u \, du $.

4Step 4: Integrate in Standard Form

The integral $ \int \sin u \, du $ is a standard form, which integrates to $ -\cos u $. Thus, we have $ \frac{1}{3} \int \sin u \, du = -\frac{1}{3} \cos u + C $.

5Step 5: Substitute Back to Original Variable

Finally, substitute $ u = 3x $ back into the antiderivative: $ -\frac{1}{3} \cos(3x) + C $.

Key Concepts

Substitution MethodTrigonometric IntegrationDifferential Calculus

Substitution Method

The substitution method is a powerful technique used in calculus for evaluating integrals. It involves changing variables to simplify an integral, making it easier to solve.

In our exercise, we employed the substitution method by setting the variable substitution as $ u = 3x $. This was suggested to transform the integral into a more manageable form. The steps involved are:

First, differentiate the new variable substitution to find the differential. Here, we calculated $ du = 3 \, dx $.
Next, solve for $ dx $, which gives $ dx = \frac{du}{3} $, allowing us to replace $ dx $ in the original integral.
Substitute both the variable $ u $ and $ dx $ in the integral, turning $ \int \sin(3x) \, dx $ into $ \frac{1}{3} \int \sin(u) \, du $.

This method effectively converts a complex integral into a standard one, which is easier to integrate. Substitution brings simplicity and often reveals the underlying structure of an integral, making it essential in calculus problem-solving.

Trigonometric Integration

Trigonometric integration is a method of integrating functions that involve trigonometric functions such as sine, cosine, tangent, and their inverses.

In the given exercise, we worked with the trigonometric function $ \sin(3x) $. After using substitution to transform the variable, we dealt with $ \sin(u) $ in the integral $ \frac{1}{3} \int \sin(u) \, du $. Here's what we typically do with integrals involving sine:

The basic integral $ \int \sin(u) \, du $ is a standard formula equal to $ -\cos(u) + C $, where $ C $ is the constant of integration.
Trigonometric integrals often rely on standard forms and identities to simplify and solve the integral.
Understanding these standard integrals and how to manipulate them is crucial for efficiently solving trig-related calculus problems.

In this problem, recognizing the standard form of $ \int \sin(u) \, du $ allowed quick integration to $ -\cos(u) $, simplifying the overall solution process.

Differential Calculus

Differential calculus is a branch of calculus focused on the concept of the derivative, which represents how a function changes. In this activity, we see differential calculus' role in substitution.

The process started by determining the derivative of $ u = 3x $. Differentiation resulted in the relationship $ du = 3 \, dx $. This was central in transforming the original integral into a different one concerning $ u $, notably:

We use differentiation to convert the problem from the original function $ \sin(3x) $ to a simpler function $ \sin(u) $.
This entails recognizing that the rate at which $ u $ changes in terms of $ x $ is described by the derivative, and that's crucial for changing variables properly.
Differential calculus provides the tools necessary to relate changes in different variables leading to an easily integrable form.

Using differential calculus effectively helps bridge the gap between complex integrals and their simpler counterparts by leveraging derivatives to transform and simplify the integrands.

Problem 7

Other exercises in this chapter

Problem 6

Write the sums without sigma notation. Then evaluate them. $$ \sum_{k=1}^{4}(-1)^{k} \cos k \pi $$

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Problem 7

Use the Substitution Formula in Theorem 7 to evaluate the integrals. $$ \text { a. } \int_{-1}^{1} \frac{5 r}{\left(4+r^{2}\right)^{2}} d r \quad \text { b. } \

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Problem 7

Evaluate the integrals $$ \int_{0}^{1}\left(x^{2}+\sqrt{x}\right) d x $$

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Problem 7

Express the limits in Exercises $1-8$ as definite integrals. \begin{equation} \lim _{\|P\| \rightarrow 0} \sum_{k=1}^{n}\left(\sec c_{k}\right) \Delta x_{k},

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