# Integration by Parts (BC Only)

> AP Calculus BC · Unit 6: Integration and Accumulation of Change
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-by-parts/

This module covers all BC-exclusive integration by parts content, including core formula derivation, LIATE u selection, repeated/cyclic integration, and the tabular shortcut, with worked examples aligned to AP exam expectations.

**Prerequisites:** Product rule for differentiation; Basic antiderivatives of common functions; u-substitution for integration

## Learning objectives

- Derive the integration by parts formula from the product rule
- Select u and dv correctly using the LIATE mnemonic
- Solve single-step, repeated, cyclic, and definite integration by parts problems
- Use tabular integration for repeated applications with polynomials
- Avoid common exam pitfalls in integration by parts calculations

## Core Integration by Parts Formula

Integration by parts reverses the product rule for differentiation, and is used to integrate products of distinct function types that cannot be solved with u-substitution alone. It splits a complex original integral into a simpler product term and an easier-to-evaluate new integral.

**Derivation:** Derive the indefinite and definite integration by parts formulas from the product rule

*Starting from:* Product rule for differentiable functions $u(x)$ and $v(x)$

1. Start with the product rule for differentiation:

   $$\frac{d}{dx}\left[u(x)v(x)\right] = u'(x)v(x) + u(x)v'(x)$$
2. Integrate both sides with respect to $x$, and simplify the left-hand side using the Fundamental Theorem of Calculus:

   $$u(x)v(x) = \int u'(x)v(x) dx + \int u(x)v'(x) dx$$
3. Rearrange to isolate the integral of $u(x)v'(x)$, substituting $dv = v'(x)dx$ and $du = u'(x)dx$ to get the indefinite integral formula:

   $$\int u \, dv = uv - \int v \, du$$
4. For definite integrals over the interval $[a,b]$, add bounds to get the modified formula:

   $$\int_a^b u \, dv = \left[uv\right]_a^b - \int_a^b v \, du$$

*Conclusion:* The most critical step in using this formula is correctly selecting which function is $u$ and which is $dv$, which is simplified by the LIATE mnemonic.

> **LIATE Mnemonic for u Selection**
>
> **LIATE**: Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential. The function that comes first in this list is always $u$, the other is $dv$. This works because we differentiate $u$, so we want to reduce complex hard-to-integrate functions like logarithms to simpler terms.

**Worked example:** Evaluate the indefinite integral $\int 3x \ln x dx$

1. By LIATE, logarithmic comes before algebraic, so set $u = \ln x$, $dv = 3x dx$.
2. Compute $du$ and $v$:

   $$du = \frac{1}{x}dx, \quad v = \int 3x dx = \frac{3}{2}x^2$$
3. Substitute into the core integration by parts formula:

   $$\int 3x \ln x dx = uv - \int v du = \frac{3}{2}x^2 \ln x - \int \frac{3}{2}x^2 \cdot \frac{1}{x} dx$$
4. Simplify and integrate the remaining term:

   $$\frac{3}{2}x^2 \ln x - \frac{3}{2}\int x dx = \frac{3}{2}x^2 \ln x - \frac{3}{4}x^2 + C$$
5. Factor for a cleaner final result:

   $$\frac{3x^2}{4}\left(2 \ln x - 1\right) + C$$

> **Exam tip:** When integrating a single function like $\ln x$ or $\arctan x$ (not an explicit product), rewrite it as $f(x) \cdot 1 dx$, set $u = f(x)$ and $dv = 1 dx$ to convert it to a standard problem.

## Repeated and Cyclic Integration by Parts

Most integrals involving higher-degree polynomials, or products of exponentials and trigonometric functions, require more than one application of integration by parts. For an $n$-th degree polynomial $u$, each application reduces the degree of $u$ by 1, so $n$ applications are needed to reduce $u$ to a constant.

A BC-exclusive special case is cyclic integration by parts, which occurs when integrating products of the form $e^{ax} \sin(bx)$ or $e^{ax} \cos(bx)$. After two applications, the original integral reappears on the right-hand side, allowing you to solve for it algebraically. This is a very common AP exam problem.

