Ultimate Guide to Integration by Parts

Introduction

If Integration by Substitution is the reverse of the Chain Rule, then Integration by Parts is the reverse of the Product Rule. When you are faced with an integral multiplying two entirely different types of functions—like an algebraic term and a trigonometric term (e.g., $x \sin x$ )—standard substitution will fail completely. In HSC Mathematics Extension 2, Integration by Parts is your ultimate weapon for breaking apart these hybrid functions.

Executive Summary

This guide covers the mechanics of Integration by Parts:

The Formula: Deriving and applying the formula $\int u \, dv = uv - \int v \, du$ .
LIATE Rule: A bulletproof mnemonic for choosing which part of the function should be ' $u$ ' and which should be ' $dv$ '.
Multiple Applications: Applying the formula two or more times when the first attempt doesn't completely clear the integral.
The "Boomerang" (Looping) Integral: Solving integrals like $e^x \sin x$ that loop back into themselves.

What is this about?

When differentiating a product $uv$ , the Product Rule is $(uv)' = u'v + uv'$ . If we integrate both sides with respect to $x$ , we get a formula that allows us to integrate a product: $\int u \, dv = uv - \int v \, du$

This formula doesn't instantly solve the integral. Instead, it trades one integral ( $\int u \, dv$ ) for a different integral ( $\int v \, du$ ). The entire goal of Integration by Parts is to choose $u$ and $dv$ strategically so that the new integral is easier to solve than the original one.

Main Content

1. The LIATE Rule (Choosing $u$ )

The hardest part of the formula is deciding which chunk of the integrand is $u$ (the part you will differentiate) and which is $dv$ (the part you will integrate). You want $u$ to be something that gets simpler when differentiated, and $dv$ to be something that is easy to integrate.

Use the LIATE mnemonic to choose $u$ . Pick the function that appears first in this list:

Logarithmic functions ( $\ln x$ )
Inverse trigonometric functions ( $\sin^{-1} x$ , $\tan^{-1} x$ )
Algebraic functions ( $x^2$ , $3x+5$ )
Trigonometric functions ( $\sin x$ , $\cos x$ )
Exponential functions ( $e^x$ )

For example, in $\int x \cos x \, dx$ , we have Algebraic ( $x$ ) and Trigonometric ( $\cos x$ ). A comes before T in LIATE, so $u = x$ , and $dv = \cos x \, dx$ .

2. Setting Up the Grid

To prevent mistakes, always set up a 2x2 grid before plugging into the formula: $u = [ \text{choose} ] \quad \rightarrow \quad du = [ \text{differentiate} ]$ $dv = [ \text{choose} ] \quad \leftarrow \quad v = [ \text{integrate} ]$

Then multiply diagonally for $uv$ , and multiply the bottom row horizontally for $\int v \, du$ .

3. Multiple Applications

Sometimes, one round of parts isn't enough. For example, in $\int x^2 e^x \, dx$ , making $u = x^2$ will reduce the new integral to $\int 2x e^x \, dx$ . The algebraic term is simpler ( $2x$ instead of $x^2$ ), but it's still a product. You simply perform Integration by Parts a second time on the new integral to destroy the $x$ entirely.

4. The Looping Integral ("Boomerang")

When integrating exponential and trigonometric hybrids, like $I = \int e^x \cos x \, dx$ , neither function ever "disappears" when differentiated. If you apply parts twice, you will end up with an equation like: $I = e^x \sin x + e^x \cos x - \int e^x \cos x \, dx$

Notice that the integral on the right is exactly the same as the original $I$ ! You can treat it like an algebra equation: $I = e^x \sin x + e^x \cos x - I$ $2I = e^x \sin x + e^x \cos x$ $I = \frac{1}{2} e^x (\sin x + \cos x) + C$

Simple Worked Example

Question: Evaluate $\int x \ln x \, dx$ .

Solution: Step 1: Choose $u$ and $dv$ using LIATE. We have Algebraic ( $x$ ) and Logarithmic ( $\ln x$ ). L comes before A, so we MUST choose $u = \ln x$ . $u = \ln x$ $dv = x \, dx$

Step 2: Create the grid. Differentiate $u$ : $du = \frac{1}{x} \, dx$ Integrate $dv$ : $v = \frac{x^2}{2}$

Step 3: Apply the formula. $\int u \, dv = uv - \int v \, du$ $\int x \ln x \, dx = (\ln x)\left(\frac{x^2}{2}\right) - \int \left(\frac{x^2}{2}\right) \left(\frac{1}{x}\right) \, dx$

Step 4: Simplify and solve the new integral. $= \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx$ $= \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx$ $= \frac{x^2}{2} \ln x - \frac{1}{2} \left(\frac{x^2}{2}\right) + C$ $= \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$

Final Answer: $\frac{x^2}{2} \ln x - \frac{x^2}{4} + C$ .

mini-FAQ page

Q: What if the integral is just $\int \ln x \, dx$ or $\int \tan^{-1} x \, dx$ ? There is no product! A: This is a classic trick. You create a product by making $dv = 1 \, dx$ . Let $u = \ln x$ and $dv = 1 \, dx$ . Then $du = \frac{1}{x}$ and $v = x$ . The formula will magically solve it!

Q: Do I have to write $+ C$ during the $v = \int dv$ step? A: No, you can ignore the constant of integration during the intermediate setup steps. Just remember to add a single $+ C$ at the very end of your final answer.

Common mistakes to avoid

Ignoring the negative sign in the formula: The formula is $uv - \int v \, du$ . Many students forget the minus sign, or lose track of it when distributing negative numbers from multiple applications of parts. Use large brackets!
Choosing the wrong $u$ : If you chose $u = x$ and $dv = \ln x$ in the example above, you would immediately get stuck trying to integrate $\ln x$ . If your new integral looks much harder than the original, you likely chose the wrong $u$ . Follow LIATE!

Practice on Vu's Maths Hub

Integration by parts requires excellent book-keeping with signs and brackets.

Practice double applications and looping integrals in the HSC Integration Booklet.
Test your integration identification skills at Vu's Maths Hub.

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