Problem 333 - Calculus Volume 1

Question

In the following exercises, find each indefinite integral by using appropriate substitutions. $$\int \frac{\cos x-x \sin x}{x \cos x} d x$$

Step-by-Step Solution

Answer

The indefinite integral is \( \ln |x \cos x| + C \).

Step 1: Identify Potential Substitution

To simplify the integration, we look for a substitution that will make the integral easier to solve. Here, let's consider using the substitution \(u = x \cos x\), as this is the expression in the denominator.

Step 2: Differentiate and Solve for dx

Once we have our substitution \(u = x \cos x\), we differentiate it to find \(\frac{du}{dx}\). Using the product rule, we get:\[ u = x \cos x \Rightarrow \frac{du}{dx} = \cos x - x \sin x \]This means that:\[ du = (\cos x - x \sin x)\,dx \]

Step 3: Substitute in the Integral

Substitute \(u = x \cos x\) and \(du = (\cos x - x \sin x)\,dx\) into the integral:\[ \int \frac{\cos x - x \sin x}{x \cos x} \, dx = \int \frac{1}{u}\, du \]

Step 4: Integrate with Respect to u

The integral of \(\frac{1}{u}\) with respect to \(u\) is \(\ln |u| + C\) (where \(C\) is the constant of integration). Thus:\[ \int \frac{1}{u}\, du = \ln |u| + C \]

Step 5: Back-Substitute

Replace \(u\) with the original expression \(x \cos x\) to express the solution in terms of \(x\). Finally, the integral becomes:\[ \ln |x \cos x| + C \]

Key Concepts

Integration by Substitution

Integration by substitution allows us to simplify integrals by transforming them into a more manageable form. Consider it as a change of variables that simplifies the function we need to integrate. In our example, the integral given is $$\int \frac{\cos x-x \sin x}{x \cos x} dx.$$
To use substitution effectively, we identify parts of the function that seem overly complex—such as the denominator here, which is 'x cos x'. We then choose our substitution:

• Let \(u = x \cos x\)

This substitution aims to simplify the fraction. The process involves differentiating our chosen \(u\) to find \(du\) in terms of \(dx\), allowing us to rewrite the integral in terms of \(u\). This transforms the integral into one that is much simpler to solve:

• \(du = (\cos x - x \sin x) \; dx\)
• Substitute into the integral: \(\int \frac{1}{u} du\)

Product Rule

The product rule is a key concept in calculus used when differentiating products of two functions. The rule states that the derivative of a product \(f(x)g(x)\) is given by:

• \((f \cdot g)' = f'g + fg'\)

In the context of our integral, we use this rule to find the differential \(du = (\cos x - x \sin x) \, dx\). Here's how it works:

• Let \(f(x) = x\), so \(f'(x) = 1\)
• Let \(g(x) = \cos x\), so \(g'(x) = -\sin x\)
• \(\frac{du}{dx} = \cos x - x \sin x \)

By applying the product rule, it becomes straightforward to substitute du in the integration. This turns a potentially tricky derivative into a simple substitution step.

Trigonometric Functions

Trigonometric functions frequently appear in calculus problems, especially integrals, due to their periodic nature and basic derivative and integration rules. In our problem, \(\cos x\) and \(\sin x\) arise as part of the integration and differentiation process. Here’s a refresher on the basic properties used:

• The derivative of \(\cos x\) is \(-\sin x\)
• The integral of \(\sin x\) is \(-\cos x\) plus a constant

By understanding how these functions behave, especially when applying them with rules like substitution and product rule, solving integrals like \(\int \frac{\cos x-x \sin x}{x \cos x} dx\) becomes intuitive. They allow for substitution methods to break complex expressions into simpler forms for easier handling.

Constant of Integration

In calculus, the constant of integration \(C\) matters because when we integrate, we're essentially "undoing" differentiation. Differentiation gives a specific result for any initial condition, and integration adds a new constant that was lost during differentiation. When you find an indefinite integral, such as from our example which concluded with \(\ln |x \cos x| + C\), this \(C\) accounts for all possible vertical shifts of the function. Without it, we would only have one particular solution to an otherwise general problem.
It’s crucial not to forget this constant, as it represents all potential antiderivatives of the function. Remember:

• Every indefinite integral must include \(+ C\)
• It symbolizes an infinite number of potential solutions

Always double-check your work for this constant to ensure completeness in your integration solutions.