Finding Indefinite Integrals


In Exercises 17–56, 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.


∫(cscθ cotθ) / 2 dθ

Verified step by step guidance

1. Recognize that the integral is \( \int \frac{\csc \theta \cot \theta}{2} \, d\theta \). Since the constant \( \frac{1}{2} \) can be factored out, rewrite the integral as \( \frac{1}{2} \int \csc \theta \cot \theta \, d\theta \).
2. Recall the derivative of \( \csc \theta \) is \( -\csc \theta \cot \theta \). This suggests that \( \csc \theta \cot \theta \) is closely related to the derivative of \( \csc \theta \).
3. Use this relationship to guess that the antiderivative of \( \csc \theta \cot \theta \) is \( -\csc \theta \), because differentiating \( -\csc \theta \) gives \( \csc \theta \cot \theta \).
4. Therefore, the integral becomes \( \frac{1}{2} \times (-\csc \theta) + C \), where \( C \) is the constant of integration.
5. Finally, verify your result by differentiating \( -\frac{1}{2} \csc \theta + C \) to ensure it matches the original integrand \( \frac{\csc \theta \cot \theta}{2} \).

Key Concepts

Indefinite Integral and Antiderivative

An indefinite integral represents the most general form of an antiderivative of a function, including a constant of integration. It reverses differentiation, finding a function whose derivative matches the integrand. Understanding this helps in solving integrals without specified limits.

Trigonometric Functions and Identities

Knowledge of trigonometric functions like cosecant (csc) and cotangent (cot), and their relationships, is essential. Recognizing identities such as the derivative of cscθ being -cscθ cotθ aids in simplifying and integrating expressions involving these functions.

Verification by Differentiation

After finding an antiderivative, differentiating it confirms the correctness of the integral. This step ensures the solution is accurate and helps identify any errors in the integration process, reinforcing understanding of the fundamental theorem of calculus.