Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indefinite integral $$ \int \sin ^{3}\left[\left(x^{2}+1\right)^{4}\right] \cos \left[\left(x^{2}+1\right)^{4}\right]\left(x^{2}+1\right)^{3} x d x $$

Step-by-Step Solution

Answer

The indefinite integral is \( \frac{1}{32} \sin^4((x^2+1)^4) + C \).

Step 1: Identify Substitution

The integral involves a complicated composition of functions, so we'll use a substitution to simplify it. Notice the expression \( u = (x^2 + 1)^4 \). This will simplify the argument of both \( \sin \) and \( \cos \).

Step 2: Differentiate Substitution

Differentiate \( u = (x^2 + 1)^4 \) with respect to \( x \). Use the chain rule: \( \frac{du}{dx} = 4(x^2 + 1)^3 \cdot 2x = 8x(x^2 + 1)^3 \). So, \( du = 8x(x^2 + 1)^3 \, dx \).

Step 3: Express Integral in Terms of u

Re-express the original integral in terms of \( u \) using the substitution from Step 1 and the derivative from Step 2. The integral becomes \( \int \sin^3(u) \cos(u) \cdot \frac{1}{8} \, du \).

Step 4: Simplify and Integrate

Factor out \( \frac{1}{8} \) from the integral. The integral is now \( \frac{1}{8} \int \sin^3(u) \cos(u) \, du \). Use the substitution \( v = \sin(u) \), leading to \( dv = \cos(u) \, du \). The integral becomes \( \frac{1}{8} \int v^3 \, dv \).

Step 5: Evaluate the Integral

Integrate with respect to \( v \): \( \frac{1}{8} \cdot \frac{v^4}{4} + C = \frac{1}{32} v^4 + C \). Replace \( v = \sin(u) \) to get \( \frac{1}{32} \sin^4((x^2 + 1)^4) + C \).

Step 6: Final Expression

The indefinite integral is \( \frac{1}{32} \sin^4((x^2 + 1)^4) + C \), where \( C \) is the constant of integration.

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus that allows us to find the antiderivative, or the integral, of a function. There are multiple techniques used to solve different types of integrals. Understanding when to use each technique is key to solving complex problems.

Some important integration techniques include:

• Substitution: This method is often used for integrals involving a composite function. It simplifies the integral by replacing a part of the original function with a new variable.
• Integration by parts: Useful for products of functions, this technique uses the product rule of differentiation in reverse.
• Partial fraction decomposition: Applied to rational functions, this technique breaks down a complex fraction into simpler parts.

Sometimes, more than one technique is needed to solve an integral, as demonstrated in the original exercise, where substitution is combined with recognizing patterns like powers and trigonometric identities.

Substitution Method

The substitution method is a powerful tool for evaluating integrals, especially when dealing with composite functions. The goal is to transform a difficult integral into an easier one.

To apply substitution:

• Choose a substitution, often represented by the variable "u," that simplifies the integral. For example, in the problem, we selected \( u = (x^2 + 1)^4 \) to simplify the expression inside the trigonometric functions.
• Differential substitution involves differentiating "u" with respect to "x." Using the chain rule, \( \frac{du}{dx} \) is calculated, providing a way to express \( dx \) in terms of \( du \).
• Substitute all occurrences of the original variable and its differential in the integral with "u" and "du," transforming the integral into a simpler form that is often easier to evaluate.

This method not only simplifies the integral but also makes the functions more recognizable and manageable.

Chain Rule

The chain rule is a pivotal concept in calculus that allows us to differentiate composite functions. It is also implicitly used in integration, especially when applying the substitution method.

The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative is given by multiplying the derivatives of the outer and inner functions: \( f'(g(x)) \cdot g'(x) \).

In the original problem, the chain rule was used when differentiating \( u = (x^2 + 1)^4 \). The derivative, \( \frac{du}{dx} = 8x(x^2 + 1)^3 \), is calculated by differentiating the outer function \( (x^2 + 1)^4 \) and multiplying by the derivative of the inner function \( x^2 + 1 \). This application enabled us to successfully change the variables, thereby setting the stage for easier integration.

Trigonometric Integrals

Trigonometric integrals involve integrals of trigonometric functions which can often be complex due to their periodic nature. When powers of trigonometric functions are involved, strategic approaches are necessary.

In the given problem, the integral involves \( \sin^3(u) \cos(u) \). After substitution, manipulating the trigonometric expression becomes feasible by using another substitution, \( v = \sin(u) \).

This substitution transforms the integral into a polynomial form \( \int v^3 \, dv \), which is much simpler to integrate than the original trigonometric expression. Here, recognizing the structure of the integral allows for the substitution and simplification that lead directly to a standard polynomial integral.

Mastering these approaches with trigonometric functions can significantly ease the process of finding their integrals.