Problem 70 - University Calculus: Early Transcendentals

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. $$\int \frac{\csc \theta}{\csc \theta-\sin \theta} d \theta$$

Step-by-Step Solution

Answer

The most general antiderivative is \( \tan \theta + C \).

Step 1: Simplify the Expression

Rewrite the integrand \( \frac{\csc \theta}{\csc \theta - \sin \theta} \).Recall that \( \csc \theta = \frac{1}{\sin \theta} \). Thus, the integrand becomes:\[ \frac{\frac{1}{\sin \theta}}{\frac{1}{\sin \theta} - \sin \theta} = \frac{1}{1 - \sin^2 \theta} \]Recognize that \( 1 - \sin^2 \theta = \cos^2 \theta \). So, the expression turns into:\[ \int \frac{1}{\cos^2 \theta} \, d \theta \] which is\[ \int \sec^2 \theta \, d \theta. \]

Step 2: Evaluate the Integral

The integral \( \int \sec^2 \theta \, d\theta \) is a standard integral that results in \( \tan \theta + C \), where \( C \) is the constant of integration.

Step 3: Verify by Differentiation

Differentiate the result \( \tan \theta + C \) to check if it equals the original integrand.The derivative of \( \tan \theta \) is \( \sec^2 \theta \), which matches the simplified integrand \( \sec^2 \theta \) from the evaluation step.Thus, the solution \( \tan \theta + C \) is verified.

Key Concepts

Antiderivative

An antiderivative is a function whose derivative gives you the original function you started with. In simple terms, it's the reverse process of differentiation. When you're asked to find the antiderivative, you're essentially looking for any functions that, when differentiated, result in the given function.
For instance, in our exercise, the goal was to find the antiderivative of the expression \( \int \frac{\csc \theta}{\csc \theta - \sin \theta} \, d\theta \). By simplifying it down to \( \int \sec^2 \theta \, d\theta \), we used known antiderivative rules to arrive at \( \tan \theta + C \).

• Antiderivative is also known as the "indefinite integral."
• The result is always plus a constant \(C\), because derivatives of constants are zero.

Antiderivatives are central to calculus and essential for solving differential equations.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for any value of the variables involved. They are crucial for simplifying expressions, especially in integration and differentiation.
In the given exercise, one key point was the use of the identity \( \csc \theta = \frac{1}{\sin \theta} \). This helped transform the integrand into a more workable form. Moreover, recognizing the Pythagorean identity \( 1 - \sin^2 \theta = \cos^2 \theta \) was instrumental in further simplifying the problem.

• Trigonometric identities help convert complex expressions into simpler forms.
• These identities are often used to recognize and simplify integrals and derivatives involving trigonometric functions.

An understanding of trigonometric identities allows one to see different expressions as interconnected in ways that ease mathematical manipulation.

Indefinite Integral

The indefinite integral of a function \( f(x) \) is written as \( \int f(x) \, dx \) and represents the family of all antiderivatives of \( f(x) \). Unlike definite integrals, which calculate a numeric value, indefinite integrals result in a general form with a constant of integration, often denoted by \( C \).
In the exercise, after expressing the original integral as \( \int \sec^2 \theta \, d\theta \), we found the indefinite integral by recognizing the standard form, giving us \( \tan \theta + C \).

• Indefinite integrals include the constant of integration, \( C \), reflecting the many possible antiderivatives.
• The process involves understanding integral forms and applying appropriate calculus techniques.

By mastering indefinite integrals, one enhances the ability to solve diverse problems that involve finding original functions from known rates of change.