AOS4 Topic 8: Integration by Parts
Definition:
Integration by parts is a technique used in calculus to evaluate the integral of a product of two functions. It is derived from the product rule for differentiation and is the counterpart of the product rule for differentiation.
The formula for integration by parts is given by:
\[ \int u \, dv = uv - \int v \, du \]
where \( u \) and \( dv \) are differentiable functions of \( x \), and \( du \) and \( v \) are their respective differentials. The goal of integration by parts is to choose \( u \) and \( dv \) in such a way that the integral on the right-hand side of the equation is easier to evaluate than the original integral.
Note: We can use integration by parts to find an integral \( \int u \, \frac{dv}{dx} dx \) if the integral \( \int v \, \frac{du}{dx} dx \) is easier to find.
Deriving the Integration by Parts Formula:
Understanding how to differentiate a product is crucial. If we have \( y = uv \), the derivative \( \frac{dy}{dx} \) can be expressed as \( u \frac{dv}{dx} + v \frac{du}{dx} \). Rearranging this yields \( u \frac{dv}{dx} = \frac{d(uv)}{dx} - v \frac{du}{dx} \). By integrating both sides, we obtain \( \int u \frac{dv}{dx} \, dx = \int \frac{d(uv)}{dx} \, dx - \int v \frac{du}{dx} \, dx \). Simplifying the first term on the right side results in \( \int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx \).
ILATE rule:
We have learned that in integration by parts, we must use the necessary formula when we are given the product of two functions. The left term is regarded as the first function and the second term as the second function when calculating the integral of the two functions. We call this technique the ILATE rule. For instance, when integrating \(x \cdot e^x\), we designate \(x\) as the first function and \(e^x\) as the second function, aligning with the ILATE Rule. This approach ensures that the chosen first function possesses a derivative that can be easily integrated.
Techniques for integration by parts
Using integration by parts more than once:
"Repetitively using integration by parts in complex integration scenarios helps untangle complex integrals, exposing their underlying structure and making evaluation easier."
Using integration by parts by solving for the unknown integral:
"When an elusive integral is encountered, using integration by parts to represent it in terms of itself frequently reveals a way to resolve it, turning ambiguity into clarity."
Using reduction formulas:
"By simplifying difficult integrals into simpler forms through the use of reduction formulae in integration, one can expedite the process and find answers that previously seemed unattainable."
Finding recursive formula:
Find a recursive formula for \( \int x^n e^x \, dx \), where \( n \in \mathbb{N} \).
Solution:
a) For each \( n \in \mathbb{N} \cup \{0\} \), define \( I_n = \int x^n e^x \, dx \).
Now let \( n \in \mathbb{N} \). Using integration by parts, we find:
- \( I_n = \int x^n e^x \, dx \)
- \( = \int x^n \frac{d}{dx}(e^x) \, dx \)
- \( = x^n e^x - \int e^x \frac{d}{dx}(x^n) \, dx \)
- \( = x^n e^x - \int e^x n x^{n-1} \, dx \)
- \( = x^n e^x - n \int x^{n-1} e^x \, dx \)
- \( = x^n e^x - nI_{n-1} \)
We have shown that \( I_n = x^n e^x - nI_{n-1} \) for all \( n \in \mathbb{N} \).
Using integration by parts for definite integrals:
"Integration by parts provides a methodical way to compute areas bounded by curves when dealing with definite integrals. It does this by utilizing the interaction between functions to extract exact answers quickly and accurately."
We can also use integration by parts to evaluate definite integrals:
\[ \int_{a}^{b} u \frac{dv}{dx} \, dx = uv \bigg|_{a}^{b} - \int_{a}^{b} v \frac{du}{dx} \, dx \]
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Example 1
Example 2
Example 3
Example 4
Example 5
\[ \begin{align*} \int \sin(2x)\cos(3x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) - 2\int \cos(2x)\cos(3x) \, dx \\ & = \frac{1}{3}\sin(2x)\sin(3x) - \frac{2}{3}\int \cos(2x)\sin(3x) \, dx \end{align*} \]
\[ \begin{align*} \int \cos(2x)\sin(3x) \, dx & = -\frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \\ & = -\frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \end{align*} \]
\[ \begin{align*} \int \sin(2x)\cos(3x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) - \frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \\ & - \frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \end{align*} \]
\[ \begin{align*} \int \sin(2x)\cos(2x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) + \frac{2}{9}\cos(2x)\cos(3x) + \frac{4}{9}\int \sin(2x)\cos(3x) \, dx \end{align*} \]
\[ \begin{align*} \frac{5}{9}\int \sin(2x)\cos(2x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) + \frac{2}{9}\cos(2x)\cos(3x) \end{align*} \]
Therefore, \[ \begin{align*} \int \sin(2x)\cos(2x) \, dx & = \frac{3}{5}\sin(2x)\sin(3x) + \frac{2}{5}\cos(2x)\cos(3x) + C, \quad C \in \mathbb{R} \end{align*} \]
Example 6
Example 7
Example 8
\[ \int x^3 e^x dx = I_3 \quad \text{(by the definition of } I_3 \text{)} \] \[ = x^3 e^x - 3I_2 \quad \text{(expressing } I_3 \text{ in terms of } I_2 \text{)} \] \[ = x^3 e^x - 3x^2 e^x - 2I_1 \quad \text{(expressing } I_2 \text{ in terms of } I_1 \text{)} \] \[ = x^3 e^x - 3x^2 e^x + 6I_1 \]
\[ = x^3 e^x - 3x^2 e^x + 6x e^x - 6I_0 \quad \text{(expressing } I_1 \text{ in terms of } I_0 \text{)} \] \[ = x^3 e^x - 3x^2 e^x + 6x e^x - 6 \int e^x dx \quad \text{(by the definition of } I_0 \text{)} \] \[ = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + c \] \[ = (x^3 - 3x^2 + 6x - 6)e^x + c \]
Example 9
Using integration by parts for definite integrals
Evaluate \( \int_{1}^{2} \ln(x) \, dx \)
Example 10
\[ \pi \int_{0}^{\pi} f(x) \, dx = -16\pi - 76\sqrt{2} - 12 \]= 6.197051864910682