Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{-\pi / 2}^{\pi / 2} z \sin ^{2}\left(z^{3}\right) \cos \left(z^{3}\right) d z\)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Analyze the Integral

Given the integral is \( \int_{-\pi / 2}^{\pi / 2} z \sin^{2}(z^{3}) \cos(z^{3}) \, dz \), our goal is to simplify it, possibly using a substitution or observation. Notice the term \( \sin^{2}(z^{3}) \cos(z^{3}) \) suggests a substitution involving a trigonometric identity or a product. Let's proceed with a substitution for simplification.

Step 2: Choose a Substitution

Identify a substitution that simplifies the expression. Here, let \( u = z^3 \), so \( du = 3z^2 \, dz \). However, to match the form \( z \, dz = \frac{1}{3} du \). Rewriting the limits for \( u \), when \( z = -\pi/2 \), \( u = (-\pi/2)^3 \) and when \( z = \pi/2 \), \( u = (\pi/2)^3 \).

Step 3: Adjust the Limits and Integral Expression

With the substitution \( u = z^3 \), we have \( z \sin^{2}(z^{3}) \cos(z^{3}) \, dz = \frac{1}{3} \sin^{2}(u) \cos(u) \, du \). Update the limits to \([-\pi^3/8, \pi^3/8] \). The integral becomes: \[ \frac{1}{3} \int_{-\pi^3/8}^{\pi^3/8} \sin^{2}(u) \cos(u) \, du \].

Step 4: Simplify Using Symmetry

The function \( \sin^2(u) \cos(u) \) is an odd function since it changes sign with \( u \) (i.e., \( f(-u) = -f(u) \)). An integral of an odd function over a symmetric interval \([-a, a]\) yields zero. Therefore, the integral \( \int_{-\pi^3/8}^{\pi^3/8} \sin^{2}(u) \cos(u) \, du = 0 \).

Step 5: Evaluate the Integral

By the symmetry property of odd functions, the evaluation gives \( \frac{1}{3} imes 0 = 0 \). Therefore, \( \int_{-\pi / 2}^{\pi / 2} z \sin^{2}(z^{3}) \cos(z^{3}) \, dz = 0 \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a handy technique in calculus used to simplify integrals involving trigonometric functions. In this context, we aim to replace complex expressions with simpler or more familiar trigonometric forms. It often involves identifying a section of the integrand (the function being integrated) that resembles a trigonometric identity or allows such substitution.

In our problem, the expression inside the integral, which is \( z \sin^2(z^3) \cos(z^3) \), is complicated. By recognizing that the problem involves a power of \( z \) multiplied by trigonometric functions, we consider substituting \( u = z^3 \). This choice is strategic as it transforms the integrand into a simpler trigonometric form, \( \sin^2(u)\cos(u) \), which is easier to handle.

• The substitution aims to transform problems into forms where basic calculus rules, like power rules or trigonometric identities, can be easily applied.
• For instance, substituting \( z = \sqrt{u} \) in some problems may be useful if the integrand includes \( \sqrt{1 - z^2} \), which directly links to the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

These concepts are not only useful in simplifying but also often in finding the anti-derivative of a function when initial direct integration seems complex or impractical.

Symmetric Interval

A symmetric interval in integration refers to an interval \([-a, a]\) where the same distance is marked on both sides of zero. This is significant because certain functions have properties that simplify evaluation over symmetric intervals.

In our problem, the limits of integration \(-\pi/2\) and \(\pi/2\) present such a symmetric form concerning 0. However, after substitution, these are transformed into new symmetric limits \([-\pi^3/8, \pi^3/8]\).

• This characteristic is useful in assessing a function's behavior, particularly when checking for symmetry properties like being odd or even.
• Simplifying work, because we need to understand what happens when the function is offsetting changes symmetrically around the origin.

By evaluating the behavior of the function and its characteristics over these intervals, we can often simplify the integration process, taking advantage of symmetry properties like the odd or even nature of the function.

Odd Function Property

An odd function is a function \( f(x) \) that satisfies \( f(-x) = -f(x) \). These functions are symmetric with respect to the origin, meaning if you flip them around both the x-axis and y-axis (rotating 180 degrees), they look the same. This attribute has a special implication when integrating over symmetric intervals.

In our context, the function inside the integral, \( \sin^2(u)\cos(u) \), is identified as an odd function. This can be verified by observing that for any \( u \), if you substitute \( -u \), the function value becomes the negative of its original value, \( -\sin^2(-u)\cos(-u) = -\sin^2(u)\cos(u) \).

• Integrating an odd function across a symmetric interval \([-a, a]\) results in zero.
• This occurs because the area under the curve on one side of the origin cancels out the area on the opposite side.

Thus, without performing any complex calculations, we conclude that the integral of our odd function over the symmetric interval is zero. This property is invaluable for solving seemingly complex integrals quickly and efficiently.