Problem 60 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-62, use the Substitution Rule for Definite Integrals to evaluate each definite integral. \(\int_{-1}^{1} x^{2} \cosh x^{3} d x\)

Step-by-Step Solution

Answer

\(\frac{2}{3} \sinh(1)\)

Step 1: Identify the substitution

Notice that the expression inside the hyperbolic cosine function is \(x^3\). This suggests a substitution where \( u = x^3 \).

Step 2: Calculate the derivative of the substitution

We differentiate \( u = x^3 \) to find \( du = 3x^2 \, dx \). This means \( dx = \frac{1}{3x^2} du \).

Step 3: Change the integration bounds

Since \( u = x^3 \), when \( x = -1 \), \( u = (-1)^3 = -1 \). And when \( x = 1 \), \( u = 1^3 = 1 \). So the integral from \( x = -1 \) to \( x = 1 \) becomes from \( u = -1 \) to \( u = 1 \).

Step 4: Substitute and simplify the integral

Substituting \( u = x^3 \) and \( dx = \frac{1}{3x^2} du \) in the integral, we get:\[\int_{-1}^{1} x^2 \cosh(x^3)\, dx = \int_{-1}^{1} x^2 \cosh(u) \frac{1}{3x^2} \, du = \frac{1}{3}\int_{-1}^{1} \cosh(u) \, du.\]Notice the \( x^2 \) terms cancel out.

Step 5: Evaluate the simplified integral

The integral \( \int \cosh(u) \, du \) is \( \sinh(u) \). Therefore:\[\frac{1}{3} \left. \sinh(u) \right|_{-1}^{1} = \frac{1}{3} \left( \sinh(1) - \sinh(-1) \right).\]

Step 6: Simplify the result using properties of hyperbolic functions

Recall that \( \sinh(-u) = -\sinh(u) \). Thus:\[\sinh(1) - \sinh(-1) = \sinh(1) + \sinh(1) = 2 \sinh(1).\]So the result is:\[\frac{1}{3} \times 2 \sinh(1) = \frac{2}{3} \sinh(1).\]

Key Concepts

Definite Integrals

Definite integrals are essential in calculus, as they help us compute the area under a curve on a specified interval. In this problem, we are evaluating the definite integral:

• hinspace ext{from} hinspace -1 hinspace ext{to} hinspace 1 of } hinspace $ hinspace$ x^2 cosh(x^3) hinspace dx.

To evaluate this, the Substitution Rule for Definite Integrals comes in handy. This rule allows us to simplify the integration process by changing variables, which eventually transforms a complicated integral into a more manageable form.
In essence, during integration, the limits of integration also need to be transformed to match the new variable. For instance, substituting $u=x^3$ means mapping $x=-1$ and $x=1$ to $u=-1$ and $u=1$. This concept allows us to focus on simpler integrals and find solutions without cumbersome computations over bounded regions.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are related to hyperbolas, not circles. In this exercise, the hyperbolic cosine function, or \(\cosh(x)\), comes into play.
These functions are significant in calculus, given their appearance in many integrals and differential equations. For example, the \(\cosh(u)\) function is defined as:

• \cosh(u) = \frac{e^u + e^{-u}}{2}

Hyperbolic functions are reminiscent of their trigonometric cousins but have unique properties too. One important property used in the solution is that the derivative of \(\cosh(u)\) is \(\sinh(u)\).
Furthermore, \(\sinh(-u) = -\sinh(u)\), which simplifies calculations. Recognizing these properties helps streamline the step-by-step process of evaluating integrals involving hyperbolic functions. Understanding these fundamentals makes handling substitutions that involve hyperbolic terms much easier.

Integral Calculus

Integral calculus focuses on the process of integration, which essentially involves finding functions that can describe accumulation—such as areas under curves.
In this exercise, integral calculus is used to evaluate a definite integral by applying substitution, one of the prevalent techniques for solving complex integrals.

• When using substitution, calculus students typically identify a function's part that can yield, through derivation, existing parts of the integrand, which then simplifies the function.
• Once substitution is applied, changing the variable of integration and adjusting the integration bounds are also necessary steps.

Following these steps, as shown, transforms \(int_{-1}^{1} x^2 \cosh(x^3) dx\) into a much simpler form \(frac{1}{3} int_{-1}^{1} \cosh(u) du\).
Integration becomes more manageable, as integral calculus effectively provides concepts and tools like the `Substitution Rule for Definite Integrals`, allowing for the transformation and simplification of complicated integrals. This method of integration not only simplifies calculations but enhances comprehension of broader mathematical principles. Understanding and effectively applying these techniques is the core of calculus applications.