Problem 40

Question

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(-\frac{\sec ^{2} x}{3}\right) d x$$

Step-by-Step Solution

Verified
Answer
The most general antiderivative is \(-\frac{1}{3} \tan x + C\).
1Step 1: Identify the Integral Form
We begin by identifying the form of the integral. The given function is \(\int -\frac{\sec ^{2} x}{3} \, dx\). This looks similar to the derivative of a trigonometric function, allowing us to anticipate the antiderivative.
2Step 2: Express in Familiar Terms
Recognize that the derivative of \( \tan x \) is \( \sec^2 x \). Therefore, our integral can be linked to \( \int \sec^2 x \, dx = \tan x \). The integral to solve becomes \( -\frac{1}{3} \int \sec^2 x \, dx \).
3Step 3: Integrate Using Known Integral
Integrating \( \sec^2 x \) gives \( \tan x \), so the integral becomes \[-\frac{1}{3} \tan x\]. Add the constant of integration, \( C \), as this is an indefinite integral. Thus, the most general antiderivative is\[-\frac{1}{3} \tan x + C\].
4Step 4: Check by Differentiating Result
Differentiate the antiderivative, \(-\frac{1}{3} \tan x + C \). The derivative of \( -\frac{1}{3} \tan x \) is \[-\frac{1}{3} \sec^2 x\], and the derivative of the constant \( C \) is 0. This confirms the original integrand.

Key Concepts

AntiderivativesTrigonometric FunctionsIntegration Techniques
Antiderivatives
Antiderivatives, also known as indefinite integrals, are functions that reverse the process of differentiation. When finding an antiderivative, you are essentially seeking a function whose derivative matches the given function.
For any function, such as the one in the exercise \(-\frac{\sec ^2 x}{3}\), finding its antiderivative involves recognizing a basic derivative pattern. Because the derivative of \(\tan x\) is \(\sec^2 x\), it can be deduced that the integral of \(\sec^2 x\) is \(\tan x\). Thus, the antiderivative of \(-\frac{1}{3} \sec^2 x\) is \(-\frac{1}{3} \tan x\) with an added constant C to represent the family of all antiderivatives.
On a practical level, always remember that indefinite integrals include a constant of integration, as there are infinitely many functions whose derivatives result in the same expression.
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, often involved in calculus problems. Specifically, \(\tan x\), \(\sec x\), and others entail their specific derivatives and antiderivatives which help simplify integral calculations.
The key aspect of trigonometric functions in integration is knowing their derivatives and antiderivatives by heart. For instance, the derivative of \(\tan x\) is \(\sec^2 x\), which indicates that the integral of \(\sec^2 x\) is \(\tan x\). These relationships are essential shortcuts that can simplify complex integration tasks.
Remember, when dealing with trigonometric functions, comparing derivatives and the given integral can rapidly guide you to the antiderivative.
Integration Techniques
Understanding the underlying integration techniques can make solving integrals much more manageable. One such technique involves recognizing patterns and matching them to known antiderivative formulas. In our exercise \(\int -\frac{\sec^2 x}{3} \, dx\), this approach helped to identify the integral form resembling the derivative of a trigonometric function.
Key integration techniques include:
  • Substitution: Useful when integrals involve a composite function, simplifying with a change of variable.
  • Integration by Parts: Useful when the integral is a product of two functions, using the formula \(\int u \, dv = uv - \int v \, du\).
  • Recognizing basic derivatives and antiderivatives: The most direct technique, which we applied in this exercise.
While the given problem did not require substitution or integration by parts, your toolbox of techniques should be ready for tackling a variety of integral challenges. Be sure to check your work by differentiating to ensure you have the correct antiderivative.