**Worked example:** Evaluate $\int e^{3x} \cos 2x dx$

1. By LIATE, trigonometric comes before exponential, so set $u = \cos 2x$, $dv = e^{3x} dx$. Compute $du$ and $v$:

   $$du = -2 \sin 2x dx, \quad v = \frac{1}{3}e^{3x}$$
2. First application of integration by parts:

   $$\int e^{3x} \cos 2x dx = \frac{1}{3}e^{3x} \cos 2x + \frac{2}{3}\int e^{3x} \sin 2x dx$$
3. Apply integration by parts a second time to the new integral, set $u = \sin 2x$, $dv = e^{3x} dx$:

   $$\frac{2}{3}\left(\frac{1}{3}e^{3x} \sin 2x - \frac{4}{3}\int e^{3x} \cos 2x dx\right) = \frac{2}{9}e^{3x} \sin 2x - \frac{4}{9}\int e^{3x} \cos 2x dx$$
4. Let $I = \int e^{3x} \cos 2x dx$, substitute back into the original equation:

   $$I = \frac{1}{3}e^{3x} \cos 2x + \frac{2}{9}e^{3x} \sin 2x - \frac{4}{9}I$$
5. Collect terms with $I$ on the left-hand side and solve:

   $$\frac{13}{9}I = \frac{e^{3x}}{9}\left(3 \cos 2x + 2 \sin 2x\right) \implies I = \frac{e^{3x}(3 \cos 2x + 2 \sin 2x)}{13} + C$$

> **Exam tip:** Always add the constant of integration $+C$ only after you have isolated the original integral $I$ on the left-hand side. Adding $C$ early will lead to an incorrect constant multiple in your final result.

## Tabular Integration (DI Method)

Tabular integration is a time-saving shortcut for repeated integration by parts that works exclusively when one function (the $u$ term) differentiates to zero after a finite number of steps, which is almost always a polynomial of any degree. The method organizes work into two columns: $D$ (differentiate $u$ repeatedly) and $I$ (integrate $dv$ repeatedly). Stop when $D$ reaches zero, then sum alternating-sign diagonal products to get the antiderivative.

> **tip**
>
> Tabular integration is fully acceptable for all AP Calculus BC exam questions, including free-response questions, as long as your table is shown if explicitly requested.

**Worked example:** Evaluate $\int (2x^2 - 5x) e^{2x} dx$

1. By LIATE, assign $u = 2x^2 - 5x$ to the $D$ column, and $dv = e^{2x} dx$ to the $I$ column. Differentiate $D$ until you reach 0, and integrate $I$ the same number of times:
2. | Row | D (Differentiate u) | I (Integrate dv) |
| --- | --- | --- |
| 1 | $2x^2 - 5x$ | $e^{2x}$ |
| 2 | $4x - 5$ | $\frac{1}{2}e^{2x}$ |
| 3 | $4$ | $\frac{1}{4}e^{2x}$ |
| 4 | $0$ | $\frac{1}{8}e^{2x}$ |
3. Alternate signs starting with $+$ for the first row: $+, - , +$. Multiply each $D$ entry by the next-row $I$ entry, multiply by the sign, then sum all terms:

   $$+(2x^2 - 5x)\left(\frac{1}{2}e^{2x}\right) - (4x - 5)\left(\frac{1}{4}e^{2x}\right) + 4\left(\frac{1}{8}e^{2x}\right) + C$$
4. Simplify by factoring out common terms:

   $$\frac{e^{2x}}{4}\left(2(2x^2 - 5x) - (4x - 5) + 2\right) + C = \frac{e^{2x}}{4}(4x^2 - 14x +7) + C$$

> **Exam tip:** Never use tabular integration for cyclic cases like $e^x \sin x$. You will never reach a zero derivative, so the method will not work and will lead to an incorrect result.

## AP Style Practice Problems

**Worked example:** Multiple Choice: Evaluate $\int_0^\pi x \sin x dx$. Which is the correct result?

1. Set $u = x$, $dv = \sin x dx$, so $du = dx$, $v = -\cos x$. Apply the definite integral integration by parts formula:

   $$\int_0^\pi x \sin x dx = \left[-x \cos x\right]_0^\pi - \int_0^\pi (-\cos x) dx$$
2. Evaluate the boundary term first, then integrate the remaining term:

   $$(-\pi \cos \pi - 0) + \int_0^\pi \cos x dx = \pi + \left[\sin x\right]_0^\pi = \pi + 0 = \pi$$
3. The correct result is $\pi$.

