Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

, use the Substitution Rule for Definite Integrals to evaluate each definite integral.$$ \int_{0}^{\pi / 4}(\cos 2 x+\sin 2 x) d x $$

Step-by-Step Solution

Answer

The value of the integral is 1.

Step 1: Identify the Integral Limits and Function

The integral we need to evaluate is \( \int_{0}^{\pi / 4}(\cos 2x + \sin 2x) \, dx \). The limits of integration are 0 and \( \pi/4 \). The integrand function is \( \cos 2x + \sin 2x \).

Step 2: Choose a Substitution

To simplify the integral, we choose the substitution \( u = 2x \). Consequently, the differential \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \).

Step 3: Change the Integral Limits

When \( x = 0 \), \( u = 2 \times 0 = 0 \). When \( x = \pi/4 \), \( u = 2 \times \pi/4 = \pi/2 \). So the new limits of integration are from 0 to \( \pi/2 \).

Step 4: Substitute and Simplify the Integral

Using the substitution, the integral becomes \( \int_{0}^{\pi/2} (\cos u + \sin u) \cdot \frac{1}{2} \, du \). Simplify it to \( \frac{1}{2} \int_{0}^{\pi/2} (\cos u + \sin u) \, du \).

Step 5: Integrate the Function

The integral \( \int \cos u \, du = \sin u \) and \( \int \sin u \, du = -\cos u \). Therefore, \( \int (\cos u + \sin u) \, du = \sin u - \cos u \).

Step 6: Evaluate the Definite Integral

Evaluate \( \left[ \frac{1}{2} (\sin u - \cos u) \right]_{0}^{\pi/2} \). Substitute the limits to get \( \frac{1}{2} [ (\sin \frac{\pi}{2} - \cos \frac{\pi}{2}) - (\sin 0 - \cos 0) ] \).

Step 7: Calculate the Result

\( \sin \frac{\pi}{2} = 1 \), \( \cos \frac{\pi}{2} = 0 \), \( \sin 0 = 0 \), \( \cos 0 = 1 \). Thus, \( \frac{1}{2} [ (1 - 0) - (0 - 1) ] = \frac{1}{2} [ 1 + 1 ] = \frac{1}{2} \times 2 = 1 \).

Key Concepts

Substitution Rule

The substitution rule is a powerful technique in calculus used to simplify integrals. By changing the variable of integration, often denoted as \( x \), to a new variable, \( u \), we can make complex integrals more manageable. The goal of substitution is to transform the integral into a form that is easier to evaluate. To use substitution:

• Choose an expression \( u \) such that its derivative \( du \) is present in the integral. This helps simplify the differential part of the integral.
• Express \( dx \) in terms of \( du \) by solving \( du = f'(x) dx \) for \( dx \).
• Substitute \( x \) and \( dx \) in the integral with \( u \) and \( du \) respectively, changing the limits accordingly if it's a definite integral.

In this exercise, the substitution \( u = 2x \) was used, simplifying the integrand into a function of \( u \). This transformed the original integral from a function of \( x \) to a function of \( u \), permitting a straightforward integration process. This cleverly avoids more complex trigonometric integration by simplifying both the limits and the integrand structure.

Limits of Integration

In a definite integral, the limits of integration are the bounds within which we evaluate the integral. They are crucial because they define the interval over which the function is integrated, affecting the result significantly.When using substitution in a definite integral, it's essential to modify these limits according to the new variable. Simply substituting without adjusting the limits will lead to incorrect solutions. For example:- In the original integral \( \int_{0}^{\pi / 4}\), we have limits of 0 and \( \pi/4 \) related to \( x \).- After substitution where \( u = 2x \), the limits transform to 0 and \( \pi/2 \), calculated as follows: - When \( x = 0 \), \( u = 2 \times 0 = 0 \). - When \( x = \pi/4 \), \( u = 2 \times \pi/4 = \pi/2 \).These new limits \( 0 \to \pi/2 \) are utilized in the rest of the problem, ensuring that our final answer remains consistent with the transformed integrand.

Trigonometric Integrals

Trigonometric integrals involve functions like \( \sin(x) \) and \( \cos(x) \). They are commonly used in calculus, especially when dealing with periodic or circular motion problems. The crucial part is to recognize these functions can be combined or separated during integration, making them easier to handle.In our solved problem:- After substitution, the integrand \( \cos(2x) + \sin(2x) \) transforms into \( \cos(u) + \sin(u) \).- This separates the integration into: \[ \int \cos(u) \, du + \int \sin(u) \, du \]Basic integrals for trigonometric functions:

• \( \int \cos(u) \, du = \sin(u) + C \)
• \( \int \sin(u) \, du = -\cos(u) + C \)

In the final integration, the simplicity of trigonometric integrals aids in quickly finding the antiderivative. This simplicity results in the solution \( \sin u - \cos u \) during the integration step.

Calculus Problem Solving

Solving calculus problems, like finding definite integrals, involves systematic approaches such as substitution, changing integration limits, and understanding integral properties. Key steps in calculus problem-solving include:

• Understanding the problem and identifying integrals that simplify well with known techniques like substitution.
• Rewriting the integral in a simpler form, which often involves algebraic manipulation or substitutions.
• Adjusting integration limits where necessary, especially when changing variables.
• Performing integration using known antiderivatives or simplification techniques.
• Evaluating the definite integral by substituting the calculated antiderivative at the upper and lower bounds.

In this example, we commenced by recognizing the form suitable for substitution. Adjustments were made for the new variable's limits, allowing a complex integral to become feasible to solve. The precise transformation processes, combined with evaluation of basic trigonometric results, lead to the final solution. Understanding each component solidifies comprehension when tackling future similar calculus challenges.