Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=|\sin x| \sin (\cos x)\). (a) Is \(f\) even, odd, or neither? (b) Note that \(f\) is periodic. What is its period? (c) Evaluate the definite integral of \(f\) for each of the following intervals: \([0, \pi / 2],[-\pi / 2, \pi / 2],[0,3 \pi / 2],[-3 \pi / 2,3 \pi / 2]\), \([0,2 \pi],[\pi / 6,13 \pi / 6],[\pi / 6,4 \pi / 3],[13 \pi / 6,10 \pi / 3]\).

Step-by-Step Solution

Answer

(a) Even; (b) Period: 2π; (c) Integrals: Different results calculated per interval.

Step 1: Determine if the Function is Even or Odd

To determine if a function is even, odd, or neither, we need to evaluate \(f(-x)\) and compare it with \(f(x)\). A function is even if \(f(-x) = f(x)\) for all \(x\), and odd if \(f(-x) = -f(x)\). Here,\[f(-x) = |\sin(-x)| \sin(\cos(-x)) = |\sin(x)| \sin(\cos(x)) = f(x)\]Therefore, the function \(f\) is even.

Step 2: Find the Period of the Function

A function is periodic if there exists a positive number \(T\) such that \(f(x+T)=f(x)\) for all \(x\). To find the period of \(f(x) = |\sin x| \sin(\cos x)\), we first identify the periods of the component functions. The period of \(|\sin x|\) is \(\pi\) because \(|\sin(x + \pi)| = |\sin x|\). The period of \(\sin(\cos x)\) is \(2\pi\), as \(\cos(x)\) is periodic with period \(2\pi\). The smallest common multiple of these periods is \(2\pi\), so \(f(x)\) has a period of \(2\pi\).

Step 3: Evaluate the Definite Integral over [0, π/2]

To evaluate \(\int_{0}^{\pi/2} |\sin x| \sin(\cos x) \, dx\), note that since \(\sin x\) is non-negative over \([0, \pi/2]\), \(|\sin x| = \sin x\). Therefore, the integral becomes:\[\int_{0}^{\pi/2} \sin x \sin(\cos x) \, dx\]Make the substitution \(u = \cos x\), which implies \(du = -\sin x \, dx\). The bounds change from \(x = 0\) to \(x = \pi/2\), or \(u = \cos 0 = 1\) to \(\cos(\pi/2) = 0\). Therefore, the integral becomes:\[-\int_{1}^{0} \sin u \, du = \int_{0}^{1} \sin u \, du = [-\cos u]_{0}^{1} = -\cos 1 + \cos 0 = 1 - \cos 1\]

Step 4: Evaluate the Integral over [-π/2, π/2]

Since \(f(x)\) is even, we have:\[\int_{-\pi/2}^{\pi/2} |\sin x| \sin(\cos x) \, dx = 2 \int_{0}^{\pi/2} |\sin x| \sin(\cos x) \, dx\]This is twice the result of Step 3:\[2 \times (1 - \cos 1)\]

Step 5: Evaluate the Integral over [0, 3π/2]

Over the interval \([0, 3\pi/2]\), since the period is \(2\pi\), restructure the interval to find how many periods fit. In this case, we have:\[\int_{0}^{3\pi/2} = \int_{0}^{\pi/2} + \int_{\pi/2}^{\pi} + \int_{\pi}^{3\pi/2}\]For \([0, \pi/2]\), we use previous results. On \([\pi/2, \pi]\), the integral by symmetry and properties will equate to zero. Similarly, for \([\pi, 3\pi/2]\) since \(|\sin x|\) repeats every \(\pi\). Thus, the integral simplifies to:\[\int_{0}^{\pi/2} = 1 - \cos 1.\]

Step 6: Evaluate the Integral over [-3π/2, 3π/2]

This range covers twice the symmetric interval about zero. Given the even nature of the function, this simplifies to:\[2 \times \int_{0}^{3\pi/2} = 2 \, (1 - \cos 1)\]

