Unit 6.11 – Integrating Using Integration by Parts

AP® Calculus BC ONLY | The Reverse Product Rule

Why This Matters: Integration by parts is the reverse of the product rule for derivatives! When u-substitution doesn't work (especially for products of different function types), integration by parts is often the key. This BC-only technique is essential for integrating products like \(x \cdot e^x\), \(x \cdot \sin x\), or \(\ln x\). Mastering this method—including the LIATE rule for choosing \(u\) and \(dv\)—is critical for success on the BC exam. It's tested frequently and is one of the most powerful integration tools!

🎯 The Integration by Parts Formula

Integration by Parts

THE FORMULA (TWO FORMS):

Form 1: Traditional

\[ \int u \, dv = uv - \int v \, du \]

Form 2: Function Notation

\[ \int u(x) \cdot v'(x) \, dx = u(x) \cdot v(x) - \int v(x) \cdot u'(x) \, dx \]

📝 Origin: This comes from the product rule for derivatives:

\[ \frac{d}{dx}[u \cdot v] = u \cdot v' + v \cdot u' \]

Integrating both sides and rearranging gives integration by parts!

📋 The LIATE Rule: Choosing u and dv

THE LIATE RULE

Choose \(u\) in order of priority (from highest to lowest):

Letter	Function Type	Examples
L	Logarithmic	\(\ln x, \log x\)
I	Inverse trig	\(\arctan x, \arcsin x, \text{arcsec } x\)
A	Algebraic (polynomials)	\(x, x^2, x^3, \sqrt{x}\)
T	Trigonometric	\(\sin x, \cos x, \tan x, \sec x\)
E	Exponential	\(e^x, 2^x, a^x\)

🔑 Key Rule: Choose \(u\) as the function that appears FIRST in LIATE, and \(dv\) gets everything else!

Why? Functions higher in LIATE get simpler when differentiated (to get \(du\)), while those lower are easy to integrate (to get \(v\)).

📐 The Integration by Parts Process

Step-by-Step Procedure:

Identify \(u\) and \(dv\):
- Use LIATE to choose \(u\) (earlier in list)
- \(dv\) is everything else (including \(dx\))
Find \(du\): Differentiate \(u\)
- \(du = u' \, dx\)
Find \(v\): Integrate \(dv\)
- \(v = \int dv\) (no +C needed here!)
Apply formula: \(\int u \, dv = uv - \int v \, du\)
Evaluate remaining integral: \(\int v \, du\)
Add +C: At the very end

📖 Comprehensive Worked Examples

Example 1: Basic Integration by Parts

Problem: Find \(\int x e^x \, dx\)

Solution:

Step 1: Choose \(u\) and \(dv\) using LIATE

\(u = x\) (Algebraic - A in LIATE)
\(dv = e^x \, dx\) (Exponential - E in LIATE)

A comes before E in LIATE, so \(u = x\) ✓

Step 2: Find \(du\)

\[ du = 1 \, dx = dx \]

Step 3: Find \(v\)

\[ v = \int e^x \, dx = e^x \]

Step 4: Apply integration by parts formula

\[ \int x e^x \, dx = x \cdot e^x - \int e^x \cdot 1 \, dx \]

\[ = xe^x - \int e^x \, dx \]

Step 5: Evaluate remaining integral

\[ = xe^x - e^x + C \]

\[ = e^x(x - 1) + C \]

Answer: \(e^x(x - 1) + C\) or \(xe^x - e^x + C\)

Example 2: Logarithmic Function

Problem: Find \(\int \ln x \, dx\)

Solution:

Step 1: Rewrite and choose \(u\), \(dv\)

Rewrite as: \(\int \ln x \cdot 1 \, dx\)

\(u = \ln x\) (Logarithmic - L in LIATE, highest priority!)
\(dv = 1 \, dx = dx\)

Step 2-3: Find \(du\) and \(v\)

\[ du = \frac{1}{x} \, dx \]

\[ v = \int 1 \, dx = x \]

Step 4-5: Apply formula and integrate

\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx \]

\[ = x \ln x - \int 1 \, dx \]

\[ = x \ln x - x + C \]

Answer: \(x \ln x - x + C\) or \(x(\ln x - 1) + C\)

