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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. $$\int \frac{\csc \theta \cot \theta}{2} d \theta$$

Short Answer

Expert verified
The antiderivative is \( -\frac{1}{2} \csc \theta + C \).

Step by step solution

01

Identify the Integrand

The integrand given is \( \frac{\csc \theta \cot \theta}{2} \). It involves the trigonometric functions \( \csc \theta \) and \( \cot \theta \). The form is reminiscent of the derivative of \( \csc \theta \), which is \( -\csc \theta \cot \theta \).
02

Simplify the Expression

Rewrite the integrand as \( \frac{1}{2} \csc \theta \cot \theta \). This makes it easier to integrate by clearly identifying the constant multiplier.
03

Guess the Antiderivative

Given that the derivative of \( \csc \theta \) is \( -\csc \theta \cot \theta \), we guess that an antiderivative might be of the form \( -\frac{1}{2} \csc \theta + C \), where \( C \) is an arbitrary constant.
04

Differentiate to Verify

Differentiate \( -\frac{1}{2} \csc \theta + C \). The derivative is \( \frac{1}{2} \csc \theta \cot \theta \), which matches our original integrand. This confirms our guessed antiderivative is correct.
05

Write the General Antiderivative

The most general antiderivative of \( \frac{\csc \theta \cot \theta}{2} \) is \( -\frac{1}{2} \csc \theta + C \), where \( C \) is any real constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
In calculus, trigonometric functions such as sine, cosine, cosecant, and cotangent play an essential role. Understanding these functions and their derivatives is crucial when dealing with integrals, especially indefinite integrals. Here, the integral involves the trigonometric functions \( \csc \theta \) (cosecant) and \( \cot \theta \) (cotangent).- **Cosecant** (\( \csc \theta \)) is the reciprocal of sine, so \( \csc \theta = \frac{1}{\sin \theta} \).- **Cotangent** (\( \cot \theta \)) is the ratio of cosine to sine, expressed as \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).These functions have certain properties and derivatives that make them useful for integration. For instance, the derivative of \( \csc \theta \) is \(-\csc \theta \cot \theta \). By recognizing the form of these derivatives, we can often identify potential antiderivatives, which simplifies solving indefinite integrals.
Antiderivative Verification
Once a guess for an antiderivative is made, verification is a crucial step in solving indefinite integrals. In our case, after simplifying the integrand, we guessed the antiderivative to be \(-\frac{1}{2} \csc \theta + C\). Verification involves differentiating this guessed antiderivative and checking whether it matches the original integrand. Here's how verification works:- **Differentiate the Antiderivative**: The derivative of the guessed antiderivative \(-\frac{1}{2} \csc \theta + C\) is calculated.- **Compare**: Ensure that the derivative obtained is \(\frac{1}{2} \csc \theta \cot \theta \), matching the original integrand.By completing these steps, we confirm that our antiderivative is correct. This methodology not only ensures the correctness of our solution but builds a deeper understanding of integrating functions.
Constant of Integration
In the realm of indefinite integrals, the constant of integration, denoted as \(C\), is an element that signifies the main difference from definite integrals. The constant arises because indefinite integration essentially reverses differentiation, which can ignore a constant shift.- **Significance of \(C\)**: Since the derivative of a constant is zero, when integrating, we cannot determine from the operation itself if a constant was involved. Hence, \(C\) represents any possible constant that could have been present.- **General Form**: This means the general antiderivative will be expressed as the specific antiderivative plus \(C\), like \(-\frac{1}{2} \csc \theta + C\).Including the constant in our expressions is crucial when dealing with indefinite integrals. It ensures that every possible original function, even those differing by a constant, is represented by our result.

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Most popular questions from this chapter

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