Unit 6.9 – Integrating Using Substitution

AP® Calculus AB & BC | U-Substitution: The Reverse Chain Rule

Why This Matters: U-substitution (also called substitution method or change of variables) is THE most important integration technique! It's the reverse of the Chain Rule for derivatives. When you have a composite function, u-substitution lets you "undo" the Chain Rule to find the antiderivative. This technique is essential for integrating most non-trivial functions. Master it completely—it appears on virtually every AP® Calculus exam and is the foundation for more advanced integration methods!

🎯 The Big Idea: Reversing the Chain Rule

U-Substitution Concept

THE CHAIN RULE IN REVERSE:

\[ \frac{d}{dx}[F(u(x))] = F'(u(x)) \cdot u'(x) \]

Therefore:

\[ \int F'(u(x)) \cdot u'(x) \, dx = F(u(x)) + C \]

Key Insight:

If the integrand has the form \(f(g(x)) \cdot g'(x)\), we can substitute \(u = g(x)\) to simplify!

📋 The U-Substitution Method

THE SUBSTITUTION FORMULA

\[ \int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \]

where \(u = g(x)\) and \(du = g'(x)\,dx\)

Step-by-Step Procedure:

Choose \(u\): Pick the "inside function" (usually the composite part)
- Look for a function whose derivative also appears in the integrand
Find \(du\): Differentiate \(u\) to get \(du = u'\,dx\)
Substitute: Replace all \(x\)'s with \(u\)'s
- Replace \(u = g(x)\)
- Replace \(du = g'(x)\,dx\)
- Should have NO \(x\)'s left!
Integrate: Find \(\int f(u)\,du\) (should be easier now)
Back-substitute: Replace \(u\) with \(g(x)\) in your answer
Add +C: Don't forget the constant of integration!

🔍 When to Use U-Substitution

Look for These Patterns:

Common U-Substitution Scenarios
Pattern	Choose \(u =\)	Example
Composite function with its derivative	Inner function	\(\int 2x(x^2+1)^5\,dx\), let \(u = x^2+1\)
Function in exponent	Exponent expression	\(\int e^{3x}\,dx\), let \(u = 3x\)
Function in denominator	Denominator	\(\int \frac{x}{x^2+1}\,dx\), let \(u = x^2+1\)
Function inside trig	Argument of trig function	\(\int \cos(5x)\,dx\), let \(u = 5x\)
Function inside root	Expression under root	\(\int x\sqrt{x^2+1}\,dx\), let \(u = x^2+1\)

📖 Comprehensive Worked Examples

Example 1: Basic Substitution

Problem: Find \(\int 3x^2(x^3 + 5)^7 \, dx\)

Solution:

Step 1: Choose \(u\)

Let \(u = x^3 + 5\) (the "inside" function)

Step 2: Find \(du\)

\[ du = 3x^2 \, dx \]

Notice: \(3x^2\,dx\) appears in the original integral! ✓

Step 3: Substitute

\[ \int 3x^2(x^3 + 5)^7 \, dx = \int u^7 \, du \]

Step 4: Integrate

\[ = \frac{u^8}{8} + C \]

Step 5: Back-substitute

\[ = \frac{(x^3 + 5)^8}{8} + C \]

Answer: \(\frac{(x^3 + 5)^8}{8} + C\)

Example 2: Adjusting Constants

Problem: Find \(\int x(x^2 + 1)^4 \, dx\)

Solution:

Step 1: Choose \(u\)

Let \(u = x^2 + 1\)

Step 2: Find \(du\)

\[ du = 2x \, dx \quad \Rightarrow \quad x\,dx = \frac{1}{2}du \]

We have \(x\,dx\) but need \(2x\,dx\), so multiply by \(\frac{1}{2}\)

Step 3: Substitute

\[ \int x(x^2 + 1)^4 \, dx = \int u^4 \cdot \frac{1}{2}\,du = \frac{1}{2}\int u^4 \, du \]

Step 4: Integrate

\[ = \frac{1}{2} \cdot \frac{u^5}{5} + C = \frac{u^5}{10} + C \]

Step 5: Back-substitute

\[ = \frac{(x^2 + 1)^5}{10} + C \]

Answer: \(\frac{(x^2 + 1)^5}{10} + C\)

Example 3: Exponential Function

Problem: Find \(\int x^2 e^{x^3} \, dx\)

Solution:

Step 1: Choose \(u\)

Let \(u = x^3\) (exponent of \(e\))

