# Integration by Parts: Key Concepts and Powerful Tricks

In the vast world of calculus, integration stands as a pillar, pivotal in solving a plethora of mathematical problems. Among its various techniques, integration by parts emerges as a particularly valuable tool, offering a unique approach to tackle integrals that might initially seem daunting. Drawing inspiration from the product rule of differentiation, integration by parts elegantly decomposes complicated expressions into simpler parts. This method not only unveils the hidden structures within certain integrals but also bridges the gap between differentiation and integration. As we delve deeper into this topic, we'll unravel the intricacies of integration by parts, exploring its foundation, application, and significance in mathematical computations.

Topics we will cover

## The Foundation: Product Rule for Differentiation

1. Statement of the Product Rule

The product rule in differentiation states that for two differentiable functions $$u(x)$$ and $$v(x)$$, the derivative of their product with respect to $$x$$ is given by:
$\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)$
where $$u'(x)$$ and $$v'(x)$$ are the derivatives of $$u$$ and $$v$$ with respect to $$x$$, respectively.

2. Brief Proof of the Product Rule

Consider the increment in the product $$u \cdot v$$ when $$u$$ and $$v$$ are incremented by small amounts $$\Delta u$$ and $$\Delta v$$ respectively:
$(u + \Delta u)(v + \Delta v) - u \cdot v = u \Delta v + v \Delta u + \Delta u \Delta v$
Dividing each side by $$\Delta x$$ and taking the limit as $$\Delta x$$ approaches zero, we get:
$\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = u'(x) \quad \text{and} \quad \lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x} = v'(x)$

Thus, the product rule is established:

$\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

3. Significance in Leading to the Formula for Integration by Parts

The product rule for differentiation is foundational for the formula of integration by parts. Integrating both sides of the product rule equation, we obtain:

$\int u'(x) \cdot v(x) \, dx + \int u(x) \cdot v'(x) \, dx = u(x) \cdot v(x)$

Rearranging this expression, we derive the standard formula for integration by parts:

$\int u \, dv = uv - \int v \, du$

Examples:

Example 1: Given $$u(x) = x$$ and $$v(x) = e^x$$. Find the derivative $$\frac{d}{dx} [x \cdot e^x]$$ using the product rule.
Using the product rule: $\frac{d}{dx} [x \cdot e^x] = 1 \cdot e^x + x \cdot e^x = e^x (x + 1)$
Example 2: For $$u(x) = \sin(x)$$ and $$v(x) = x^2$$. Compute $$\frac{d}{dx} [\sin(x) \cdot x^2]$$ with the product rule.
By the product rule: $\frac{d}{dx} [\sin(x) \cdot x^2] = \cos(x) \cdot x^2 + \sin(x) \cdot 2x = x^2 \cos(x) + 2x \sin(x)$

## The Integration by Parts Formula

Statement of the Theorem:

The method of integration by parts is derived from the product rule of differentiation. If $$u$$ and $$v$$ are differentiable functions of $$x$$, then the formula for integration by parts is given by:
$\int u \, dv = uv - \int v \, du$

Where:

- $$u$$ is the function we choose to differentiate.
- $$dv$$ is the function we choose to integrate.
The resulting terms after differentiation and integration are $$du$$ and $$v$$ respectively.

Understanding the Formula:

The formula for integration by parts is reminiscent of the product rule for differentiation but in the integral form. It provides a method to handle integrals that involve products of functions, especially when the integral of the product is challenging or impossible to find directly. By splitting the functions into $$u$$ and $$dv$$, we reduce the complexity of the problem by handling smaller, more straightforward integrals.

Application:

To apply the integration by parts formula:

1. Identify the functions $$u$$ and $$dv$$ in the integral.
2. Differentiate $$u$$ to get $$du$$.
3. Integrate $$dv$$ to get $$v$$.
4. Apply the formula $$\int u \, dv = uv - \int v \, du$$.

NOTE:

Choosing $$u$$ and $$dv$$ appropriately is crucial for the simplification of the problem. A common mnemonic used to decide which function to differentiate and which one to integrate is "LIATE", which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions, in that order of preference for $$u$$.

## Proof of the Integration by Parts Formula

Starting from the product rule of differentiation

Recall the product rule for differentiation: If $$u(x)$$ and $$v(x)$$ are both differentiable functions of $$x$$, then the derivative of their product is given by:
$\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Where:

- $$u'(x)$$ is the derivative of $$u$$ with respect to $$x$$.
- $$v'(x)$$ is the derivative of $$v$$ with respect to $$x$$.

