Problem 70 - Calculus: Early Transcendental Functions

Question

Derive the formulas \(\int \sec ^{2} x d x=\tan x+c\) and \(\int \sec x \tan x \, d x=\sec x+c\)

Step-by-Step Solution

Answer

The derivatives of tangent and secant functions are, respectively, \( \sec^{2} x\) and \( \sec x \tan x\), so when we reverse the operation (integration), we get the integrals as: \[\int \sec^{2} x \, dx=\tan(x)+c\] and \[\int \sec(x) \tan(x) \, dx=\sec(x)+c\].

Step 1: Integrate the function sec²(x)

The integral of \( \sec^{2} x\) with respect to \(x\) is \( \tan x\). Therefore, \[\int \sec^{2} x \, dx=\tan(x)+c\]. The expression \(c\) represents the constant of integration as indefinite integrals include an arbitrary constant.

Step 2: Integrate the function sec(x)tan(x)

The integral of \(\sec x \tan x\) with respect to \(x\) is \( \sec x\). Therefore, \[ \int \sec x \tan x \, dx=\sec(x)+c\]. The expression \(c\) denotes the constant of integration.

Key Concepts

Indefinite Integrals

Indefinite integrals are a fundamental concept in calculus. They represent the antiderivative of a function. This means finding a function whose derivative is the original function given. Unlike definite integrals, indefinite integrals do not evaluate to a specific number; instead, they provide a family of functions.
An indefinite integral includes a constant of integration, often denoted as "\(c\)." The reason for this is that the derivative of a constant is zero, and thus there are infinitely many functions with the same derivative. For example, if \(F(x)\) is an antiderivative of \(f(x)\), then \(F(x) + c\) is also an antiderivative of \(f(x)\) for any constant \(c\).
The notation for an indefinite integral is \(\int f(x) \, dx\). Here, \(f(x)\) is the integrand, and \(dx\) indicates the variable of integration. Evaluating indefinite integrals is often a step in solving differential equations and understanding function behavior.

Integration Techniques

When it comes to integration, there are several techniques used to solve different types of integrals. These techniques simplify complex integrals or make them solvable when they appear unsolvable at first glance.

• **Basic Integration:** Direct integration is possible for straightforward expressions, relying primarily on known antiderivatives.
• **Substitution:** This method works similarly to the chain rule in differentiation. It involves changing variables to simplify the integration process and is especially useful when dealing with compositions of functions.
• **Integration by Parts:** This technique uses the product rule for differentiation in reverse. It's often employed when the integrand is a product of two different functions.
• **Trigonometric Identities:** Certain integrals, particularly those involving trigonometric functions, require the use of trigonometric identities to transform the integrand into a simpler form.

Each technique has its own set of scenarios where it is most effective. In the exercise we looked at, recognizing the antiderivative directly from known integrals of trigonometric functions helped to solve it efficiently.

Trigonometric Integrals

Trigonometric integrals involve integrands that include trigonometric functions like sine, cosine, secant, and tangent. These integrals often require specific strategies to solve, leveraging the properties and identities of trigonometric functions.
For example, when faced with the integral of \(\sec^2 x\), it's useful to remember that the derivative of \(\tan x\) is \(\sec^2 x\). This connection is a standard result in calculus, making \(\tan x\) the antiderivative.
Similarly, the integral \(\int \sec x \tan x \, dx\) leads to \(\sec x + c\) because the derivative of \(\sec x\) is \(\sec x \tan x\).
Solving these types of integrals often involves routine knowledge of derivative and antiderivative pairs for trigonometric functions.

• Having a strong grasp of basic derivative rules for these functions is beneficial.
• Memorizing integral results for common trigonometric functions aids in quick resolution of integrals.
• Understanding the relationships between different trigonometric functions can simplify seemingly complex integrals.

Building confidence in tackling trigonometric integrals comes from practice and familiarity with these relationships.