Example 3: Trigonometric Product

Problem: Find \(\int x \sin x \, dx\)

Solution:

Step 1: Choose \(u\) and \(dv\)

\(u = x\) (Algebraic - A)
\(dv = \sin x \, dx\) (Trigonometric - T)

A comes before T in LIATE ✓

Step 2-3: Find \(du\) and \(v\)

\[ du = dx \]

\[ v = \int \sin x \, dx = -\cos x \]

Step 4-5: Apply and integrate

\[ \int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx \]

\[ = -x \cos x + \int \cos x \, dx \]

\[ = -x \cos x + \sin x + C \]

Answer: \(-x \cos x + \sin x + C\)

Example 4: Repeated Integration by Parts

Problem: Find \(\int x^2 e^x \, dx\)

Solution: Need to apply integration by parts TWICE!

First application:

\(u = x^2\), \(dv = e^x \, dx\)
\(du = 2x \, dx\), \(v = e^x\)

\[ \int x^2 e^x \, dx = x^2 e^x - \int e^x \cdot 2x \, dx = x^2 e^x - 2\int x e^x \, dx \]

Second application on \(\int x e^x \, dx\):

\(u = x\), \(dv = e^x \, dx\)
\(du = dx\), \(v = e^x\)

\[ \int x e^x \, dx = xe^x - \int e^x \, dx = xe^x - e^x \]

Combine:

\[ \int x^2 e^x \, dx = x^2 e^x - 2(xe^x - e^x) + C \]

\[ = x^2 e^x - 2xe^x + 2e^x + C \]

\[ = e^x(x^2 - 2x + 2) + C \]

Answer: \(e^x(x^2 - 2x + 2) + C\)

Example 5: Cyclic Integration by Parts

Problem: Find \(\int e^x \sin x \, dx\)

Solution: This requires a special technique!

Let \(I = \int e^x \sin x \, dx\)

First application: \(u = \sin x\), \(dv = e^x \, dx\)

\[ I = \sin x \cdot e^x - \int e^x \cos x \, dx \]

Second application on \(\int e^x \cos x \, dx\):

\(u = \cos x\), \(dv = e^x \, dx\)

\[ \int e^x \cos x \, dx = \cos x \cdot e^x - \int e^x(-\sin x) \, dx \]

\[ = e^x \cos x + \int e^x \sin x \, dx = e^x \cos x + I \]

Substitute back:

\[ I = e^x \sin x - (e^x \cos x + I) \]

\[ I = e^x \sin x - e^x \cos x - I \]

\[ 2I = e^x \sin x - e^x \cos x \]

\[ I = \frac{e^x(\sin x - \cos x)}{2} + C \]

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

📊 Tabular Method (Repeated Integration by Parts)

Tabular Integration

When you need to apply integration by parts multiple times (like \(\int x^n e^x \, dx\) or \(\int x^n \sin x \, dx\)), the tabular method is a shortcut!

Steps for Tabular Method:

Make two columns: one for \(u\) (differentiate repeatedly) and one for \(dv\) (integrate repeatedly)
Continue until \(u\) becomes zero (for polynomials) or a pattern emerges
Draw diagonal arrows alternating signs (+, −, +, −, ...)
Multiply along arrows and sum

Example: \(\int x^3 e^x \, dx\) using tabular method

Sign	\(u\) and derivatives	\(dv\) and integrals
+	\(x^3\)	\(e^x\)
−	\(3x^2\)	\(e^x\)
+	\(6x\)	\(e^x\)
−	\(6\)	\(e^x\)
+	\(0\)	\(e^x\)

\[ \int x^3 e^x \, dx = x^3e^x - 3x^2e^x + 6xe^x - 6e^x + C \]

\[ = e^x(x^3 - 3x^2 + 6x - 6) + C \]

💡 Essential Tips & Strategies

✅ LIATE Mastery:

Always use LIATE: It works 95% of the time
Logarithms always go first: They simplify when differentiated
Polynomials before trig/exp: They eventually become zero
If both are trig or both exp: Either choice usually works
Check your choice: If \(\int v \, du\) is harder than original, try switching \(u\) and \(dv\)

🔥 Special Cases:

Cyclic integrals: (\(e^x \sin x\), \(e^x \cos x\)) — apply parts twice, solve for original integral
Repeated parts: For \(x^n e^x\) or \(x^n \sin x\), use tabular method
Definite integrals: Apply formula first, then evaluate at bounds
Don't add +C: Until the very end!

