Problem 40

Question

a, b, and $c$ are constants and $g(x)$ is a continuous function whose derivative $g^{\prime}(x)$ is also continuous. Use substitution to evaluate the indefinite integrals. $$ \int g^{\prime}(x) \sin [g(x)] d x $$

Step-by-Step Solution

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Answer

The integral is $ -\cos[g(x)] + C $.

1Step 1: Choose the Substitution

Recognize that the inner function in the integral is $ g(x) $. We use substitution to simplify the problem. Set $ u = g(x) $. This choice exploits the fact that the derivative $ g'(x) $ is present in the integrand, suitable for substitution.

2Step 2: Determine the Differential

Differentiate $ u = g(x) $ with respect to $ x $ to find the differential of $ u $: $ du = g'(x) \, dx $. This differential matches part of the integrand.

3Step 3: Rewrite the Integral in Terms of $ u $

Substitute $ u = g(x) $ and $ du = g'(x) \, dx $ into the integral. This transforms the integral to $ \int \sin(u) \, du $.

4Step 4: Integrate with Respect to $ u $

Compute the integral $ \int \sin(u) \, du $. The antiderivative of $ \sin(u) $ is $ -\cos(u) + C $, where $ C $ is the constant of integration.

5Step 5: Substitute Back in Terms of $ x $

Replace $ u $ with $ g(x) $ to express the antiderivative in terms of the original variable: $ -\cos(g(x)) + C $.

Key Concepts

Understanding the Substitution MethodGrasping the AntiderivativeThe Chain Rule Connection

Understanding the Substitution Method

The substitution method is like a strategy to simplify integrals. It involves replacing the complex part of an equation with a single variable, often labeled as "$ u $." This allows us to turn a complicated integral into something much simpler.

First, identify a part of the function inside the integral that can be called $ u $. In this case, it’s $ g(x) $.
Next, differentiate $ u $ with respect to $ x $ to find $ du = g'(x) \, dx $. Note how the $ g'(x) \, dx $ matches a part of the given integral.
We replace all instances of the identified function and its differential with $ u $ and $ du $, respectively, to get a new and simpler integral.

This method transforms a potentially difficult problem into one you already know how to solve. Once the simpler integral is computed, you substitute back the original expression for $ u $. This final step is important as it ensures the solution is in terms of the original variable.

Grasping the Antiderivative

An antiderivative is essentially the reverse process of taking a derivative. While derivatives explain how functions change, antiderivatives help us find functions whose rate of change fits a given pattern. Integrating a function yields its antiderivative.In the exercise, after substituting and simplifying the integral to $ \int \sin(u) \, du $, we seek to find the antiderivative of $ \sin(u) $. This just means determining a function that, when differentiated, results in $ \sin(u) $.

The well-known antiderivative for $ \sin(u) $ is $ -\cos(u) + C $, where $ C $ is the constant of integration.
This constant $ C $ reflects the fact that shifting a function up or down does not affect its rate of change, thus any number added to the antiderivative still fulfills the condition upon differentiation.
In practical terms, integrating and finding the antiderivative helps us reverse a worked-out change to find the original paths or outcomes.

The Chain Rule Connection

The chain rule is a critical concept in calculus, particularly when dealing with compositions of functions like $ g(x) $ inside another function. While it’s most commonly used to differentiate such compositions, it’s closely linked to the substitution method in integration.The chain rule states that when you have a composition of two functions, the derivative can be found by multiplying the derivative of the outer function by the derivative of the inner function. Think about it as peeling an onion where each layer requires attention.For integrals, we apply this concept backwards:

In integration, the presence of $ g'(x) $ along with another function of $ g(x) $ suggests that substitution might simplify it.
In the exercise example, $ g(x) $ being the inside function means the substitution $ u = g(x) $ mirrors what the chain rule would do when the roles are reversed.
This backwards application allows the integral to be simplified such that the inner and outer functions align as much as possible.

Understanding the chain rule and its application in reverse through substitution not only simplifies problems but also deepens comprehension of function relationships in calculus.

Problem 40

Other exercises in this chapter

Problem 39

$$ \text { In Problems } , \text { evaluate each definite integral. } $$ $$ \int_{0}^{1} \frac{x}{x^{2}+1} d x $$

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Problem 40

First make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals. $$ \int \sin \sqrt{x} d x $$

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Problem 40

Find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{1}^{\infty} \frac{1}{\sqrt{1+x^{6}}} d x $$

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Problem 40

$$ \text { In Problems } , \text { evaluate each definite integral. } $$ $$ \int_{1}^{2} \frac{x^{2}+1}{x} d x $$

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