Problem 39 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int\left(-3 \csc ^{2} x\right) d x $$

Step-by-Step Solution

Answer

The indefinite integral is \(3 \cot x + C\).

Step 1: Identify the Antiderivative Rule

To find the antiderivative of a trigonometric function like \(-3 \csc^{2} x\), recall that the derivative of \(-\cot x\) is \-\csc^{2} x\. Therefore, the antiderivative of \(-3 \csc^{2} x\) is \(-3 (-\cot x)\).

Step 2: Calculate the Antiderivative

Apply the antiderivative rule from Step 1 to compute the indefinite integral: \[ \int (-3 \csc^{2} x) \, dx = -3 (-\cot x) + C = 3 \cot x + C \]where \(C\) is the constant of integration.

Step 3: Differentiate Your Answer to Verify

To ensure our antiderivative is correct, differentiate \(3 \cot x + C\). Remember that the derivative of \(\cot x\) is \-\csc^{2} x\. Thus, \[ \frac{d}{dx}(3 \cot x + C) = 3(-\csc^{2} x) + 0 = -3 \csc^{2} x \] This matches the original integrand, confirming our solution is correct.

Key Concepts

Antiderivative

Finding an antiderivative is like running a movie backward; you are retracing the steps of differentiation. The antiderivative of a function provides the original function before differentiation. It's the reverse process of finding derivatives. When tackling the antiderivative, you want to identify which function could have led to the given derivative.In our exercise, we need to find the antiderivative of \(-3 \csc^{2} x\). By recalling that the derivative of \(-\cot x\)is \(-\csc^{2} x\), we establish that the antiderivative of \(-3 \csc^{2} x\) is \(3 \cot x\). This step requires us to apply our knowledge of derivatives to reverse-engineer the original function.

• To check our work, differentiate \(3 \cot x\) to ensure it matches the original function \(-3 \csc^{2} x\).
• Verification like this ensures mathematical accuracy.

After finding the antiderivative, we mustn't forget to add the constant \(C\), since the derivative of a constant is zero and could have existed in the original function.Next, let’s look into specific types of functions, such as those involving trigonometric elements.

Trigonometric Integration

Integration often involves a plethora of functions, and when dealing explicitly with trigonometric ones, it can seem daunting at first. Trigonometric integration focuses on integrating functions that involve trigonometric functions, such as \(\sin x\), \(\cos x\), \(\tan x\), and their reciprocals like \(\csc x\).In the context of our problem, we integrated \(-3 \csc^{2} x\), a function not directly covered by simple trigonometric derivatives like those of \(\sin\) or \(\cos\). This highlights the importance of familiarity with various trigonometric derivatives to deduce the correct antiderivative effectively. Simply put, knowing these derivatives helps in finding their antiderivatives.

• The function \(\csc^{2} x\) is the derivative of \(-\cot x\).
• Thus knowing this helps flip it during integration to find \(\int -\csc^{2} x \, dx = \cot x + C\).

These types of integrations teach valuable lessons in pattern recognition and the power of memorizing common derivative and antiderivative pairs.

Indefinite Integral

An indefinite integral represents the collection of all possible functions whose derivative matches the integrand. Unlike a definite integral, it doesn’t have upper and lower limits, so there isn't a specific value. Instead, it provides a family of functions expressed with a constant \(C\). This arises because differentiation eliminates constants, necessitating their inclusion during integration.The indefinite integral of our function, \(\int (-3 \csc^{2} x) \, dx\), produces \(3 \cot x + C\). This expression indicates that \(3 \cot x\) is a member of all functions that could yield \(-3 \csc^{2} x\) upon differentiation.

• The \(C\) represents any constant term that vanished during differentiation.
• Each different value of \(C\) corresponds to a different function, yet all satisfy the derived condition.

Understanding the concept of an indefinite integral is crucial in Integral Calculus, as it lays the groundwork for mastering definite integrals, where specific bounds and evaluations occur.