Question

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. \(\int_{-\pi / 3}^{\pi / 3} \sqrt{\sec ^{2} x-1} d x\)

Short answer

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

Step-by-step solution

Recognize the Integral

Identify the integral expression: \[\int_{-\pi / 3}^{\pi / 3} \sqrt{\sec ^{2} x-1} \, dx\]Recognize that the expression under the square root, \(\sec^2 x - 1\), is a trigonometric identity.

Simplify Using Trigonometric Identity

Recall the identity: \[\sec^2 x - 1 = \tan^2 x\]Thus, the integral becomes:\[\int_{-\pi / 3}^{\pi / 3} |\tan x| \, dx\]The absolute value is important because the tangent function can be negative.

Evaluate Absolute Value Over the Interval

Since the tangent function is negative in the interval \([-\pi/3, 0)\) and positive in the interval \((0, \pi/3]\), the integral splits into:\[\int_{-\pi / 3}^{0} (-\tan x) \, dx + \int_{0}^{\pi / 3} \tan x \, dx\]This allows us to handle the absolute value properly.

Integrate Each Interval Separately

For the negative interval:\[\int_{-\pi / 3}^{0} -\tan x \, dx = -\left[ \ln |\sec x| \right]_{-\pi / 3}^{0}\]For the positive interval:\[\int_{0}^{\pi / 3} \tan x \, dx = \left[ \ln |\sec x| \right]_{0}^{\pi / 3}\]The integration of \(\tan x\) gives \(\ln |\sec x|\).

Calculate Each Integral

Calculate the definite integral for the "negative" part:\[-\left( \ln |\sec 0| - \ln |\sec(-\pi/3)| \right) = -(0 - \ln 2) = \ln 2\]Calculate the definite integral for the "positive" part:\[\ln |\sec (\pi/3)| - \ln |\sec 0| = \ln 2 - 0 = \ln 2\]

Add the Results

Add the results from both intervals:\[\ln 2 + \ln 2 = 2 \ln 2\]This is the value of the definite integral. Hence, the integral evaluates to \(2 \ln 2\).

Key concepts

Trigonometric Identities

Trigonometric identities are fundamental tools used to simplify and solve complex trigonometric expressions. One such identity is \( \sec^2 x - 1 = \tan^2 x \). This identity helps to transform the square root of \( \sec^2 x - 1 \) into a simpler expression, \( |\tan x| \). This transformation makes the integration process much more straightforward. Understanding trigonometric identities like these allows us to rewrite expressions in ways that are easier to integrate or differentiate. In practice, recognizing these identities enables us to deconstruct complex expressions, resulting in a more manageable form to tackle with techniques of calculus.

Integral Calculus

Integral Calculus plays a crucial role in determining the area under the curve of functions over a given interval. The definite integral we are dealing with, \( \int_{-\pi / 3}^{\pi / 3} \sqrt{\sec^2 x - 1} \, dx \), requires evaluating the area where the function \( | \tan x | \) exists between the limits of integration. Definite integrals are vital as they provide a precise value and are used frequently to calculate physical quantities in the sciences. This involves two main components: recognizing how the function behaves over the interval and ensuring calculations of the area account for such behavior, particularly when absolute values are involved. Through definite integration, we not only solve for area but also apply significant concepts such as the Fundamental Theorem of Calculus, which links derivatives and integrals in a meaningful way.

Integration Techniques

Integration Techniques extend our capability to evaluate challenging integrals. In our example, we split the definite integral \( \int_{-\pi/3}^{\pi/3} |\tan x| \, dx \) into two intervals: \([-\pi/3, 0)\) and \((0, \pi/3]\). Since \( \tan x \) changes sign at \( x = 0 \), recognizing this allows us to deal with the absolute value of the function correctly. By evaluating \( \int_{-\pi/3}^{0} -\tan x \, dx \) and \( \int_{0}^{\pi/3} \tan x \, dx \) separately, we accurately account for the behavior of \( \tan x \) across these intervals. Integration by parts may be necessary for different functions, but in this scenario, recognizing the direct antiderivative of \( \tan x \), which is \( \ln |\sec x| \), is the key. Mastery of such techniques allows us to overcome hurdles presented by more complex integrals across varying functions.