Problem 9 - Essential Calculus Early Transcendentals

Question

Find the most general antiderivative of the function (Check your answer by differentiation $$h(\theta)=2 \sin \theta-\sec ^{2} \theta$$

Step-by-Step Solution

Answer

The antiderivative is \(-2\cos\theta - \tan\theta + C\).

Step 1: Identify the Function Components

The given function is \( h(\theta) = 2\sin\theta - \sec^2\theta \). We need to find the most general antiderivative of this function. Identify the components: \( 2\sin\theta \) and \( -\sec^2\theta \).

Step 2: Antiderivative of Each Component

Compute the antiderivative of each component separately. For \( 2\sin\theta \), the antiderivative is \(-2\cos\theta\) and for \( -\sec^2\theta \), the antiderivative is \(- \tan\theta \). Thus, the antiderivative of \( h(\theta) \) is \(-2\cos\theta - \tan\theta + C \), where \( C \) is the constant of integration.

Step 3: Combine the Antiderivatives

Combine the antiderivatives obtained from each component to get the most general antiderivative of \( h(\theta) \). This results in the expression \(-2\cos\theta - \tan\theta + C \).

Step 4: Verify by Differentiation

Differentiate \( F(\theta) = -2\cos\theta - \tan\theta + C \). The derivative \( \frac{d}{d\theta}[-2\cos\theta] = 2\sin\theta \) and \( \frac{d}{d\theta}[-\tan\theta] = -\sec^2\theta \), confirming that \( F'(\theta) = 2\sin\theta - \sec^2\theta \), which matches the original function \( h(\theta) \).

Key Concepts

Trigonometric Functions

Trigonometric functions are an integral part of calculus, especially when dealing with periodic or wave-like phenomena. The basic trigonometric functions include \( \sin\theta \), \( \cos\theta \), and \( \tan\theta \), along with their reciprocals: \( \csc\theta \), \( \sec\theta \), and \( \cot\theta \).

• The sine function, \( \sin \theta \), describes the y-coordinate of a point on the unit circle.
• The cosine function, \( \cos \theta \), describes the x-coordinate and is closely related to sine through the equation \( \sin^2\theta + \cos^2\theta = 1 \).
• The tangent function, \( \tan \theta = \frac{\sin\theta}{\cos\theta} \), represents the slope of the line joining a point to the origin on the Cartesian plane.

Understanding these functions is crucial as they frequently appear in various forms in calculus problems. Identifying these components, like \( 2\sin\theta \) and \( -\sec^2\theta \) in our given function, is the first step in solving integration and differentiation problems involving trigonometric functions. It's important to be familiar with their properties and transformations.

Integration

Integration is the process of finding the antiderivative or integral of a function, which is essentially the reverse process of differentiation. In simple terms, it helps us determine the original function from its derivative. When solving calculus problems like the one given, we must break down the function into manageable parts and find the antiderivative of each component separately. For example, consider the function \( h(\theta) = 2\sin\theta - \sec^2\theta \):- The antiderivative of \( 2\sin\theta \) is \( -2\cos\theta \).- The antiderivative of \( -\sec^2\theta \) is \( -\tan\theta \).Adding these results gives us the general antiderivative: \( -2\cos\theta - \tan\theta + C \), where \( C \) is a constant of integration.By mastering integration techniques, such as recognizing common antiderivatives and using substitution, students greatly enhance their problem-solving capabilities in calculus.

Calculus Problem Solving

Solving calculus problems requires a strategic approach of both understanding and applying mathematical techniques. In problems where integration of trigonometric functions is required, it's essential to properly identify each component of the given function. Every step should build on fundamental concepts of calculus, like differentiation and integration.

• Start by clearly identifying the components of the function that need to be integrated.
• Find the antiderivative of each component before combining them to find the general solution.
• Always remember to include a constant of integration \( C \), as it represents the family of all possible solutions.

Verification through differentiation is the final checkpoint. In our example, differentiating \( F(\theta) = -2\cos\theta - \tan\theta + C \) confirms that the original function is retrieved. This process not only checks accuracy but also reinforces understanding by revisiting and confirming each step. Therefore, effective problem-solving in calculus involves a strong grasp of techniques, structured planning, and consistent practice to become proficient in mathematics.