Problem 449 - Calculus Volume 1

Question

Evaluate the following integrals. $$\int_{\pi / 3}^{\pi / 2} 2 \sec (2 \theta) \tan (2 \theta) d \theta$$

Step-by-Step Solution

Answer

The evaluated integral is 1.

Step 1: Identify the integral to be solved

The integral to solve is \( \int_{\pi / 3}^{\pi / 2} 2 \sec (2 \theta) \tan (2 \theta) \, d\theta \). This integral involves the trigonometric functions \( \sec(2\theta) \) and \( \tan(2\theta) \).

Step 2: Use substitution

Use the substitution method to simplify the integral. Let \( u = 2\theta \), which means \( du = 2 \cdot d\theta \) or \( d\theta = \frac{1}{2} du \). Adjust the limits of integration: when \( \theta = \frac{\pi}{3} \), \( u = \frac{2\pi}{3} \); when \( \theta = \frac{\pi}{2} \), \( u = \pi \).

Step 3: Rewrite the integral in terms of u

Substitute \( u = 2\theta \) and \( d\theta = \frac{1}{2} du \) into the integral:\[\int_{\frac{2\pi}{3}}^{\pi} \sec(u) \tan(u) \cdot du\]

Step 4: Integrate the transformed integral

The integral \( \int \sec(u) \tan(u) \cdot du \) is recognized as one of the basic trigonometric integrals, which equals \( \sec(u) + C \). Therefore,\[\int_{\frac{2\pi}{3}}^{\pi} \sec(u) \tan(u) \, du = \left[ \sec(u) \right]_{\frac{2\pi}{3}}^{\pi}\]

Step 5: Substitute back and compute the definite integral

Evaluate \( \sec(u) \) at the bounds \( u = \pi \) and \( u = \frac{2\pi}{3} \):\[ \sec(\pi) = -1 \\sec\left(\frac{2\pi}{3}\right) = -2 \]Now, find the definite integral:\[\sec(\pi) - \sec\left(\frac{2\pi}{3}\right) = -1 - (-2) = 1\]

Step 6: Final answer

Therefore, the evaluated integral is 1.

Key Concepts

Substitution Method

The substitution method is a powerful tool used in calculus to simplify the process of integration, especially for integrals involving complex expressions. By substituting a part of the integral with a new variable, we can transform the original integral into a simpler form which is easier to solve. In the context of our original exercise, the substitution method involves the following steps:

• Identify the expression that can be replaced with a single variable (in this case, let \( u = 2\theta \)).
• Differentiate the substitution equation to express \( d\theta \) in terms of \( du \). Here, \( du = 2 \, d\theta \) so \( d\theta = \frac{1}{2} du \).
• Change the limits of integration from \( \theta \)-bounds to \( u \)-bounds. For example, when \( \theta = \frac{\pi}{3} \), \( u = \frac{2\pi}{3} \) and when \( \theta = \frac{\pi}{2}, u = \pi \).
• Finally, substitute everything back into the integral to solve the modified integral.

This approach reduces integrals to more basic forms and leverages known antiderivatives, making the solution process manageable and systematic.

Definite Integrals

Definite integrals provide a way to calculate the area under a curve, from one point to another along the x-axis. Unlike indefinite integrals, which include a constant of integration \( C \), definite integrals evaluate the difference between the values of an antiderivative at two specific points, known as the bounds.In our original exercise, we calculate a definite integral:

• Substitute to find the bounds \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{\pi}{2} \), which change to \( u = \frac{2\pi}{3} \) and \( u = \pi \) after substitution.
• After finding the antiderivative, use the bounds to find the difference: \( F(b) - F(a) \), where \( F \) is the antiderivative function.
• Compute the value using the formula \[ F(b) - F(a) \] to get the area under the curve or the total accumulated change. In this problem, the outcome is \( 1 \).

Understanding definite integrals is essential for solving problems related to areas, total changes, and accumulations in real-world contexts.

Trigonometric Functions

Trigonometric functions, like sine, cosine, tangent, and secant, are fundamental in calculus and play a crucial role in solving integrals, particularly within contexts involving periodic or oscillatory behaviors. When working with trigonometric integrals, it’s important to understand the properties and relationships between these functions.For the exercise at hand, we worked with \( \sec(2\theta) \) and \( \tan(2\theta) \). Key insights include:

• The integral \( \int \sec(u) \tan(u) \, du \) is a well-known trigonometric integral that simplifies directly to \( \sec(u) + C \).
• Understanding the identities, such as \( \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} \) and \( \sec(2\theta) = \frac{1}{\cos(2\theta)} \), helps in recognizing patterns and simplifying expressions.
• Knowing how these functions interact allows us to create substitutions that simplify integrals considerably.

Mastering trigonometric functions and their properties is vital for tackling a wide variety of problems in calculus and engineering.