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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. $$\int \frac{1}{2}\left(\csc ^{2} x-\csc x \cot x\right) d x$$

Short Answer

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

Step by step solution

01

Simplify the Integrand

The integrand is \( \frac{1}{2}(\csc^2 x - \csc x \cot x) \). Break it into two separate integrals: \( \frac{1}{2} \int \csc^2 x \, dx - \frac{1}{2} \int \csc x \cot x \, dx \).
02

Find Antiderivative for the First Term

The antiderivative of \( \csc^2 x \) is \(-\cot x\). Thus, for the first term: \[ \frac{1}{2} \int \csc^2 x \, dx = -\frac{1}{2} \cot x + C_1 \] where \( C_1 \) is a constant of integration.
03

Find Antiderivative for the Second Term

The antiderivative of \( \csc x \cot x \) is \(-\csc x\). Thus, for the second term: \[ -\frac{1}{2} \int \csc x \cot x \, dx = \frac{1}{2} \csc x + C_2 \] where \( C_2 \) is another constant of integration.
04

Combine Results

Combine the results from Steps 2 and 3. The most general antiderivative is:\[ -\frac{1}{2} \cot x + \frac{1}{2} \csc x + C \] where \( C = C_1 + C_2 \) is a general constant.
05

Check by Differentiation

Differentiate the result from Step 4: \[ \frac{d}{dx}\left(-\frac{1}{2}\cot x + \frac{1}{2}\csc x + C\right) = \frac{1}{2} \csc^2 x - \frac{1}{2} \csc x \cot x \] which is the original integrand, confirming the antiderivative is correct.

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Key Concepts

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

Definite Integrals
Definite integrals represent the accumulation of quantities, such as areas under curves, over specified intervals. However, in the context of this exercise, we're focusing on a different concept, the *indefinite integral*, which relates to finding the most general form of an antiderivative. Remember, while a definite integral computes a number or total quantity, an indefinite integral results in a function—the antiderivative.
When solving indefinite integrals, the result includes a constant of integration, denoted usually as "C". This constant represents all possible vertical displacements of the antiderivative since differentiation of a constant yields zero.
- Indefinite integrals are about finding a general form rather than a specific numerical answer. - Don't forget the constant of integration! It's crucial to denote the full family of potential solutions.
Trigonometric Functions
Trigonometric functions such as sine, cosine, tangent, cosecant (csc), secant, and cotangent (cot) form the foundation of many calculus problems. In our exercise, we're particularly concerned with less common functions:
  • **Cosecant** (\(\csc x\)): the reciprocal of sine (\(\csc x = \frac{1}{\sin x}\))
  • **Cotangent** (\(\cot x\)): the reciprocal of tangent, or the quotient of cosine over sine (\(\cot x = \frac{\cos x}{\sin x}\))
These functions have specific derivatives:
  • The derivative of \(\csc^2 x\) is related to tangents and cosecants.
  • The derivative of \(\csc x \cot x\) is straightforward \(- \csc^2 x\), thus direct when reversed in solving integrals.
Understanding the derivatives helps find antiderivatives, which is central in tackling this exercise.
Integration Techniques
To solve the given integral, some handy integration techniques are at play. Splitting an integrand into separate parts is a common first step. Let's break it down:
  • **Splitting the Integrand:** This involves separating a complex expression into simpler integrals, which can be solved individually, then combined. In our solution, \(\frac{1}{2}(\csc^2 x - \csc x \cot x)\) becomes two separate integrals.
  • **Applying Known Antiderivatives:** It's crucial to know the antiderivatives of standard functions. For example, the antiderivative of \(\csc^2 x\) is known as \(-\cot x\). Recognizing these patterns allows quick solving.
  • **Combining Results:** Once individual parts are integrated, they are combined back together, always including the constant of integration to reflect all possible solutions.
These techniques simplify the solution process, ensuring accurate results verified by differentiation.

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