Problem 26 - Calculus

Question

Use the method of substitution to find each of the following indefinite integrals. $$ \int x^{2} \cos \left(x^{3}+5\right) d x $$

Step-by-Step Solution

Answer

\( \frac{1}{3} \sin(x^3 + 5) + C \).

Step 1: Identify the Inner Function

First, we need to identify the inner function in the composite expression inside the integral, which is suitable for substitution. Here, the inner function is the expression inside the cosine function: \( u = x^3 + 5 \).

Step 2: Compute the Derivative of the Inner Function

Next, we differentiate the inner function with respect to \( x \): \( \frac{du}{dx} = 3x^2 \). This will help us determine what to substitute for \( dx \).

Step 3: Express dx in Terms of du

From the derivative, we can solve for \( dx \): \( du = 3x^2 \, dx \). Therefore, \( dx = \frac{du}{3x^2} \).

Step 4: Substitute in the Integral

Substitute \( u = x^3 + 5 \) and \( dx = \frac{du}{3x^2} \) into the integral: \[ \int x^{2} \cos(u) \cdot \frac{du}{3x^2} \]. The \( x^2 \) terms cancel out, simplifying it to \( \frac{1}{3} \int \cos(u) \, du \).

Step 5: Integrate with Respect to u

Compute the integral with respect to \( u \): \( \int \cos(u) \, du = \sin(u) + C \). Therefore, \( \frac{1}{3} \int \cos(u) \, du = \frac{1}{3} \sin(u) + C \).

Step 6: Substitute Back to Original Variable

Finally, substitute back \( u = x^3 + 5 \) to return to the original variable: \( \frac{1}{3} \sin(x^3 + 5) + C \). This is the indefinite integral of the given function.

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus that helps simplify the process of finding integrals, especially when dealing with complex expressions. It is essentially the reverse of the chain rule used in differentiation. The idea involves replacing a part of the integral with a new variable to make the equation simpler and easier to integrate.

Here's how it works:

• Identify the Inner Function: The first step is to recognize the "inner function," often found inside another function or operation, like inside a cosine or a square root. In our example, the inner function is the expression inside the cosine function, i.e., \( u = x^3 + 5 \).
• Take the Derivative: Differentiate the inner function with respect to \( x \) to find \( \frac{du}{dx} \). This step is necessary for the substitution process as it helps in expressing \( dx \) in terms of \( du \).
• Replace the Variables: Substitute all instances of the inner function in the integrand with a new variable \( u \), and convert \( dx \) to the new differential \( du \), aligning everything to the new variable.

These steps transform a difficult integral into a simpler form that is often straightforward to solve. After integration, don't forget to substitute back the original variable to express the final answer in terms of the initial variable.

Trigonometric Integration

Trigonometric integration involves applying various trigonometric identities and formulas to find the integral of a trigonometric function. It's particularly useful for dealing with functions such as sine, cosine, and tangent that arise naturally in many calculus problems.

In our exercise, the function \( \cos(x^3 + 5) \) is integrated using the substitution method, reducing the problem to an integral involving \( \cos(u) \). The integral of the cosine function is a standard calculation:

• Integral of Cosine: The integral of \( \cos(u) \) is \( \sin(u) + C \), where \( C \) is the constant of integration.
• Apply Substitution: Substitute \( u \) with the inner function \( x^3 + 5 \) once the integration with respect to \( u \) is completed.

Understanding how to handle trigonometric functions within integrals is crucial, as they appear frequently in both theoretical and application-based problems in calculus.

Composite Functions

Composite functions combine two or more functions into one by applying one function to the results of another. In integration, they often show up as complex expressions, where substitution can simplify the process. The idea here is to focus on one part of the composite function as a separate entity.

Consider our original function \( \int x^2 \cos(x^3 + 5) \, dx \):

• Recognize the Composition: The composition happens because \( \cos(x^3 + 5) \) is a cosine trigonometric function applied to the polynomial \( x^3 + 5 \).
• Focus on the Inner Function: By identifying the polynomial \( x^3 + 5 \) as our point of substitution, we effectively separate and simplify the complex expression.

Composite functions are a common source of complexity in calculus, but by recognizing and managing this structure effectively, they can be greatly simplified, making integration feasible.