Problem 20 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(15-22\). $$ \int_{-\pi / 4}^{\pi / 4} \sqrt{\sec ^{2} x-1} d x $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Expression Inside the Integral

The expression inside the integral is \( \sqrt{\sec^2 x - 1} \). Recall that \( \sec^2 x = 1 + \tan^2 x \). So, \( \sec^2 x - 1 \) simplifies to \( \tan^2 x \). Therefore, \( \sqrt{\sec^2 x - 1} = \sqrt{\tan^2 x} = \lvert \tan x \rvert \), provided \( \tan x \) is defined.

Step 2: Set Up the Integral with Absolute Value

Now that we have \( \lvert \tan x \rvert \), we need to address the absolute value when evaluating the integral. Since \( x \) ranges from \(-\pi/4\) to \(\pi/4\), and in this interval, \( \tan x \) is non-negative, the absolute value can be removed, i.e. \( \int_{-\pi/4}^{\pi/4} \lvert \tan x \rvert \, dx = \int_{-\pi/4}^{\pi/4} \tan x \, dx \).

Step 3: Evaluate the Antiderivative

The antiderivative of \( \tan x \) is \(-\ln |\cos x|\). So, we need to evaluate \([-\ln |\cos x|] \) from \(-\pi/4\) to \(\pi/4\).

Step 4: Evaluate the Definite Integral

Substitute the bounds into the antiderivative: \[-\ln |\cos(\pi/4)| + \ln |\cos(-\pi/4)|\]Since \( \cos(\pi/4) = \cos(-\pi/4) = \frac{\sqrt{2}}{2} \), the expression becomes \[-2 \ln \left( \frac{\sqrt{2}}{2} \right) = -2 \ln(\sqrt{2}/2) = -2 ( \ln(\sqrt{2}) - \ln(2) ) = -2 \left( \frac{1}{2} \ln(2) - \ln(2) \right) = \ln(2)\].

Step 5: Compute the Final Result

The evaluated integral is thus \( \ln(2) \).

Key Concepts

Definite Integral Evaluation

Calculating a definite integral involves finding the antiderivative of a function and then using specified limits to determine the exact area under the curve between those limits. In the given exercise, we are evaluating \[ \int_{-\pi / 4}^{\pi / 4} \sqrt{\sec ^{2} x-1} \, dx \]Understanding what the expression represents is crucial. The function inside the integral is simplified as \( \sqrt{\tan^2 x} = |\tan x| \), which needs careful attention because of the absolute value.
The limits \(-\pi/4\) and \(\pi/4\) guide us to substitute these values into the antiderivative of the function to find this definite integral.

Trigonometric Integration

Trigonometric integration involves solving integrals that include trigonometric functions. A common strategy is using identities to simplify the functions. In our problem:

• We start with \( \int_{-\pi / 4}^{\pi / 4} \sqrt{\sec ^{2} x-1} \, dx \)
• Using the identity \( \sec^2 x = 1 + \tan^2 x \), simplifies the expression inside the square root to \( \tan^2 x \).
• This eventually simplifies to \( |\tan x|\), which is much easier to integrate, especially within the specified limits.

Knowing these identities can greatly simplify the integration process, making it easier to proceed with finding antiderivatives.

Absolute Value in Integration

Handling absolute values in integration requires careful consideration of the intervals over which the function is being integrated. For this exercise, after simplifying \( \sqrt{\sec^2 x - 1} \) to \( |\tan x| \):

• We need to assess whether the absolute value impacts the evaluation. Since \( \tan x \) is non-negative between \(-\pi/4\) and \(\pi/4\), the absolute value doesn't alter the function.
• This allows us to write \( |\tan x| \) simply as \( \tan x \) within these bounds.

This consideration simplifies the integration process since \( \tan x \) behaves predictably within the limits, permitting us to integrate without added complexity.

Antiderivative Calculation

Calculating an antiderivative is a fundamental step in integration. For the function \( \tan x \):

• The antiderivative is \(-\ln |\cos x| \).
• After determining the antiderivative, you must evaluate it at a series of specific limits to compute a definite integral.
  For this exercise, the limits are \(-\pi/4\) and \(\pi/4\).

The step-by-step evaluation involves:

• Substituting these bounds into the antiderivative, \([-\ln |\cos x|] \)
• Simplifying the expression by calculating \(-2 \ln (\sqrt{2}/2)\) which further reduces to \(\ln(2)\).

The careful calculation of the antiderivative and the subsequent steps provide us with the final result, showing the area under the curve as \(\ln(2)\).