Problem 14 - Thomas Calculus

Question

In Exercises \(1-16,\) find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\csc ^{2} x \quad\) b. \(-\frac{3}{2} \csc ^{2} \frac{3 x}{2} \quad\) c. \(1-8 \csc ^{2} 2 x\)

Step-by-Step Solution

Answer

Antiderivatives are: a. \(-\cot x + C\), b. \(-\cot \left(\frac{3x}{2}\right) + C\), c. \(x - 4 \cot 2x + C\).

Step 1: Understanding Antiderivative

An antiderivative of a function is a function whose derivative is the original function. The process of finding an antiderivative is essentially the reverse of differentiation.

Step 2: Finding Antiderivative for a. \(\csc^2 x\)

The function \(\csc^2 x\) is the derivative of \(-\cot x\). Therefore, the antiderivative of \(\csc^2 x\) is \(-\cot x + C\), where \(C\) is a constant of integration.

Step 3: Finding Antiderivative for b. \(-\frac{3}{2} \csc^2 \frac{3x}{2}\)

Recognize that \(\csc^2 u\) has the antiderivative \(-\cot u\). Let \(u = \frac{3x}{2}\), then \(du = \frac{3}{2} dx\), and rearranging gives \(dx = \frac{2}{3} du\). Therefore, the antiderivative is:\[-\frac{3}{2} \int \csc^2 u \cdot \frac{2}{3} du = -\int \csc^2 u \, du =-\cot u + C = -\cot\left(\frac{3x}{2}\right) + C.\]

Step 4: Finding Antiderivative for c. \(1 - 8 \csc^2 2x\)

The antiderivative of \(1\) is \(x\). For \(-8 \csc^2 2x\), notice the antiderivative of \(\csc^2 2x\) is \(-\frac{1}{2} \cot 2x\). Therefore, multiply by \(-8\) and apply the chain rule, the result becomes \(-4 \cot 2x\). Thus, the full antiderivative is \(x - 4 \cot 2x + C\).

Step 5: Verifying by Differentiation

Differentiate each antiderivative to ensure the result is the original function. The derivative of \(-\cot x \) is \(\csc^2 x\), for \(- \cot \left(\frac{3x}{2}\right) \) the derivative is \(-\frac{3}{2} \csc^2 \frac{3x}{2}\), and for \(x - 4 \cot 2x\), the derivative is \(1 - 8 \csc^2 2x\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. It is the process of finding the derivative, which represents the gradient or slope of the function's graph at a point. Differentiation can be applied to both algebraic expressions and trigonometric functions. - For example, if you differentiate the function \(f(x) = x^2\), you get \(f'(x) = 2x\), indicating that the slope varies linearly with \(x\). - Differentiation allows us to understand how quantities change relative to each other.In the context of finding antiderivatives, differentiation is used as a verification tool. Once you find an antiderivative, you can differentiate it to see if you obtain the original function.

Chain Rule

The chain rule is an essential technique in differentiation that allows us to differentiate composite functions. A composite function is one where you have a function within another function, like \(h(x) = f(g(x))\).- To apply the chain rule, differentiate the outer function with respect to the inner function, and then multiply the result by the derivative of the inner function.For example, consider the function \((h(x) = \cos(3x)\). To differentiate it, apply the chain rule:- The derivative of the outer function \(\cos(u)\) is \(-\sin(u)\). - Multiply by the derivative of the inner function \(g(x) = 3x\), which is \(3\).- Thus, replacing the derivative gives \(-\sin(3x) imes 3 = -3\sin(3x)\). When finding antiderivatives, you will often use the reverse of the chain rule, as seen in problems involving functions like \(-\frac{3}{2} \csc^2\left(\frac{3x}{2}\right)\).

Constant of Integration

When finding antiderivatives, a common practice is to add a constant of integration, denoted as \(C\). This is because the process of integration, which is the opposite of differentiation, cannot determine the original constant term that was lost during differentiation.- Consider the function \(F(x) = x^2 + C\) as an antiderivative. Differentiating this function will always lead back to \(2x\), regardless of the value of \(C\).- The constant of integration is crucial because it indicates there are infinitely many possible antiderivatives for a given function.In our antiderivatives exercise, each solution involves adding \(C\) at the end, like in the solutions \(-\cot x + C\). It is a reminder that when integrating, the constant term lost during differentiation is unknown.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and cosecant, are periodic functions that are often used in calculus to model oscillatory behavior. They hold special requirements during differentiation and integration.- Important derivatives to remember include: derivative of \(\sin x = \cos x\), and \(\csc^2 x\) is the derivative of \(-\cot x\).- Similarly, integrating \(\csc^2 x\) yields \(-\cot x + C\).When finding antiderivatives, recognizing the basic derivatives and antiderivatives of trigonometric functions is essential. For instance, knowing that \(\csc^2 x\) leads to \(-\cot x\) is used directly in problems like those given in the exercise, allowing for efficient calculation of the antiderivative.Remember, trigonometric functions can transform across multiple steps using identities and rules like the chain rule, greatly impacting the solutions in calculus problems.