Common Integration by Parts Patterns:

Quick Reference Patterns
Integral Type	Choose \(u\)	Choose \(dv\)
\(\int x^n e^x \, dx\)	\(x^n\)	\(e^x \, dx\)
\(\int x^n \sin x \, dx\)	\(x^n\)	\(\sin x \, dx\)
\(\int x^n \cos x \, dx\)	\(x^n\)	\(\cos x \, dx\)
\(\int \ln x \, dx\)	\(\ln x\)	\(1 \, dx\)
\(\int \arctan x \, dx\)	\(\arctan x\)	\(1 \, dx\)
\(\int e^x \sin x \, dx\)	\(\sin x\) or \(e^x\)	\(e^x \, dx\) or \(\sin x \, dx\)

❌ Common Mistakes to Avoid

Mistake 1: Choosing \(u\) and \(dv\) incorrectly (not following LIATE)
Mistake 2: Forgetting to include \(dx\) in \(dv\)
Mistake 3: Sign errors when evaluating \(\int v \, du\)
Mistake 4: Adding +C to \(v\) when finding it (don't do this!)
Mistake 5: For cyclic integrals, forgetting to solve algebraically for \(I\)
Mistake 6: Not simplifying \(\int v \, du\) before integrating
Mistake 7: Forgetting negative signs from trig integrals (\(\int \sin x = -\cos x\))
Mistake 8: Giving up after first application when repeated parts needed
Mistake 9: Not checking answer by differentiating
Mistake 10: Forgetting +C at the very end

📝 Practice Problems

Set A: Basic Integration by Parts

\(\int x \cos x \, dx\)
\(\int x e^{2x} \, dx\)
\(\int \ln(2x) \, dx\)

Answers:

\(x \sin x + \cos x + C\)
\(\frac{e^{2x}(2x-1)}{4} + C\)
\(x\ln(2x) - x + C\)

Set B: Advanced

\(\int x^2 \sin x \, dx\)
\(\int e^x \cos x \, dx\)

Answers:

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

✏️ AP® Exam Success Tips

What AP® BC Graders Look For:

Clearly label \(u\), \(du\), \(v\), \(dv\): Show all four components
Write the formula: \(\int u \, dv = uv - \int v \, du\)
Show substitution: Explicitly substitute values into formula
Evaluate remaining integral: Don't leave \(\int v \, du\) unevaluated
For repeated parts: Show each application clearly
Include +C: For indefinite integrals
Simplify final answer: Factor when possible

⚡ Ultimate Quick Reference

INTEGRATION BY PARTS ESSENTIALS

The Formula:

\[ \int u \, dv = uv - \int v \, du \]

LIATE Priority (choose \(u\) first):

\[ \text{L} \rightarrow \text{I} \rightarrow \text{A} \rightarrow \text{T} \rightarrow \text{E} \]

Logarithmic → Inverse trig → Algebraic → Trig → Exponential

The Process:

Choose \(u\) and \(dv\) (LIATE)
Find \(du = u' \, dx\)
Find \(v = \int dv\)
Apply: \(uv - \int v \, du\)
Add +C

Master Integration by Parts! This BC-only technique reverses the product rule: \(\int u \, dv = uv - \int v \, du\). The key is choosing \(u\) and \(dv\) correctly using the LIATE rule: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Choose \(u\) as the function appearing first in LIATE (it simplifies when differentiated), and \(dv\) gets everything else (easy to integrate). Common applications: \(\int x^n e^x\), \(\int x^n \sin x\), \(\int \ln x\), \(\int \arctan x\). For repeated applications (like \(\int x^2 e^x\)), either apply parts multiple times or use the tabular method. For cyclic integrals like \(\int e^x \sin x\), apply parts twice and solve algebraically for the original integral. Always show \(u\), \(du\), \(v\), \(dv\) clearly on exams. Verify by differentiating your answer—you should get the original integrand. Don't add +C to \(v\), only at the very end! This is tested frequently on BC exams—practice until automatic! 🎯✨