Question

Set up the integral to compute the arc length of the function on the given interval. Do not evaluate the integral. \(f(x)=\sec x\) on \([-\pi / 4, \pi / 4]\)

Short answer

The integral to compute the arc length is \( L = \int_{-\pi/4}^{\pi/4} \sec^2 x \, dx \).

Step-by-step solution

Identify the Formula for Arc Length

The formula for the arc length of the function \( f(x) \) from \( x = a \) to \( x = b \) is given by \( \int_{a}^{b} \sqrt{1 + \left( f'(x) \right)^2} \, dx \). First, we need to differentiate \( f(x) = \sec x \).

Differentiate the Function

The derivative of \( \sec x \) is \( f'(x) = \sec x \tan x \). This result will be used in the arc length formula.

Plug the Derivative into the Arc Length Formula

Substituting \( f'(x) = \sec x \tan x \) into the arc length formula gives: \[ \int_{-\pi/4}^{\pi/4} \sqrt{1 + (\sec x \tan x)^2} \, dx \].

Simplify the Expression Under the Square Root

The expression under the square root is:\[ \sqrt{1 + \sec^2 x \tan^2 x} \].Using the identity \( \tan^2 x = \sec^2 x - 1 \), we have:\[ \sec^2 x \tan^2 x = \sec^2 x ( \sec^2 x - 1) = \sec^4 x - \sec^2 x \].Therefore, \[ 1 + \sec^2 x \tan^2 x = 1 + \sec^4 x - \sec^2 x = \sec^4 x - \sec^2 x + 1 \].Notice that this simplifies to another trigonometric identity which reduces to:\[ \sqrt{\sec^2 x ( \sec^2 x) } = \sec^2 x \].

Final Integral Setup

The arc length integral is now simpler:\[ L = \int_{-\pi/4}^{\pi/4} \sec^2 x \, dx \].This is the final integral setup to compute the arc length.

Key concepts

Derivative of Trigonometric Functions

When dealing with the derivative of trigonometric functions, it's crucial to remember their specific forms and rules. For the trigonometric function \(f(x) = \sec x\), the derivative needs special attention because it's not as straightforward as polynomials or basic exponentials. The secant function can be expressed in terms of cosine: \(\sec x = \frac{1}{\cos x}\). The derivative of \(\sec x\) is given by \(f'(x) = \sec x \tan x\).

• To differentiate \(\sec x\), you can use the quotient rule or recall the derivative directly if it's memorized.
• Using the product \(\sec x \times \tan x\), we simplify the process to find derivative accuracy quickly.

Understanding these derivatives is key for completing different calculus problems involving trigonometric functions, especially those concerning arc lengths.

Trigonometric Identities

Trigonometric identities play a vital role in simplifying expressions. One essential identity is \(\tan^2 x = \sec^2 x - 1\). This identity is crucial when simplifying expressions within calculations, such as those involved in arc length formulas.

• Simplifying \(\sec^2 x \tan^2 x\) into more manageable expressions depends heavily on applying these identities.
• For example, rearranging terms like \(\tan^2 x = \sec^2 x - 1\) allows us to break down more complex expressions into simpler forms.

In the given problem, utilizing these identities helps to transform the expression into a simpler form \(\sec^4 x - \sec^2 x + 1\). Mastery of such identities is indispensable for anyone looking to solve trigonometric calculus problems effectively.

Integral Setup

Setting up the integral correctly is essential in calculating arc length. For an arc of \(f(x) = \sec x\) from \([-\pi/4, \pi/4]\), the process starts with forming the arc length integral formula:\[L = \int_{a}^{b} \sqrt{1 + \left( f'(x) \right)^2} \, dx\]Using the derivative found earlier, \(f'(x) = \sec x \tan x\), we plug this into the formula, giving us:\[\int_{-\pi/4}^{\pi/4} \sqrt{1 + (\sec x \tan x)^2} \, dx\]

• Simplify the square root expression taking advantage of trigonometric identities, leading us to \(\sec^2 x\).
• The key step is the simplification, resulting in a considerably easier integral \(\int_{-\pi/4}^{\pi/4} \sec^2 x \, dx\).

This integral setup is refined and ready for evaluation, a crucial step for determining the arc length without missteps. Properly forming and simplifying the integral is often the hardest part of such problems, making understanding each setup step vital.