Integration by parts

Welcome to Integration by Parts!

In your calculus journey so far, you have learned how to integrate basic functions and how to use substitution. But what happens when you encounter two different types of functions multiplied together, like \( \int x \cos(x) dx \)? Just like we have the Product Rule for differentiation, we have Integration by Parts for integration. Don't worry if it looks intimidating at first; once you learn the "rhythm" of the formula, it becomes much easier!

Think of Integration by Parts as a way to "trade" a difficult integral for a simpler one. By the end of these notes, you will be able to tackle products of algebraic, trigonometric, and exponential functions with confidence.

1. The Formula: Where does it come from?

The Integration by Parts formula is actually just the Product Rule for differentiation written in reverse. If you remember that \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \), then by integrating both sides and rearranging, we get the standard formula:

\( \int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx \)

In some textbooks, you might see it written in a shorter form: \( \int u dv = uv - \int v du \). Both mean exactly the same thing!

Key Terms to Remember:

u: The part of the function you choose to differentiate.
dv/dx: The part of the function you choose to integrate.

Key Takeaway: Integration by parts is used when you have a product of two functions that cannot be easily integrated using simple methods or substitution.

2. How to Choose 'u' (The LATE Rule)

The biggest challenge in this chapter is deciding which part of your integral should be \( u \) and which should be \( dv/dx \). If you choose the wrong way around, your integral might actually get harder!

To make this easy, we use the LATE mnemonic. You should choose your \( u \) based on which function appears first in this list:

L - Logarithmic functions (e.g., \( \ln x \))
A - Algebraic functions (e.g., \( x^2, 3x, x \))
T - Trigonometric functions (e.g., \( \sin x, \cos x \))
E - Exponential functions (e.g., \( e^x \))

Example: In the integral \( \int x e^x dx \), \( x \) is Algebraic and \( e^x \) is Exponential. Since 'A' comes before 'E' in LATE, we choose \( u = x \).

Quick Review: Choosing \( u \) to be a function that "disappears" or becomes simpler when differentiated (like \( x \)) is usually the best strategy!

3. Step-by-Step Guide to Integration by Parts

Let's look at how to solve \( \int x \cos(x) dx \) step-by-step.

Step 1: Choose your \( u \) and \( \frac{dv}{dx} \).
Using LATE, \( x \) is Algebraic and \( \cos(x) \) is Trigonometric. So:
\( u = x \)
\( \frac{dv}{dx} = \cos(x) \)

Step 2: Differentiate \( u \) and Integrate \( \frac{dv}{dx} \).
\( \frac{du}{dx} = 1 \)
\( v = \sin(x) \) (Remember: the integral of \( \cos \) is \( \sin \))

Step 3: Plug everything into the formula.
\( \int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx \)
\( \int x \cos(x) dx = (x)(\sin x) - \int (\sin x)(1) dx \)

Step 4: Solve the final, simpler integral.
\( \int x \cos(x) dx = x \sin x - (-\cos x) + C \)
\( \int x \cos(x) dx = x \sin x + \cos x + C \)

Did you know? The second integral in the formula is designed to be easier than the first one. If it looks harder, go back and check if you picked the wrong \( u \)!

4. Special Case: Integrating \( \ln x \)

One of the most famous tricks in A Level Mathematics is using Integration by Parts to integrate \( \ln x \). On its own, \( \ln x \) doesn't look like a product, but we can pretend it is!

To integrate \( \int \ln x dx \), we write it as \( \int 1 \cdot \ln x dx \).

1. Identify parts: Using LATE, Logarithm comes first. So:
\( u = \ln x \)
\( \frac{dv}{dx} = 1 \)

2. Calculate the rest:
\( \frac{du}{dx} = \frac{1}{x} \)
\( v = x \)

3. Use the formula:
\( \int \ln x dx = (\ln x)(x) - \int (x)(\frac{1}{x}) dx \)
\( \int \ln x dx = x \ln x - \int 1 dx \)
\( \int \ln x dx = x \ln x - x + C \)

Key Takeaway: Whenever you see \( \ln x \) on its own in an integral, think: "Parts with \( u = \ln x \) and \( dv/dx = 1 \)."

5. Doing it Twice (Repeated Integration by Parts)

Sometimes, the "simpler" integral you get is still a product. Don't panic! This just means you need to apply the Integration by Parts method a second time to that new integral.

This typically happens when you have \( x^2 \) multiplied by a trig or exponential function, like \( \int x^2 e^x dx \). The first time you use parts, \( x^2 \) becomes \( 2x \). The second time, \( 2x \) becomes \( 2 \), and you can finally finish the problem.

Don't worry if this seems tricky at first! Just stay organized. Use brackets to keep your signs clear, especially with that minus sign in the middle of the formula.

6. Common Mistakes to Avoid

1. Forgetting the Minus Sign: The formula is \( uv - \int v du \). A very common error is changing that minus to a plus by accident.

2. Wrong Choice of \( u \): If you pick \( u = e^x \) and \( dv/dx = x \), you will end up with \( v = \frac{1}{2}x^2 \). Now your integral has an \( x^2 \) in it—it's getting worse! If the powers of \( x \) are increasing, swap your \( u \) and \( dv \).

3. Forgetting \( +C \): Like all indefinite integrals, don't forget your constant of integration at the very end.

4. Differentiation/Integration Mix-ups: Be careful not to accidentally integrate \( u \) and differentiate \( v \). Label your four parts (\( u, v, du, dv \)) clearly in a little box on the side of your page.

Summary Checklist

1. Is this a product of two different types of functions? If yes, use Parts.
2. Use LATE to pick \( u \).
3. Write down \( u, \frac{du}{dx}, \frac{dv}{dx}, \) and \( v \) clearly.
4. Plug into \( uv - \int v \frac{du}{dx} dx \).
5. Simplify and integrate the final part.
6. Add \( +C \) and celebrate!