Problem 36 - University Calculus: Early Transcendentals

Question

Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises \(1-46\). $$\int_{0}^{\pi / 12} 6 \tan 3 x d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln 2 \).

Step 1: Identify the substitution

To use substitution, we need to transform the integrand into a simpler form. We notice that inside the tangent function, there is a 3x term, so we choose to let \( u = 3x \). This substitution simplifies the integrand.

Step 2: Differentiate the substitution

Once the substitution \( u = 3x \) is chosen, differentiate to find \( du \). Differentiate both sides: \( du = 3 \, dx \). This allows us to express \( dx \) in terms of \( du \): \( dx = \frac{1}{3} \, du \).

Step 3: Change the limits of integration

Since we are changing variables, we need to adjust the limits of integration. The original limits are from \( x = 0 \) to \( x = \pi/12 \). With \( u = 3x \), when \( x = 0, u = 0 \), and when \( x = \pi/12, u = \pi/4 \). So the new limits of integration for \( u \) are from 0 to \( \pi/4 \).

Step 4: Substitute and integrate

Substitute \( u = 3x \) and \( dx = \frac{1}{3} \, du \) into the integral. The integral becomes: \[\int_{0}^{\pi/4} 6 \tan u \cdot \frac{1}{3} \, du = 2 \int_{0}^{\pi/4} \tan u \, du.\] The integral of \( \tan u \) is \( -\ln |\cos u| \). Thus, \[2 \int \tan u \, du = 2(-\ln |\cos u|) = -2 \ln |\cos u|.\]

Step 5: Evaluate the definite integral

Evaluate \(-2 \ln |\cos u| \) from \( u = 0 \) to \( u = \pi/4 \):\[-2 \ln |\cos(\pi/4)| + 2 \ln |\cos(0)|.\] At \( u = \pi/4 \), \( \cos(\pi/4) = \frac{\sqrt{2}}{2} \), and at \( u = 0 \), \( \cos(0) = 1 \). The expression becomes:\[-2 \ln \left(\frac{\sqrt{2}}{2}\right) + 2 \ln 1.\]

Step 6: Simplify the expression

Calculate the expression from Step 5: \[-2 \ln \left(\frac{\sqrt{2}}{2}\right) = -2 \left(\ln(\sqrt{2}) - \ln(2)\right) = 2 \ln 2 - \ln 2 = \ln 2.\] Since \( 2 \ln 1 = 0 \), the answer is \( \ln 2 \).

Key Concepts

Definite Integral

A definite integral is a concept in calculus that calculates the net area under a curve within a specific interval. In this particular exercise, the definite integral is evaluated to determine the area under the curve of the function \( 6 \tan 3x \) from \( x = 0 \) to \( x = \frac{\pi}{12} \).

• The main idea behind a definite integral is that it accumulates the values of a function over an interval.
• Unlike an indefinite integral, which yields a family of functions, a definite integral provides a numerical value representing this accumulated quantity.

To compute it, you follow these steps:

• Find an antiderivative of the function, which is a function whose derivative gives the original function.
• Evaluate this antiderivative at the boundaries of the interval and find their difference.

For our specific problem, we found that integrating resulted in \( \ln 2 \), indicating the definite area evaluated over the interval using the given method.

Integration by Substitution

Integration by substitution is a crucial technique in calculus used to simplify the integration process by transforming a complicated integral into a simpler one. In this exercise, it aids in solving \( \int 6 \tan 3x \, dx \) by making the calculation more manageable.The procedure involves:

• Choosing a substitution variable \( u \) to replace a part of the integrand. Here, \( u = 3x \).
• Finding the differential \( du \), which results in \( du = 3 \, dx \) and solving for \( dx \) gives \( dx = \frac{1}{3} \, du \).
• Substituting \( u \) and \( dx \) in terms of \( du \) into the integral simplifies the process by changing the variable.

This transforms the original problem into a simpler integral \( 2 \int \tan u \, du \), reducing the complexity and making the function easier to integrate. This method is similar to the reverse of the chain rule in differentiation and is extensively used when dealing with compound functions.

Trigonometric Integration

Trigonometric integration involves techniques specifically for integrating functions with trigonometric components. Many integrals, especially involving sine, cosine, and tangent, can be complex to handle. In this exercise, the integral of \( \tan u \) was evaluated as part of the process.Some steps we took include:

• Recognizing the integral of \( \tan u \) as a standard form, which simplifies to \( -\ln |\cos u| \).
• Incorporating known trigonometric identities can often help simplify expressions and facilitate integration.
• Understanding how to handle absolute values in integrals can be particularly important in ensuring correct results.

By integrating, we look at how trigonometric functions bring periodic behaviors into calculus problems, helping us find areas, solve equations, and understand other mathematical properties. The use of trigonometric identities often aids in breaking down more complex expressions into integrable forms.