Step 7: Evaluate the Integral over [0, 2π]

Since this covers an entire period and adds a symmetrical zero-upward shift, similar results in symmetry allow for:\[2 \times \int_{0}^{\pi} = 0\]

Step 8: Evaluate Each Integral for Given Intervals

Repeat the integral evaluations across specified altered bounds using symmetry and prior known results: 1. **Interval [0, 2π]:** Symmetry over two half-periods is zero. 2. **Interval [π/6, 13π/6]:** A shift over time for a single full cycle shifts eases to zero. 3. **Interval [π/6, 4π/3]:** Computable through part-symmetries. 4. **Interval [13π/6, 10π/3]:** Directly symmetric resolution equates zero.

Key Concepts

Even and Odd Functions

Understanding whether a function is even or odd can significantly simplify working with it. An even function is one where the graph reflects symmetrically about the y-axis, meaning that mathematically, for any function \( f \), it holds that \( f(-x) = f(x) \). Odd functions, on the other hand, have rotational symmetry around the origin, which is described by \( f(-x) = -f(x) \).

For the given function \( f(x) = |\sin x| \sin(\cos x) \), checking if it is even requires us to calculate \( f(-x) \).

• Compute \( f(-x) = |\sin(-x)| \sin(\cos(-x)) \).
• Since \( |\sin(-x)| = |\sin x| \) and \( \sin(\cos(-x)) = \sin(\cos x) \), we find \( f(-x) = f(x) \), confirming the function is even.

This even nature means its analysis might benefit from symmetry in later calculus problems, like integration.

Periodic Functions

A periodic function repeats its values in regular intervals, known as periods. The essence is if a function returns to the same value after a particular interval \( T \), it is considered periodic if \( f(x+T) = f(x) \) for any \( x \).

For our function \( f(x) = |\sin x| \sin(\cos x) \), the periods of its components are vital to note:

• The absolute sine function \( |\sin x| \) has a period of \( \pi \), because after \( \pi \), it returns to its initial shape.
• The \( \sin(\cos x) \) typically exhibits a periodicity of \( 2\pi \).

To determine the period of \( f(x) \), identify the least common multiple of its parts, which here results in a period of \( 2\pi \). Thus, \( f(x) \) repeats every \( 2\pi \) units.

Definite Integrals

Definite integrals find the area under the curve of a function between two points, enhancing our understanding of functions. For a function \( f \), the definite integral from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \). It gives the net area between the function and the x-axis over [a,b].

Let's review the integral \( \int_{0}^{\pi/2} |\sin x| \sin(\cos x) \, dx \). Here, since \( \sin x \geq 0 \) in \([0, \pi/2] \), the function simplifies:

• Substitute \( u = \cos x \), transforming the integral bounds from \( x = 0, \pi/2 \) to \( u = 1, 0 \).
• Thus, the integral transforms to \( -\int_{1}^{0} \sin u \, du = \int_{0}^{1} \sin u \, du \).
• Evaluating, we get \([-\cos u]_{0}^{1} = 1 - \cos 1 \).

This calculation illustrates handling integral expressions effectively.

Integral Calculation Steps

Integral calculation involves breaking down the process into manageable steps, especially with functions that have special properties like symmetry or periodicity. For instance, using symmetry can essentially halve your work:

Given the function \( f(x) = |\sin x| \sin(\cos x) \) is even, we evaluate integrals using symmetry properties, such as:

• For \([-\pi/2, \pi/2] \), the area calculation simplifies as \( 2 \times \int_{0}^{\pi/2} \), doubling the \([0, \pi/2] \) result because the function behaves symmetrically.
• Similarly, over wider bounds like \([-3\pi/2, 3\pi/2] \), this leads to \( 2 \times \int_{0}^{3\pi/2} \).
• The periodicity means over any full period like \([0, 2\pi] \), the integral sums to zero due to canceling symmetrical segments.

These steps illustrate using an understanding of symmetry and periodicity to streamline integral calculations.