Problem 71 - Thomas Calculus

Question

Use reduction formulas to evaluate the integrals. \(\int 16 x^{3}(\ln x)^{2} d x\)

Step-by-Step Solution

Answer

The evaluated integral is \( 4x^4 (\ln x)^2 - 2x^4 \ln x + \frac{1}{2}x^4 + C \).

Step 1: Identify the Integral Structure

The given integral is \( \int 16 x^{3}(\ln x)^{2} \, dx \). This integral features a polynomial term \( x^3 \) and a logarithmic term \((\ln x)^2\). It can be solved using integration by parts with a specific reduction strategy.

Step 2: Apply Integration by Parts Formula

Integration by parts is given by \( \int u \, dv = uv - \int v \, du \). Let's choose \( u = (\ln x)^2 \) and \( dv = 16x^3 \, dx \). Then \( du = 2\ln x \frac{1}{x} \, dx \) and \( v = \frac{16}{4}x^4 = 4x^4 \).

Step 3: Compute the First Application of Parts

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int 16 x^3 (\ln x)^2 \, dx = (\ln x)^2 \cdot 4x^4 - \int 4x^4 \cdot 2\ln x \cdot \frac{1}{x} \, dx. \] Simplify to get:\[ 4x^4 (\ln x)^2 - 8 \int x^3 \ln x \, dx. \]

Step 4: Apply Integration by Parts Again

Now use integration by parts on \( \int x^3 \ln x \, dx \) with \( u = \ln x \) and \( dv = x^3 \, dx \). Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^4}{4} \). Reapply the formula:\[ \int x^3 \ln x \, dx = \ln x \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \frac{1}{x} \, dx \]Simplifies to:\[ \frac{x^4 \ln x}{4} - \frac{1}{4} \int x^3 \, dx. \]

Step 5: Integrate the Remaining Polynomial

Now integrate \( \int x^3 \, dx = \frac{x^4}{4} + C \, \) to get the polynomial integral: \( \frac{x^4}{4} \). Substituting back, we have: \[ \frac{x^4 \ln x}{4} - \frac{x^4}{16}. \]

Step 6: Combine All Parts

Substitute this result back into the integration by parts steps from Step 3:\[ 4x^4 (\ln x)^2 - 8 \left( \frac{x^4 \ln x}{4} - \frac{x^4}{16} \right). \] Simplify to complete the integral:\[ 4x^4 (\ln x)^2 - 2x^4 \ln x + \frac{1}{2}x^4 + C. \]

Key Concepts

Polynomial Integration

Polynomial integration is the process of integrating expressions involving powers of a variable, such as \(x^n\).
It's straightforward because polynomials have predictable patterns.

• For any term like \(x^n\), the integral becomes \(\frac{x^{n+1}}{n+1} + C\).
• The constant \(C\) represents the constant of integration, indicating any constant value could be added to the function.

In the context of our exercise, we had to integrate \(x^3\).
Applying the rule, this became \(\frac{x^4}{4} + C\). Remember, polynomial integration is often applied in the last steps of reduction formulas or integration by parts, after dealing with more complex terms.

Logarithmic Integration

Logarithmic integration involves integrating functions with logarithmic expressions, such as \(\ln x\).
This can be more challenging than integrating polynomials due to the behavior and structure of logarithms.When the integral contains \(\ln x\), we typically use integration by parts.
The natural logarithm \(\ln x\) is particularly suited to integration by parts because its derivative, \(\frac{1}{x}\), is straightforward, and its presence allows for simplification in the integral. In our example, the integral \((\ln x)^2\) required this approach since there's no simple direct formula for integrating \((\ln x)^2\).
Choosing \(u = (\ln x)^2\) simplified the problem by breaking it down through differentiation and substitution. This demonstrates the usefulness of understanding both the properties of logarithmic functions and how to apply integration by parts strategically.

Reduction Formulas

Reduction formulas are formulas used to simplify the integration of complex expressions by reducing them to simpler parts.
They're particularly useful when dealing with integrals involving repeated patterns or products of different types of functions.In principle, a reduction formula expresses an integral in terms of a simpler or related integral.
These formulas are recursively used to break down integrals by applying rules like integration by parts multiple times, just as we did in the step-by-step solution.

• The given integral \(\int 16 x^3(\ln x)^2 \, dx\) was reduced by applying integration by parts twice, which is a common pattern in reduction strategies.
• By organizing integrals into smaller, more manageable components, complex integrals become accessible.

Reduction formulas and strategies are powerful, enabling the tackling of otherwise complex or unsolvable integrals by systematic breakdown and reduction through known simple integrals.