Problem 57 - Calculus

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) The function is odd. (b) The period is \( \pi \). (c) Integrals are 0, due to odd symmetry and periodicity.

Step 1: Understanding Function Symmetry

To determine if \( f(x) = |\sin x| \sin (\cos x) \) is even, odd, or neither, check both evenness and oddness conditions. An even function satisfies \( f(-x) = f(x) \), and an odd function satisfies \( f(-x) = -f(x) \). Calculate \( f(-x) = |\sin(-x)| \sin(\cos(-x)) = |\sin x| (-\sin(\cos x)) = -|\sin x| \sin(\cos x) = -f(x) \). Since \( f(-x) = -f(x) \), the function is odd.

Step 2: Identify the Periodicity

The function \( f(x) = |\sin x| \sin(\cos x) \) has two periodic functions: \( |\sin x| \), which has a period \( \pi \), and \( \sin(\cos x) \), which does not have any standard periodicity in conventional trigonometric sense. Since the period of \( |\sin x| \) (the bounding behavior) restricts the total periodicity of \( f \), \( f \) is periodic with period \( \pi \).

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

Integrate \( f(x) \) over \([0, \pi / 2]\): \( \int_{0}^{\pi/2} |\sin x| \sin(\cos x) \, dx = \int_{0}^{\pi/2} \sin x \sin(\cos x) \, dx \). Simplify using substitution \( u = \cos(x) \), \( du = -\sin(x) \, dx \). This gives \( \int_{1}^{0} \sin(u) (-du) = \int_{0}^{1} \sin(u) \, du = -[\cos(u)]_{0}^{1} = -[\cos(1) - \cos(0)] = 1 - \cos(1). \)

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

Due to the odd nature of \( f \), the integral \( \int_{-\pi/2}^{\pi/2} f(x) \, dx = 0 \). For odd functions, integrals over symmetric intervals around the origin cancel to zero.

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

Split the integral as \([0, \pi]\) and \([\pi, 3\pi/2]\) for ease: \[ \int_{0}^{3\pi/2} f(x) \, dx = \int_{0}^{\pi/2} f(x) \, dx + \int_{\pi/2}^{\pi} f(x) \, dx + \int_{\pi}^{3\pi/2} f(x) \, dx \]. Each \( \int_{0}^{\pi/2} f(x) \, dx \) and \( \int_{\pi}^{3\pi/2} f(x) \, dx \) yield \( 1 - \cos(1) \) from previous calculations. Overall result: \( 2(1 - \cos(1)) \).

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

Leverage odd symmetry: \( \int_{-3\pi/2}^{3\pi/2} f(x) \ dx = 0 \), since odd functions symmetrically integrate to zero over such spans.

Step 7: Definite Integral over [0, 2π]

Owing to the periodicity of \( f \), \( \int_{0}^{2\pi} f(x) \, dx = 0 \). This applies due to full periodic cancelation of contributions over two \( \pi \)-length periods.

Step 8: Definite Integral over Custom Intervals

\([\pi/6, 13\pi/6]\), \([\pi/6, 4\pi/3]\), and \([13\pi/6, 10\pi/3]\) span full periods of \( f \). For any interval fully spanning \( n\pi \) (where \( n \) is an integer), these integrate to zero for the same periodic reason encompassing a complete set of periods.

Key Concepts

Periodic Functions

Periodic functions are fundamental in calculus and mathematics. These functions repeatedly cycle through values at regular intervals. This property stems from trigonometric functions like sine, cosine, and tangent, each exhibiting periodic behavior.

A function is periodic if there is a non-zero constant \( T \), called the period, such that for every value of \( x \) in the function's domain, \( f(x + T) = f(x) \).

• The function \(|\sin x|\) has a period of \(\pi\). This means every \(\pi\) units in the x-direction, the pattern of \(|\sin x|\) repeats itself.
• Compared to a standard sine function, the absolute value function doesn't change the periodicity but affects the waveform's shape.

In our problem, the given function \( f(x) = |\sin x| \sin(\cos x) \), was determined to primarily inherit its periodicity from \(|\sin x|\). Therefore, it has a period of \( \pi \). Understanding the periods helps simplify calculations involving integrals over such functions, as it reduces the number of unique value intervals we must assess.

Definite Integrals

In calculus, definite integrals calculate the net area under a curve within specific bounds, efficiently providing exact values. They're represented by the expression \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the lower and upper bounds, respectively.

For periodic functions, definite integrals over complete period intervals can sometimes yield zero. This happens when the positive and negative areas cancel each other out.

• In the problem, the integral of \( f(x) = |\sin x| \sin(\cos x) \) over symmetric intervals centered around zero ended up being zero due to symmetry.
• For more complex functions, evaluating definite integrals might require substitution. For instance, substituting \( u = \cos x \) simplifies the integral \( \int_{0}^{\pi/2} |\sin x| \sin(\cos x) \, dx \) since it transforms the problem into one involving simpler trigonometric functions.

Effectively understanding and calculating definite integrals enables precise determinations of values linked to areas or rates of change, especially pivotal in physics and engineering.

Function Symmetry

Symmetry in functions generally simplifies analysis by offering predictable behavior across segments of their domain. Two primary symmetries are especially important: even and odd.

• An even function satisfies \( f(-x) = f(x) \). Graphically, they mirror at the y-axis. Classic examples include \( x^2 \) or \( |x| \).
• An odd function satisfies \( f(-x) = -f(x) \). These functions exhibit rotational symmetry about the origin. Example: \( x^3 \) or \( \sin x \).

For the given problem, the function \( f(x) = |\sin x| \sin(\cos x) \) was shown to be odd. This was verified by checking the condition: \( f(-x) = -f(x) \).
Graphically, this tells us if one side of the graph reflects rotational steps around the origin to the other, the segments cancel out over symmetric limits in integrals — a feature utilized in calculating definite integrals over such span.