Problem 74 - Thomas Calculus

Question

Evaluate each integral in Exercises \(71-82\) by using any technique you think is appropriate. $$ \int\left(1+\frac{1}{x}\right) \cot (x+\ln x) d x $$

Step-by-Step Solution

Answer

\( \ln |\sin(x+\ln x)| + C \)

Step 1: Simplify the Integrand

First, let's understand the structure of the integrand: \(\left(1+\frac{1}{x}\right) \cot (x+\ln x)\). Notice that \(1+\frac{1}{x}\) can be related to the derivative of \(x + \ln x\), as the derivative of \(x+\ln x\) is \(1+\frac{1}{x}\). This suggests substitution may simplify our problem.

Step 2: Use Substitution Method

Let \( u = x + \ln x \). Then the differential \( du = \left(1+\frac{1}{x}\right)dx \). This substitution fits perfectly because our integrand includes \( \left(1+\frac{1}{x}\right) \). So, the integral becomes \( \int \cot u \, du \).

Step 3: Integrate \( \cot u \)

Recall the integral of \( \cot u \), which is \( \int \cot u \, du = \ln |\sin u| + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back

Replace \( u \) back with \( x+\ln x \) to return to the original variable. Therefore, \( \int \left(1+\frac{1}{x}\right) \cot(x+\ln x) \, dx = \ln |\sin(x+\ln x)| + C \).

Key Concepts

Substitution Method

The Substitution Method is a powerful tool in calculus, especially useful in integration. It's like a strategic shortcut for complex problems. Essentially, you replace a part of the integral with a new variable that simplifies the expression.

Here's the basic process:

• Identify a part of the integrand that can be replaced or transformed into a simpler form.
• Choose a substitution variable, often denoted as 'u', and express it in terms of the original variable (like 'x').
• Calculate the differential of the substitution variable to match the differential in the integral.

In the exercise, by letting \( u = x + \ln x \), the integral simplifies beautifully. Recognizing this was key to making the problem easier to solve. Using substitution turns the original daunting expression into a much simpler one that's straightforward to integrate.

Trigonometric Integrals

Trigonometric Integrals are integrals that involve trigonometric functions like sine, cosine, and cotangent. Solving these requires understanding trigonometric identities and properties.

In this exercise, the integral involved \( \cot u \). The cotangent function is the reciprocal of the tangent function, and its integral is one of the standard results you should remember:

• The integral of \( \cot u \) is \( \ln |\sin u| + C \). The '+ C' here is the constant of integration, which we always include in indefinite integrals.

Recognizing trigonometric forms and remembering their integrals can save a lot of time and effort in solving such problems. Trigonometric integrals are common in calculus, so being comfortable with them is important.

Integral Calculus

Integral Calculus is a branch of calculus focused on the concept of the integral. It is essentially the reverse process of differentiation, allowing us to find areas, volumes, and other concepts under curves. When you integrate a function, you're summing up infinitesimal differences to find the total amount something changes.

In our original problem, we integrated a complex function by simplifying it first. This highlights the crucial skill in Integral Calculus: recognizing when and how to simplify the function before integrating. Approaches like substitution and recognizing known integral forms are key techniques used for simplifying expressions.

Integral calculus is not just an academic exercise; it has real-world applications – from physics to engineering, it's used to calculate quantities like work, charge density, or even the mass of an object with variable density.

Integration Techniques

There are a variety of Integration Techniques available to tackle different kinds of integrals. Selecting the right one often depends on the specific form and structure of the function you're integrating. Here's a quick overview of some common techniques:

• Substitution: Simplifies the integral by changing variables. Perfect for cases like our exercise where recognition of a derivative led to simplification.
• Integration by Parts: Useful when the integrand is a product of two functions.
• Partial Fractions: Great for rational functions, dividing them into simpler parts before integrating.
• Trigonometric Identities: Helpful when dealing with complex trigonometric functions.

In our problem, Substitution was the best choice due to the presence of a derivative inside the integral. Factors like these guide the selection of a technique, turning a challenging integral into a manageable one. Mastery of these methods will boost your problem-solving toolkit in calculus.