Deriving the Formula:

Integrate both sides of the above equation with respect to $$x$$:
$\int \frac{d}{dx} [u(x) \cdot v(x)] \, dx = \int [u'(x) \cdot v(x) + u(x) \cdot v'(x)] \, dx$

Given that the integral of a derivative is just the original function (Fundamental Theorem of Calculus), the left side becomes:

$u(x) \cdot v(x) = \int u'(x) \cdot v(x) \, dx + \int u(x) \cdot v'(x) \, dx$

Rearrange this to isolate one of the integrals:

$\int u(x) \cdot v'(x) \, dx = u(x) \cdot v(x) - \int u'(x) \cdot v(x) \, dx$
If we let $$dv = v'(x) \, dx$$ and $$du = u'(x) \, dx$$, this formula takes the familiar form:
$\int u \, dv = uv - \int v \, du$

Examples:

1. Example with $$e^x$$ and $$\sin(x)$$:

Given the integral:

$\int e^x \sin(x) \, dx$

We can choose:

$$u = e^x$$ (which gives $$du = e^x \, dx$$) and $$dv = \sin(x) \, dx$$ (which integrates to $$v = -\cos(x)$$).

Using the formula:

$\int e^x \sin(x) \, dx = -e^x \cos(x) - \int -e^x \cos(x) \, dx$
And then you can further integrate the term $$-e^x \cos(x)$$.
This is just one of many examples, and the method can be applied to a variety of functions. The key lies in aptly choosing $$u$$ and $$dv$$ for simplification.

## What is the ILATE Rule?

Choosing $$u$$ and $$dv$$: The ILATE Rule
When performing integration by parts, the choice of $$u$$ and $$dv$$ is crucial for simplifying the integral. The ILATE rule provides a guideline on which function to pick as $$u$$. The rule stands for:
• I - Inverse trigonometric functions
• L - Logarithmic functions
• A - Algebraic functions (like polynomials)
• T - Trigonometric functions
• E - Exponential functions
The rule suggests the order of preference for choosing $$u$$ when multiple functions are present in the integrand.

Rationale Behind the ILATE Rule

The reason for this order is to simplify the resulting integral. Typically, differentiating a function from the start of the list and integrating one from the end of the list tends to make the integral simpler or at least no more complicated than the original.

Examples Using the ILATE Rule

Example 1: Given the integral:

$\int x \sin(x) \, dx$
Here, we have an algebraic function $$x$$ and a trigonometric function $$\sin(x)$$. According to ILATE, we choose the algebraic function as $$u$$.

Choosing:

$$u = x$$ (gives $$du = dx$$) and $$dv = \sin(x) \, dx$$ (which integrates to $$v = -\cos(x)$$)

Using the integration by parts formula:

$\int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx = -x \cos(x) + \sin(x) + C$

Example 2: Given the integral:

$\int e^x \ln(x) \, dx$
Here, we have an exponential function $$e^x$$ and a logarithmic function $$\ln(x)$$. According to ILATE, we choose the logarithmic function as $$u$$.

Choosing:

$u = \ln(x)$ (this gives $$du = \frac{1}{x} dx$$) $dv = e^x dx$ (which integrates to $$v = e^x$$)

Using the integration by parts formula, we get:

$\int e^x \ln(x) \, dx = e^x \ln(x) - \int e^x \cdot \frac{1}{x} dx$
Now, the new integral $$\int e^x \cdot \frac{1}{x} dx$$ may still require special techniques to solve, or it could be expressed in terms of special functions. It's used here to illustrate the process and show that sometimes integration by parts may need to be applied more than once or in combination with other techniques.

It's essential to understand that the ILATE rule is a guideline. In some contexts, another choice might lead to a simpler result or might be more suitable based on the subsequent steps or techniques being applied.

## Simple examples to introduce the technique

Here are some examples to provide a comprehensive understanding of the integration by parts method:

1. Integration of $$x \sin(x)$$

Consider the integral:

$\int x \sin(x) \, dx$

To solve this using integration by parts, we recall our formula:

$\int u \, dv = uv - \int v \, du$

Choosing:

$$u = x$$ (which gives $$du = dx$$) \\ $$dv = \sin(x) \, dx$$ (which integrates to $$v = -\cos(x)$$)

Substituting these values into our formula, we get:

$\int x \sin(x) \, dx = -x \cos(x) - \int (-\cos(x)) \, dx$ $= -x \cos(x) + \int \cos(x) \, dx$ $= -x \cos(x) + \sin(x) + C$ Where $$C$$ is the constant of integration.
2. Integration of $$x e^{-x}$$

Consider the integral:

$\int x e^{-x} \, dx$

Using the integration by parts formula:

Choosing:

$$u = x$$ (which gives $$du = dx$$) \\ $$dv = e^{-x} \, dx$$ (which integrates to $$v = -e^{-x}$$)

Substituting these values in:

$\int x e^{-x} \, dx = -x e^{-x} - \int (-e^{-x}) \, dx$ $= -x e^{-x} - e^{-x} + C$ Where $$C$$ is the constant of integration.