**Worked example:** Free Response: Let $f(x) = x^2 \cos x$ on $[0, 2\pi]$. (a) Evaluate $\int x^2 \cos x dx$ (b) Evaluate $\int_0^{2\pi} x^2 \cos x dx$

1. (a) Use tabular integration: $u = x^2$, $dv = \cos x dx$. Signs alternate starting with $+$, sum terms to get:

   $$x^2 \sin x + 2x \cos x - 2 \sin x + C$$
2. (b) Evaluate the antiderivative at the bounds $0$ and $2\pi$:

   $$\left[x^2 \sin x + 2x \cos x - 2 \sin x\right]_0^{2\pi} = (0 + 2(2\pi)(1) - 0) - 0 = 4\pi$$

## Common pitfalls

- **Wrong:** Choosing $u = x$ and $dv = \ln x dx$ for $\int x \ln x dx$
  - Why it fails: Students mix up LIATE order, needing to integrate $\ln x$ for $v$ which unnecessarily complicates the problem
  - Correct: Follow LIATE: logarithms come before algebraic functions, so set $u = \ln x$, $dv = x dx$
- **Wrong:** After getting $I = ... - kI$ in cyclic integration, leaving $-kI$ on the right-hand side
  - Why it fails: Students forget the original integral reappears and must be solved for algebraically
  - Correct: Assign the original integral the placeholder $I$ at the start, then collect like terms to isolate $I$
- **Wrong:** For definite integration by parts, failing to evaluate the boundary term $[uv]_a^b$, leaving it as a function of $x$
  - Why it fails: Students focus on integrating the $\int v du$ term and forget the boundary term entirely
  - Correct: Evaluate and simplify $[uv]_a^b$ first, before integrating the second term
- **Wrong:** Starting alternating signs with a negative first term in tabular integration
  - Why it fails: Students confuse the sign flip from the $- \int v du$ term in the core formula
  - Correct: Write the sign explicitly next to each row of the table, starting with $+$ for the first non-zero $D$ term
- **Wrong:** Claiming $\int \arctan x dx$ can't be solved with integration by parts because it is not a product
  - Why it fails: Students incorrectly assume integration by parts only works for explicit products
  - Correct: Rewrite the integral as $\int 1 \cdot \arctan x dx$, set $u = \arctan x$, $dv = 1 dx$, then proceed

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Indefinite Integration by Parts | $\int u \, dv = uv - \int v \, du$ | Core formula from reversing product rule |
| Definite Integration by Parts | $\int_a^b u \, dv = \left[ u(x)v(x) \right]_a^b - \int_a^b v \, du$ | Evaluate $[uv]_a^b = u(b)v(b) - u(a)v(a)$ first |
| LIATE u Selection | Log > Inverse Trig > Algebra > Trig > Exp | First in list = $u$, other = $dv$ |
| Single Function Integral | $\int f(x) dx = \int f(x) \cdot 1 \, dx, \, u=f(x), dv=1 dx$ | For $\ln x$, $\arctan x$, etc. |
| Cyclic Integration | Let $I =$ original integral, solve for $I$ algebraically | Add $+C$ only after isolating $I$ |
| Tabular Integration | Sum = $\sum (\text{sign}) \times D_i \times I_{i+1}$ | Use only when $u$ differentiates to zero (polynomials) |
| Tabular Sign Order | $+, -, +, -,...$ starting with first $D$ term | Starting with negative is the most common tabular error |

## What's next

Integration by parts is a foundational technique for all advanced integration methods you will learn next in Unit 6 of AP Calculus BC, including integration by partial fractions and evaluation of improper integrals. Many multi-step AP exam problems require combining integration by parts with u-substitution or partial fractions, so mastering u/dv selection and avoiding common sign errors is critical to solving these problems efficiently and correctly. Beyond integration, integration by parts is used to derive reduction formulas for power integrals (which occasionally appear on multiple-choice questions) and is required for solving many applied differential equation problems tested on the BC exam. A solid command of this technique is essential to earning a high score on the AP Calculus BC exam.

- [Improper Integrals](https://www.owlsprep.com/study/ap-calculus-bc-u6-improper-integrals/)
- [Integration using partial fractions (BC only)](https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-using-partial-fractions/)
- [Selecting techniques for antidifferentiation](https://www.owlsprep.com/study/ap-calculus-bc-u6-selecting-techniques-for-antidifferentiation/)

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