Problem 69 - Thomas Calculus

Question

Evaluate each integral in Exercises \(63-70\) by eliminating the square root. $$ \int_{-\pi / 4}^{\pi / 4} \sqrt{1+\tan ^{2} y} d y $$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Simplify the Expression Under the Square Root

Recognize that \( 1 + \tan^2 y = \sec^2 y \). This is a trigonometric identity. Thus, the integral becomes \( \int_{-\pi/4}^{\pi/4} \sqrt{\sec^2 y} \, dy \).

Step 2: Evaluate the Square Root

Since \( \sqrt{\sec^2 y} = |\sec y| \), and for \( y \) in \([-\pi/4, \pi/4]\), \( \sec y \geq 0 \). Thus, \( \sqrt{\sec^2 y} = \sec y \). Now the integral simplifies to \( \int_{-\pi/4}^{\pi/4} \sec y \, dy \).

Step 3: Integrate the Secant Function

The integral of \( \sec y \) is \( \ln |\sec y + \tan y| + C \). So, evaluate the integral using this antiderivative: \( \left[ \ln |\sec y + \tan y| \right]_{-\pi/4}^{\pi/4} \).

Step 4: Evaluate the Definite Integral

Compute \( \ln |\sec(\pi/4) + \tan(\pi/4)| \) and \( \ln |\sec(-\pi/4) + \tan(-\pi/4)| \). For both, \( \sec(\pi/4) = \sqrt{2} \) and \( \tan(\pi/4) = 1 \) which yield \( \ln(\sqrt{2} + 1) \). Thus, \( \int_{-\pi/4}^{\pi/4} \sec y \, dy = \ln(\sqrt{2} + 1) - \ln(\sqrt{2} + 1) = 0 \).

Key Concepts

Trigonometric Identities

Trigonometric identities are crucial in simplifying complex expressions involving trigonometric functions. In the given exercise, recognizing the identity \( 1 + \tan^2 y = \sec^2 y \) was key to simplifying the integral. This identity is derived from the Pythagorean identity \( \sin^2 y + \cos^2 y = 1 \), and it helps us express functions involving the tangent and secant in simpler terms.

Using this identity, we transformed \( \sqrt{1 + \tan^2 y} \) into \( \sqrt{\sec^2 y} \), which is \( |\sec y| \). However, within the given range \([-\pi/4, \pi/4]\), the secant function \( \sec y \) remains positive, allowing us to drop the absolute value and directly evaluate the square root as \( \sec y \).

Understanding and recognizing trigonometric identities can greatly simplify integrals and other mathematical expressions. It reduces complexity and enables us to apply standard techniques more effectively.

Definite Integrals

Definite integrals calculate the net area under a curve within a specific interval. In the exercise, we were tasked to evaluate the integral from \(-\pi/4\) to \(\pi/4\). A definite integral measures the accumulation of quantities, accounting for both positive and negative areas.

Definite integrals differ from indefinite integrals as they yield specific numerical values rather than a general function plus a constant \( C \). After finding an antiderivative, definite integrals are computed using the Fundamental Theorem of Calculus, which involves evaluating the antiderivative at the upper and lower bounds and finding the difference.

In our example, the antiderivative was determined as \( \ln |\sec y + \tan y| \). We then evaluated it at the boundaries \(-\pi/4\) and \(\pi/4\). Strikingly, both evaluations resulted in identical expressions, \( \ln(\sqrt{2} + 1) \), effectively simplifying the integral's result to zero.

Integration Techniques

In calculus, integration techniques help solve integrals that cannot be directly evaluated. A variety of methods exist to tackle integrals, with the choice depending on the function type involved.

• Substitution: A method used when an integral consists of a composite function, rewriting it in terms of a simpler variable.
• Integration by parts: Useful when integrating products of functions. It’s derived from the product rule for differentiation.
• Trigonometric substitution: Applied when an integral contains radical expressions related to the Pythagorean identity.

In the provided exercise, we used the identity \( 1 + \tan^2 y = \sec^2 y \) to simplify the integral. After obtaining \( \int \sec y \, dy \), we applied knowledge of the secant function's antiderivative. The integral \( \int \sec y \, dy \) involves a less common technique. Its antiderivative is \( \ln |\sec y + \tan y| + C \), which some may memorize, but others may derive using integration by parts or considering geometric interpretations.

Using appropriate integration techniques is essential for accurately solving intricate integrals, often turning complex indefinite integrals into manageable algebraic or logarithmic expressions.