## Integration by Parts with Definite Integrals

Handling the Limits of Integration in Integration by Parts

When we're integrating a function over a specific interval, i.e., definite integration, the procedure with integration by parts remains similar, but we must be careful to apply the limits of integration at each step.

Formula with Limits

The integration by parts formula for definite integration over the interval $$[a, b]$$ is: $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$ Where $$[uv]_a^b$$ denotes evaluating the product $$uv$$ at $$b$$ and subtracting its value at $$a$$.
Example: Integration of $$x \sin(x)$$ from 0 to $$\pi$$

Consider the integral:

$\int_0^{\pi} x \sin(x) \, dx$

Solution:

Choose:

$$u = x$$ (so $$du = dx$$) \\ $$dv = \sin(x) \, dx$$ (which gives $$v = -\cos(x)$$)

Applying our formula, we get:

$\int_0^{\pi} x \sin(x) \, dx = [ -x \cos(x) ]_0^{\pi} + \int_0^{\pi} \cos(x) \, dx$

Evaluating the expressions at the limits:

$[-\pi \cos(\pi) + 0 \cos(0)] + [\sin(x)]_0^{\pi}$ $= \pi + 0 = \pi$

## Practice Problems on Integration by Parts

1. Compute the integral:

$\int x \ln(x) \, dx$

Solution:

$$u = \ln(x)$$ leading to $$du = \frac{1}{x} \, dx$$ \\ $$dv = x \, dx$$ which gives $$v = \frac{x^2}{2}$$

Using the formula:

$\int x \ln(x) \, dx = x^2 \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = x^2 \ln(x) - \frac{x^2}{2} + C$

2. Evaluate:

$\int x e^x \, dx$

Solution:

Choose:

$$u = x$$ leading to $$du = dx$$ \\ $$dv = e^x \, dx$$ which gives $$v = e^x$$

Using the formula:

$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C$

3. Find the value of:

$\int_0^{\pi} x \cos(x) \, dx$

Solution:

Choose:

$$u = x$$ leading to $$du = dx$$ \\ $$dv = \cos(x) \, dx$$ which gives $$v = \sin(x)$$

Using the formula, we get:

$\int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx$ $= x \sin(x) + \cos(x)$
Now, evaluate between 0 and $$\pi$$:
$[\pi \sin(\pi) + \cos(\pi)] - [0 \sin(0) + \cos(0)] = -1 - 1 = -2$

4. Calculate:

$\int e^x \sin(x) \, dx$

Solution:

Choose:

$$u = e^x$$ leading to $$du = e^x \, dx$$ \\ $$dv = \sin(x) \, dx$$ which gives $$v = -\cos(x)$$

Using the formula:

$\int e^x \sin(x) \, dx = -e^x \cos(x) - \int (-e^x \cos(x)) \, dx$

The resulting integral will also require integration by parts. The final answer, after some simplification, is:

$\frac{e^x (\sin(x) - \cos(x))}{2} + C$

5. Evaluate the integral:

$\int \ln(x) \cos(x) \, dx$

Solution:

This one is a bit tricky and can be done in multiple ways. One common way is:

Choose:

$$u = \ln(x)$$ leading to $$du = \frac{1}{x} \, dx$$ \\ $$dv = \cos(x) \, dx$$ which gives $$v = \sin(x)$$

Using the formula, we get a new integral which can be solved using the integration by parts once more. The final result is:

$x \ln(x) \sin(x) + \text{Si}(x) + C$

where Si is the sine integral.

6. Find the value of:

$\int_0^1 x^2 e^{x^3} \, dx$

Solution:

Let:

$u = x^3 \Rightarrow du = 3x^2 \, dx$

Make a substitution:

$\int_0^1 x^2 e^{x^3} \, dx = \frac{1}{3} \int_0^1 e^u \, du = \frac{1}{3} [e^u]_0^1 = \frac{1}{3} (e - 1)$

## Conclusion

Integration by parts is a powerful technique in calculus that provides a method to tackle integrals that at first may seem complex or non-straightforward. Rooted in the product rule of differentiation, this method requires a strategic choice of parts from the integrand to simplify the problem. Through the examples and problems provided, it's evident that practice is key. As with many concepts in calculus, familiarity and experience will enhance one's ability to quickly identify the best strategy for a given integral.

For those looking to deepen their understanding, it's essential to practice consistently, challenge oneself with advanced problems, and engage with various resources to see different problem-solving approaches.

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