Problem 29
Question
Use integration by parts to prove the reduction formula. \(\int(\ln x)^{n} d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} d x\)
Step-by-Step Solution
Verified Answer
Use integration by parts with \( u = (\ln x)^n \) and \( dv = dx \) to derive the reduction formula.
1Step 1: Identify Parts for Integration by Parts
In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). For the given integral \( \int (\ln x)^n \, dx \), let \( u = (\ln x)^n \) and \( dv = dx \).
2Step 2: Differentiate and Integrate Parts
Differentiate \( u \) to get \( du = n (\ln x)^{n-1} \frac{1}{x} \, dx \), and integrate \( dv \) to get \( v = x \).
3Step 3: Apply the Integration by Parts Formula
Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \[ \int (\ln x)^n \, dx = x(\ln x)^n - \int x \left(n (\ln x)^{n-1} \frac{1}{x} \right) \, dx \].
4Step 4: Simplify the Integral
Simplify the expression inside the integral: \[ \int x \left(n (\ln x)^{n-1} \frac{1}{x} \right) \, dx = n \int (\ln x)^{n-1} \, dx \].
5Step 5: Write the Reduction Formula
Substitute the simplified integral back into the expression from Step 3 to get the reduction formula: \[ \int (\ln x)^n \, dx = x(\ln x)^n - n \int (\ln x)^{n-1} \, dx \], which matches the given formula in the problem.
Key Concepts
Integration by PartsReduction FormulaCalculus Problem Solving
Integration by Parts
Integration by Parts is a powerful technique in calculus used to find the integral of products of functions. It is especially useful when one of the functions becomes simpler through differentiation. With the mnemonic "ILATE" (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), we choose which function to differentiate (\(u\)) and which to integrate (\(dv\)).
For the integral \(\int (\ln x)^n \, dx\), we select \(u = (\ln x)^n\) because logarithms are a priority, and \(dv = dx\). Differentiating \(u\) gives us \(du = n (\ln x)^{n-1} \frac{1}{x} \, dx\), and integrating \(dv\) gives \(v = x\).
By applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we express the original integral in terms of a simpler one. This process involves substituting back part of the original integral, which is a common theme in integration techniques.
For the integral \(\int (\ln x)^n \, dx\), we select \(u = (\ln x)^n\) because logarithms are a priority, and \(dv = dx\). Differentiating \(u\) gives us \(du = n (\ln x)^{n-1} \frac{1}{x} \, dx\), and integrating \(dv\) gives \(v = x\).
By applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we express the original integral in terms of a simpler one. This process involves substituting back part of the original integral, which is a common theme in integration techniques.
Reduction Formula
A Reduction Formula is a mathematical expression that connects an integral in terms of a higher power to an integral with a lower power. It allows us to handle intricate integrals by breaking them down recursively into simpler parts.
In our problem, we show how the integral \(\int(\ln x)^n \, dx\) reduces to \(x(\ln x)^n - n \int(\ln x)^{n-1} \, dx\). Each step involves letting an integral with a higher power relate to one with a lower power, minimizing complexity.
This reduction method simplifies the integration process by transforming a potentially challenging logarithmic integration problem into a recursion of easier problems, using integration by parts as the engine.
Reduction formulas are immensely beneficial in calculus because they enable systematic problem-solving by tactically simplifying the original problem.
In our problem, we show how the integral \(\int(\ln x)^n \, dx\) reduces to \(x(\ln x)^n - n \int(\ln x)^{n-1} \, dx\). Each step involves letting an integral with a higher power relate to one with a lower power, minimizing complexity.
This reduction method simplifies the integration process by transforming a potentially challenging logarithmic integration problem into a recursion of easier problems, using integration by parts as the engine.
Reduction formulas are immensely beneficial in calculus because they enable systematic problem-solving by tactically simplifying the original problem.
Calculus Problem Solving
Calculus Problem Solving often requires a strategic combination of techniques, such as Integration by Parts or using a Reduction Formula. When solving calculus problems, like the one given, a structured approach is crucial.
The initial step involves identifying the components of the problem and deciding which technique to apply. For complex integrals, breaking down the problem—like rewriting or simplifying expressions—can aid in managing the solution process.
Effective calculus problem-solving is like puzzle-solving. Tools include:
The initial step involves identifying the components of the problem and deciding which technique to apply. For complex integrals, breaking down the problem—like rewriting or simplifying expressions—can aid in managing the solution process.
Effective calculus problem-solving is like puzzle-solving. Tools include:
- Understanding the relationships between functions.
- Identifying patterns that can lead to the simplification of calculations.
- Utilizing foundational calculus rules.
Other exercises in this chapter
Problem 29
Evaluate the indefinite integral. \(\int \sec ^{3} x \tan x d x\)
View solution Problem 29
29-34 Evaluate the integral by interpreting it in terms of areas. $$\int_{0}^{3}\left(\frac{1}{2} x-1\right) d x$$
View solution Problem 30
If \(\int_{-\infty}^{\infty} f(x) d x\) is convergent and \(a\) and \(b\) are real numbers, show that $$\int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x=\
View solution Problem 30
Evaluate the indefinite integral. \(\int x^{2} \sqrt{2+x} d x\)
View solution