Problem 73 - Applied Calculus

Question

Find the indefinite integrals. $$\int\left(\frac{2}{x}+\pi \sin x\right) d x$$

Step-by-Step Solution

Answer

\( 2 \ln |x| - \pi \cos x + C \)

Step 1: Identify the Integral Components

The given integral is \( \int \left( \frac{2}{x} + \pi \sin x \right) \, dx \). This integral can be split into two separate integrals: \( \int \frac{2}{x} \, dx \) and \( \int \pi \sin x \, dx \). We will solve each part individually.

Step 2: Solve the First Integral

Focus on \( \int \frac{2}{x} \, dx \). This is a standard integral form of \( \int \frac{a}{x} \, dx = a \ln |x| + C \). Thus, the integral of \( \frac{2}{x} \) is \( 2 \ln |x| + C_1 \).

Step 3: Solve the Second Integral

Now, solve \( \int \pi \sin x \, dx \). The constant \( \pi \) can be factored out: \( \pi \int \sin x \, dx \). The integral of \( \sin x \) is \(- \cos x + C \), so \( \pi \int \sin x \, dx = -\pi \cos x + C_2 \).

Step 4: Combine Results

Combine the results from the two integrals. We have \( 2 \ln |x| + C_1 \) from the first integral and \(-\pi \cos x + C_2 \) from the second integral. Thus, the indefinite integral is \( 2 \ln |x| - \pi \cos x + C \), where \(C\) is the combined constant of integration.

Key Concepts

Integration Techniques

Understanding integration techniques is fundamental for solving indefinite integrals. To integrate a composite expression, like the one given in the exercise, we often separate it into simpler parts. This approach allows us to handle each part individually using known integral formulas.

For example, the given integral \( \int \left( \frac{2}{x} + \pi \sin x \right) \, dx \) can be split into \( \int \frac{2}{x} \, dx \) and \( \int \pi \sin x \, dx \).

Each of these components can be tackled separately using specific techniques:

• For \( \int \frac{2}{x} \, dx \), utilize the formula for logarithmic integrals related to \( \int \frac{1}{x} \, dx \).
• For \( \int \pi \sin x \, dx \), use knowledge of trigonometric integrals to solve it easily.

Mastering these techniques involves recognizing which formula or method fits the integral at hand. Keep in mind that constants, like \( \pi \), can be factored out, simplifying the integration process.

Learning to decompose mixed integrals is a practical integration technique that enhances computational efficiency and understanding.

Trigonometric Integrals

Trigonometric integrals involve expressions that include trigonometric functions such as sine and cosine. In this exercise, we encounter a trigonometric integral part: \( \int \pi \sin x \, dx \).

To manage this, note these simple steps:

• Recognize that \( \int \sin x \, dx \) equates to \( -\cos x + C \).
• Factor out constants like \( \pi \) from the integral, resulting in \( \pi \int \sin x \, dx \).

Solving this integral requires applying the rule for \( \int \sin x \, dx \), yielding the result of \( -\pi \cos x + C \).

These functions are cyclical, and their integrals reflect this by involving other trigonometric functions. Understanding the basic integral forms of sine, cosine, and their derivatives is essential.

Integrals like these are ubiquitous in various calculus problems and useful in fields such as physics, where wave functions and oscillations play significant roles.

Logarithmic Integrals

Logarithmic integrals are those integral expressions that lead to logarithm functions upon integration. In our given problem, one part involves a logarithmic integral: \( \int \frac{2}{x} \, dx \).

The formula \( \int \frac{1}{x} \, dx = \ln |x| + C \) is a crucial tool here. This expresses how the integral of \( \frac{1}{x} \) translates directly into the natural logarithm function. The coefficient \( 2 \) in \( \frac{2}{x} \) means you multiply the logarithmic result by 2.

Here's the calculation step-by-step:

• Identify the integral form \( \int \frac{2}{x} \, dx \).
• Use the basic formula and include the constant multiplier, resulting in \( 2 \ln |x| + C \).

Logarithmic integrals are foundational in calculus and often appear in problems dealing with growth and decay, such as problems in biology and economics.

Understanding how to transform simple rational expressions into logarithmic results makes tackling such integrals much simpler and straightforward.