Integration by Parts (BC Only)

AP Calculus BC· AP Calculus BC CED — Integration and Accumulation of Change· 14 min read

1. Core Integration by Parts Formula★★☆☆☆⏱ 3 min

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

Goal:

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 $d v = v^{'} (x) d x$ and $d u = u^{'} (x) d x$ 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:
$\int_a^b u \, dv = \left[uv\right]_a^b - \int_a^b v \, du$

Result:

The most critical step in using this formula is correctly selecting which function is $u$ and which is $d v$ , which is simplified by the LIATE mnemonic.

📐 Worked Example

Evaluate the indefinite integral $\int 3 x ln x d x$

1
By LIATE, logarithmic comes before algebraic, so set $u = ln x$ , $d v = 3 x d x$ .
2
Compute $d u$ 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 $f (x) \cdot 1 d x$ , set $u = f (x)$ and $d v = 1 d x$ to convert it to a standard problem.

2. Repeated and Cyclic Integration by Parts★★★☆☆BC only⏱ 4 min

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 $e^{a x} sin (b x)$ or $e^{a x} cos (b x)$ . 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 $\int e^{3 x} cos 2 x d x$

1
By LIATE, trigonometric comes before exponential, so set $u = cos 2 x$ , $d v = e^{3 x} d x$ . Compute $d u$ 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 $u = sin 2 x$ , $d v = e^{3 x} d x$ :
$\frac{2}{3}\left(\frac{1}{3}e^{3x} \sin 2x - \frac{4}{3}\int e^{3x} \cos 2x dx\right) = \frac{2}{9}e^{3x} \sin 2x - \frac{4}{9}\int e^{3x} \cos 2x dx$
4
Let $I = \int e^{3 x} cos 2 x d x$ , 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.

3. Tabular Integration (DI Method)★★★☆☆BC only⏱ 3 min

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 $d v$ repeatedly). Stop when $D$ reaches zero, then sum alternating-sign diagonal products to get the antiderivative.

📐 Worked Example

Evaluate $\int (2 x^{2} - 5 x) e^{2 x} d x$

1
By LIATE, assign $u = 2 x^{2} - 5 x$ to the $D$ column, and $d v = e^{2 x} d x$ 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 $2 x^{2} - 5 x$ $e^{2 x}$
2 $4 x - 5$ $\frac{1}{2} e^{2 x}$
3 $4$ $\frac{1}{4} e^{2 x}$
4 $0$ $\frac{1}{8} e^{2 x}$
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:
$\frac{e^{2x}}{4}\left(2(2x^2 - 5x) - (4x - 5) + 2\right) + C = \frac{e^{2x}}{4}(4x^2 - 14x +7) + C$

Row	D (Differentiate u)	I (Integrate dv)
1	$2 x^{2} - 5 x$	$e^{2 x}$
2	$4 x - 5$	$\frac{1}{2} e^{2 x}$
3	$4$	$\frac{1}{4} e^{2 x}$
4	$0$	$\frac{1}{8} e^{2 x}$

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.

4. AP Style Practice Problems★★★★☆⏱ 4 min

📐 Worked Example

Multiple Choice: Evaluate $\int_{0}^{π} x sin x d x$ . Which is the correct result?

1
Set $u = x$ , $d v = sin x d x$ , so $d u = d x$ , $v = - cos x$ . 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 $π$ .

📐 Worked Example

Free Response: Let $f (x) = x^{2} cos x$ on $[0, 2 π]$ . (a) Evaluate $\int x^{2} cos x d x$ (b) Evaluate $\int_{0}^{2 π} x^{2} cos x d x$

1
(a) Use tabular integration: $u = x^{2}$ , $d v = cos x d x$ . 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 $2 π$ :
$\left[x^2 \sin x + 2x \cos x - 2 \sin x\right]_0^{2\pi} = (0 + 2(2\pi)(1) - 0) - 0 = 4\pi$

5. Common Pitfalls

Wrong move:

Choosing $u = x$ and $d v = ln x d x$ for $\int x ln x d x$

Why:

Students mix up LIATE order, needing to integrate $ln x$ for $v$ which unnecessarily complicates the problem

Correct move:

Follow LIATE: logarithms come before algebraic functions, so set $u = ln x$ , $d v = x d x$

Wrong move:

After getting $I = ... - k I$ in cyclic integration, leaving $- k I$ on the right-hand side

Why:

Students forget the original integral reappears and must be solved for algebraically

Correct move:

Assign the original integral the placeholder $I$ at the start, then collect like terms to isolate $I$

Wrong move:

For definite integration by parts, failing to evaluate the boundary term $[uv]_{a}^{b}$ , leaving it as a function of $x$

Why:

Students focus on integrating the $\int v d u$ term and forget the boundary term entirely

Correct move:

Evaluate and simplify $[uv]_{a}^{b}$ first, before integrating the second term

Wrong move:

Starting alternating signs with a negative first term in tabular integration

Why:

Students confuse the sign flip from the $- \int v d u$ term in the core formula

Correct move:

Write the sign explicitly next to each row of the table, starting with $+$ for the first non-zero $D$ term

Wrong move:

Claiming $\int arctan x d x$ can't be solved with integration by parts because it is not a product

Why:

Students incorrectly assume integration by parts only works for explicit products

Correct move:

Rewrite the integral as $\int 1 \cdot arctan x d x$ , set $u = arctan x$ , $d v = 1 d x$ , then proceed

6. Quick Reference Cheatsheet

Category	Formula	Notes
Indefinite Integration by Parts	$\int u d v = uv - \int v d u$	Core formula from reversing product rule
Definite Integration by Parts	$\int_{a}^{b} u d v = [u (x) v (x)]_{a}^{b} - \int_{a}^{b} v d u$	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 = $d v$
Single Function Integral	$\int f (x) d x = \int f (x) \cdot 1 d x, u = f (x), d v = 1 d x$	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 = $\sum (sign) \times D_{i} \times I_{i + 1}$	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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Indefinite integration by parts
2022 · FRQ
Cyclic integration by parts

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.