Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-62, use the Substitution Rule for Definite Integrals to evaluate each definite integral. \(\int_{-\pi / 2}^{\pi / 2}(\cos 3 x+\sin 5 x) d x\)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understanding the Problem

We have a definite integral \( \int_{-\pi / 2}^{\pi / 2}(\cos 3x + \sin 5x) \, dx \) that we need to evaluate using the Substitution Rule. In this problem, we'll deal with two different terms inside the integral: \( \cos 3x \) and \( \sin 5x \). Each term will involve its own substitution.

Step 2: Split the Integral

The integral consists of two parts: \( \int_{-\pi / 2}^{\pi / 2} \cos 3x \, dx \) and \( \int_{-\pi / 2}^{\pi / 2} \sin 5x \, dx \). We will evaluate each part separately and then sum the results.

Step 3: Substitution for the First Integral

For the integral \( \int_{-\pi / 2}^{\pi / 2} \cos 3x \, dx \), let \( u = 3x \). Then \( du = 3 \, dx \) or \( dx = \frac{1}{3} \ du \). The limits of integration change accordingly: when \( x = -\pi/2 \), \( u = -3\pi/2 \), and when \( x = \pi/2 \), \( u = 3\pi/2 \). The integral becomes: \[ \int_{-3\pi/2}^{3\pi/2} \cos u \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{-3\pi/2}^{3\pi/2} \cos u \, du. \]

Step 4: Evaluate First Integral

The integral \( \frac{1}{3} \int_{-3\pi/2}^{3\pi/2} \cos u \, du \) can be directly evaluated since the integral of \( \cos u \) is \( \sin u \). Therefore, \( \frac{1}{3} \left[ \sin u \right]_{-3\pi/2}^{3\pi/2} = \frac{1}{3} (\sin (3\pi/2) - \sin (-3\pi/2)) \, du = \frac{1}{3} (0 - 0) = 0.\)

Step 5: Substitution for the Second Integral

For the integral \( \int_{-\pi / 2}^{\pi / 2} \sin 5x \, dx \), let \( v = 5x \). Then \( dv = 5 \, dx \) or \( dx = \frac{1}{5} \ dv \). The limits of integration change accordingly: when \( x = -\pi/2 \), \( v = -5\pi/2 \), and when \( x = \pi/2 \), \( v = 5\pi/2 \). The integral becomes: \[ \int_{-5\pi/2}^{5\pi/2} \sin v \cdot \frac{1}{5} \, dv = \frac{1}{5} \int_{-5\pi/2}^{5\pi/2} \sin v \, dv. \]

Step 6: Evaluate Second Integral

The integral \( \frac{1}{5} \int_{-5\pi/2}^{5\pi/2} \sin v \, dv \) can be solved since the integral of \( \sin v \) is \( -\cos v \). Therefore, \( \frac{1}{5} [-\cos v]_{-5\pi/2}^{5\pi/2} = \frac{1}{5} (-\cos(5\pi/2) + \cos(-5\pi/2)) = \frac{1}{5} (0 + 0) = 0. \) Both terms are zero because \( \cos(5\pi/2) \) and \( \cos(-5\pi/2) \) are both equivalent to \( \cos(\pi/2) \), which equals zero.

Step 7: Combine the Results

Adding both results from the integrals: \( 0 + 0 = 0. \) Thus, the original definite integral \( \int_{-\pi / 2}^{\pi / 2} (\cos 3x + \sin 5x) \, dx = 0 \).

Key Concepts

Definite Integral Evaluation

In calculus, evaluating a definite integral means calculating the net area under the curve of a function between specified limits. For our problem, we have a definite integral between \(-\pi/2\) and \(\pi/2\): \(\int_{-\pi/2}^{\pi/2} (\cos 3x + \sin 5x) \, dx\). This expression combines the sine and cosine functions, two fundamental trigonometric functions.
To solve, you divide the integral into simpler parts:

• \(\int_{-\pi/2}^{\pi/2} \cos 3x \, dx\)
• \(\int_{-\pi/2}^{\pi/2} \sin 5x \, dx\)

These parts are evaluated separately and then added together for the final result.

The procedure of evaluating each part accurately involves applying substitution, a method that makes complex integrals manageable by simplifying the integrand.

Trigonometric Integrals

Trigonometric integrals involve evaluating integrals that include trigonometric functions like sine or cosine. They often require specialized techniques because these functions behave periodically.

In our problem, the trigonometric integrals are \(\int_{-\pi/2}^{\pi/2} \cos 3x \, dx\) and \(\int_{-\pi/2}^{\pi/2} \sin 5x \, dx\).

• The first integral involves \(\cos 3x\). Substitution helps because trigonometric functions often reappear in easy-to-calculate forms after substitution.
• The second integral uses \(\sin 5x\) and similar trigonometric identities can be applied. Trigonometric functions are special as they contribute consistent results due to their symmetrical properties across defined intervals.

Understanding these periodic and symmetrical properties helps predict outcomes of integrals without exhaustive computations, leading to results like zero in balanced intervals.

Calculus Techniques

Many calculus problems involve a toolbox of techniques, of which substitution is a key tool. The Substitution Rule helps to transform an integrand into a simpler form, easing the process of integration.

**Substitution Rule:**- Choose a substitution (e.g., \(u = 3x\) for \(\cos 3x\) or \(v = 5x\) for \(\sin 5x\)).- Change the differential accordingly (\(du = 3 \, dx\) or \(dv = 5 \, dx\)).- Adjust the integration limits to align with the new variable.By carefully executing substitutions, both integrals simplify drastically, sometimes even leading to straightforward evaluations.

Substitution not only reduces complexity but also harnesses the cyclical nature of trigonometric functions to arrive at quick, accurate results, as seen with our problem's zero sums. Mastering these techniques equips students to tackle a wide range of integral problems effectively.