Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int x^{2} \sinh x^{3} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{3} \cosh(x^3) + C \).

Step 1: Identify the Integration Technique

Notice the integral involves a polynomial and a hyperbolic sine function. This suggests the use of substitution since the polynomial's exponent is compatible with a variable substitution.

Step 2: Choose an Appropriate Substitution

Let's try the substitution\[ u = x^3 \]This implies\[ \frac{du}{dx} = 3x^2 \quad \text{or} \quad du = 3x^2 dx \]

Step 3: Solve for dx

From the substitution, rearrange to solve for the differential:\[ dx = \frac{du}{3x^2} \]

Step 4: Substitute Back into the Integral

Substitute into the integral, transforming all expressions in terms of \(u\):\[ \int x^2 \sinh(x^3) \, dx = \int x^2 \sinh(u) \cdot \frac{du}{3x^2} = \frac{1}{3} \int \sinh(u) \, du \]

Step 5: Integrate with Respect to u

Integrate \( \sinh(u) \) with respect to \(u\):\[ \int \sinh(u) \, du = \cosh(u) + C \]

Step 6: Substitute Back the Original Variable

Return to the original variable by replacing \(u\) with \(x^3\):\[ \frac{1}{3} \int \sinh(u) \, du = \frac{1}{3}(\cosh(u) + C) = \frac{1}{3}(\cosh(x^3) + C) \]

Step 7: Final Step: Conclude the Integral Solution

The solution to the integral is:\[ \frac{1}{3} \cosh(x^3) + C \]

Key Concepts

Substitution Method

The substitution method is a popular integration technique often used to simplify the process of finding integrals. It involves changing variables to make the integral easier to solve. This technique is particularly useful when you notice that the integral includes composite functions or expressions.

To use substitution:

• Identify a part of the integral that can be substituted with a new variable. In our example, the polynomial expression within the hyperbolic function was suitable, leading to the choice of substituting \( u = x^3 \).
• Differentiate your chosen expression to relate the differential of the new variable \( du \) to the differential in the integral \( dx \).
• Use this relation to rearrange and substitute all parts of the integral in terms of the new variable, including substituting \( dx \) with equivalent expressions in terms of \( du \).
• After integration, don't forget to substitute back using the original variable for the final solution.

This method can significantly simplify complex integrals, making them more manageable.

Hyperbolic Functions

Hyperbolic functions such as \( \sinh(x) \) and \( \cosh(x) \) are analogues to the trigonometric sine and cosine functions. However, they involve combinations of exponential functions. A key property of hyperbolic functions is that even though they are defined similarly to their trigonometric counterparts, they behave differently and have unique properties.

For instance:

• The hyperbolic sine function, \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• The hyperbolic cosine function, \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions appear often in calculus and are important in integration, as they have straightforward antiderivatives. Integrating \( \sinh(x) \) yields \( \cosh(x) + C \), offering a smooth transition from one function to another.

In our example, the direct integration of \( \sinh(u) \) using \( du \) resulted in a simple expression, making hyperbolic functions a useful tool in calculus.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial. An indefinite integral, as seen in the given exercise, represents a family of functions and includes a constant \( C \), symbolizing an infinite set of functions rather than a specific numeric value.

Key points about indefinite integrals:

• They do not have specified limits for integration and represent a general form of the antiderivative.
• The constant \( C \) reflects the fact that several functions could have the same derivative.

On the other hand, definite integrals involve upper and lower limits of integration and result in a specific value representing the area under the curve between these limits.

While our focus here was on solving an indefinite integral, it's important to recognize the broader context to correctly apply integration techniques as needed in various scenarios.