Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \csc x(\cot x-\csc x) d x$$

Short answer

The indefinite integral of the function $$\csc x(\cot x-\csc x)$$ is $$-\csc x -\cot x + C$$, where C is the constant of integration.

Step-by-step solution

Apply the distributive property

First, distribute the \(\csc x\) within the parentheses to make two separate terms. This way, the integral can be split into two simpler integrals. $$\int \csc x(\cot x - \csc x) dx = \int (\csc x \cot x - \csc^2 x) dx$$

Separate the integral

Break down the integral into two separate integrals. This makes it easier to evaluate each integral term individually. $$\int (\csc x \cot x - \csc^2 x) dx = \int \csc x \cot x dx - \int \csc^2 x dx$$

Evaluate the first integral

Recall that the integral of \(\csc x \cot x\) is \(-\csc x\): $$\int \csc x \cot x dx = -\csc x$$

Evaluate the second integral

Likewise, the integral of \(\csc^2 x\) is \(-\cot x\): $$\int \csc^2 x dx = -\cot x$$

Combine the results

Add the results of the two integrals together to obtain the final answer, remember to add a constant of integration, denoted as \(C\). $$\int \csc x(\cot x - \csc x) dx = -\csc x -\cot x + C$$

Check the result by differentiation

Differentiate the result \(-\csc x - \cot x + C\) with respect to \(x\) to confirm that it matches the original integrand. $$ \frac{d}{dx}(-\csc x - \cot x + C) = \csc x \cot x - \csc^2 x = \csc x(\cot x - \csc x) $$ The differentiation matches the original integrand, so the final answer is correct: $$ \int \csc x(\cot x - \csc x) dx = -\csc x -\cot x + C$$

Key concepts

Integration Techniques

Understanding how to navigate through various integration techniques is essential for solving indefinite integrals in calculus. An integral represents the accumulation of quantities and is the inverse process of differentiation. When faced with a complex expression, such as \(\int \csc x(\cot x - \csc x) dx\), the first step is to simplify the problem using algebraic methods before applying integration rules.

For our example, using the distributive property effectively breaks down the integral into simpler expressions. This process allows us to separate the indefinite integral into two parts, which can then be approached individually using basic integral formulas or trigonometric identities. Knowing fundamental antiderivatives, such as \(\int \csc x \cot x dx = -\csc x\) and \(\int \csc^2 x dx = -\cot x\), is crucial for rapid evaluation. Practice with various integrals over time helps in not only memorizing these formulas but also developing intuition for approaching new integrals.

Additionally, it is important to understand the properties of indefinite integrals, such as linearity, which allows the integral of a sum to be split into the sum of integrals. The example problem demonstrates this property clearly. Effective use of these techniques requires thorough practice and a solid grasp of foundational calculus concepts.

Trigonometric Integrals

When working with trigonometric integrals, familiarity with various trigonometric identities and their integrals is key. The original exercise involves integrals of the form \(\int \csc x \cot x dx\) and \(\int \csc^2 x dx\), which are standard forms in trigonometric integration.

For the successful evaluation of trigonometric integrals, it’s essential to remember the basic antiderivatives associated with trigonometric functions. These are often derived from the derivatives of the trigonometric functions—another insight into the deep connection between integration and differentiation in calculus. For instance, knowing that \(\frac{d}{dx}\cot x = -\csc^2 x\) allows us to easily determine that \(\int \csc^2 x dx = -\cot x\).

Common Trigonometric Integrals

• \(\int \sin x dx = -\cos x + C\)
• \(\int \cos x dx = \sin x + C\)
• \(\int \tan x dx = -\ln |\cos x| + C\)
• \(\int \cot x dx = \ln |\sin x| + C\)

These integrals form the backbone of solving more complex problems involving trigonometric functions. Integrating trigonometric functions often involves algebraic manipulation, the use of trigonometric identities, or sometimes applying advanced techniques such as integration by parts and trigonometric substitution.

Calculus Differentiation

The process of calculus differentiation is used to check the correctness of an indefinite integral. Differentiation is the inverse operation of integration, and it provides the rate of change of a function. When we differentiate the result of an indefinite integral, we expect to retrieve the original function being integrated—if we do not, this indicates an error in the integration process.

In the given exercise, the differentiation of the integral result \(\frac{d}{dx}(-\csc x - \cot x + C)\) is performed to confirm accuracy. This step is akin to reverse checking in arithmetic—just as you might use division to check the result of a multiplication problem. The rules of differentiation are applied to \(\csc x\) and \(\cot x\) to give \(\csc x \cot x - \csc^2 x\), demonstrating that the integral operation was conducted correctly.

Key Points in Calculus Differentiation:

• Differentiate term-by-term using known derivatives of functions.
• Chain rule, product rule, and quotient rule are essential in differentiating complex functions.
• The derivative of a constant is zero, which is why the integration constant \(C\) disappears upon differentiation.

The practice of checking your integration by differentiation is an excellent habit that reinforces understanding of both integral and derivative calculus concepts, providing a full circle view of the fundamental theorem of calculus.