Problem 15 - Thomas Calculus in SI Units

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\csc x \cot x\) b. \(-\csc 5 x \cot 5 x\) c. \(-\pi \csc \frac{\pi x}{2} \cot \frac{\pi x}{2}\)

Step-by-Step Solution

Answer

Antiderivatives are: (a) \( -\csc x + C \), (b) \( -\frac{1}{5}\csc 5x + C \), and (c) \( -2\csc \frac{\pi x}{2} + C \).

Step 1: Understand the Problem

We need to find the antiderivative (integral) of each function given. The antiderivative, or indefinite integral, of a function is a function whose derivative is the given function. This means we need to identify a function whose derivative results in the given expressions.

Step 2: Antiderivative for (a): \( \csc x \cot x \)

The antiderivative of \( \csc x \cot x \) is \( -\csc x + C \), where \( C \) is the constant of integration. This is because the derivative of \( -\csc x \) is \( \csc x \cot x \).

Step 3: Antiderivative for (b): \( -\csc 5x \cot 5x \)

To find this integral, use the substitution method. Let \( u = 5x \), then \( du = 5 dx \). The integral becomes \( \int -\csc u \cot u \frac{du}{5} = \int -\frac{1}{5}\csc u \cot u \,du \). Antidifferentiating gives \( -\frac{1}{5}\csc u + C = -\frac{1}{5}\csc 5x + C \).

Step 4: Antiderivative for (c): \( -\pi \csc \frac{\pi x}{2} \cot \frac{\pi x}{2} \)

Similarly, use the substitution method. Let \( v = \frac{\pi x}{2} \), then \( dv = \frac{\pi}{2} dx \), which gives \( dx = \frac{2}{\pi} dv \). The integral turns into \( \int -\pi \csc v \cot v \frac{2}{\pi} \, dv = \int -2 \csc v \cot v \, dv \). Antidifferentiating results in \( -2 \csc v + C = -2 \csc \frac{\pi x}{2} + C \).

Step 5: Verification through Differentiation

Differentiate each antiderivative to confirm correctness. For (a), differentiating \( -\csc x \) returns \( \csc x \cot x \), as needed. For (b), differentiating \(-\frac{1}{5}\csc 5x \) using the chain rule returns \(-\csc 5x \cot 5x \). For (c), differentiating \( -2 \csc \frac{\pi x}{2} \) using the chain rule gives \( -\pi \csc \frac{\pi x}{2} \cot \frac{\pi x}{2} \), confirming the correctness for all.

Key Concepts

Differentiation

Differentiation is the process of finding the derivative of a function. It tells us the rate at which a quantity changes with respect to another quantity. When applied to a function of one variable, it can be interpreted as the slope of the function at any given point.
Important points to remember about differentiation include:

• The derivative of a constant is zero.
• The derivative of a power function, such as \(x^n\), is \(nx^{n-1}\).
• Trigonometric functions like sine or cosine have specific derivatives: \(\sin x\) becomes \(\cos x\), while \(\cos x\) becomes \(-\sin x\).

Differentiation is widely used to verify the correctness of antiderivatives. By differentiating an antiderivative, we should return to the original function. For example, as shown in the solution, differentiating \(-\csc x\) returns \(\csc x \cot x\). This confirms we found the correct antiderivative.

Substitution Method

The substitution method is a useful technique for finding antiderivatives, especially when dealing with more complex expressions. It involves changing variables to simplify an integral.
This is similar to using the chain rule in reverse during differentiation. The process typically follows these basic steps:

• Select a substitution variable, usually denoted by \(u\) or \(v\), that simplifies the integrand.
• Express the differential \(dx\) in terms of the new variable by using the substitution. For instance, if \(u = 5x\), then \(du = 5dx\), or \(dx = \frac{1}{5}du\).
• Rewrite the integral using these new expressions and solve it.
• Finally, substitute back any terms to return to the original variable.

In the solution provided, substitution was used to handle the trigonometric integrals. For example, in antiderivative (b), letting \(u = 5x\) simplifies the integration process, as it allows us to integrate terms like \(\csc u \cot u\) more straightforwardly.

Trigonometric Functions

Trigonometric functions are functions of angles and play a fundamental role in calculus, particularly in the context of differentiation and integration.
Key trigonometric functions include sine, cosine, and tangent, along with their reciprocals cosecant, secant, and cotangent.
Essential properties of these functions in calculus include:

• The cyclic nature of these functions, which means they repeat their values in regular intervals (known as periods).
• Common derivatives: \(\sin x\) becomes \(\cos x\) and \(\cos x\) becomes \(-\sin x\). The derivative of \(\csc x\) is \(-\csc x \cot x\).
• Their integrals are often less intuitive than their derivatives, requiring methods like substitution for integration.

In the solution, trigonometric functions like \(\csc x \cot x\) were used. The antiderivatives were confirmed by differentiating, ensuring each returned to the original function demonstrate the cyclicity and usefulness of these functions in integral calculus.