Step 2: Find \(du\)

\[ du = 3x^2 \, dx \quad \Rightarrow \quad x^2\,dx = \frac{1}{3}du \]

Step 3: Substitute

\[ \int x^2 e^{x^3} \, dx = \int e^u \cdot \frac{1}{3}\,du = \frac{1}{3}\int e^u \, du \]

Step 4-5: Integrate and back-substitute

\[ = \frac{1}{3}e^u + C = \frac{1}{3}e^{x^3} + C \]

Answer: \(\frac{1}{3}e^{x^3} + C\)

Example 4: Trigonometric Function

Problem: Find \(\int \sin^3 x \cos x \, dx\)

Solution:

Step 1: Choose \(u\)

Let \(u = \sin x\)

Step 2: Find \(du\)

\[ du = \cos x \, dx \]

Perfect! We have \(\cos x\,dx\) in the integral.

Step 3: Substitute

\[ \int \sin^3 x \cos x \, dx = \int u^3 \, du \]

Step 4-5: Integrate and back-substitute

\[ = \frac{u^4}{4} + C = \frac{\sin^4 x}{4} + C \]

Answer: \(\frac{\sin^4 x}{4} + C\)

Example 5: Logarithmic Function

Problem: Find \(\int \frac{2x}{x^2 + 1} \, dx\)

Solution:

Step 1: Choose \(u\)

Let \(u = x^2 + 1\) (denominator)

Step 2: Find \(du\)

\[ du = 2x \, dx \]

Excellent! We have exactly \(2x\,dx\) in the numerator.

Step 3: Substitute

\[ \int \frac{2x}{x^2 + 1} \, dx = \int \frac{1}{u} \, du \]

Step 4-5: Integrate and back-substitute

\[ = \ln|u| + C = \ln|x^2 + 1| + C = \ln(x^2 + 1) + C \]

(Note: \(x^2 + 1 > 0\) always, so absolute value not needed)

Answer: \(\ln(x^2 + 1) + C\)

Example 6: Definite Integral with Substitution

Problem: Evaluate \(\int_0^2 x(x^2 + 1)^3 \, dx\)

Solution:

Step 1-2: Choose \(u\) and find \(du\)

Let \(u = x^2 + 1\), then \(du = 2x\,dx\), so \(x\,dx = \frac{1}{2}du\)

Step 3: Change limits of integration

When \(x = 0\): \(u = 0^2 + 1 = 1\)
When \(x = 2\): \(u = 2^2 + 1 = 5\)

New limits: \(u\) goes from 1 to 5

Step 4: Substitute

\[ \int_0^2 x(x^2 + 1)^3 \, dx = \int_1^5 u^3 \cdot \frac{1}{2} \, du = \frac{1}{2}\int_1^5 u^3 \, du \]

Step 5: Integrate and evaluate

\[ = \frac{1}{2}\left[\frac{u^4}{4}\right]_1^5 = \frac{1}{8}[u^4]_1^5 \]

\[ = \frac{1}{8}(5^4 - 1^4) = \frac{1}{8}(625 - 1) = \frac{624}{8} = 78 \]

Answer: 78

💡 Essential Tips & Strategies

✅ Choosing \(u\) Successfully:

Look for "inside" functions: Usually the argument of a composite function
Its derivative should appear: Or be proportional to something in the integrand
Common choices: Exponents, denominators, arguments of trig/exp/log functions
Try it! If it doesn't work, try a different \(u\)
LIATE doesn't apply: That's for integration by parts (later topic)

🔥 Common Patterns to Memorize:

Quick U-Substitution Patterns
Integral Form	Let \(u =\)	Result
\(\int f'(x)[f(x)]^n\,dx\)	\(u = f(x)\)	\(\frac{[f(x)]^{n+1}}{n+1} + C\)
\(\int \frac{f'(x)}{f(x)}\,dx\)	\(u = f(x)\)	\(\ln\|f(x)\| + C\)
\(\int f'(x)e^{f(x)}\,dx\)	\(u = f(x)\)	\(e^{f(x)} + C\)
\(\int f'(x)\sin(f(x))\,dx\)	\(u = f(x)\)	\(-\cos(f(x)) + C\)
\(\int f'(x)\cos(f(x))\,dx\)	\(u = f(x)\)	\(\sin(f(x)) + C\)

Definite Integrals: Two Methods

Method 1: Change Limits (Recommended)

Convert limits from \(x\)-values to \(u\)-values
Integrate in terms of \(u\)
Evaluate—NO need to back-substitute!

Method 2: Back-Substitute

Keep original limits
Find antiderivative in terms of \(u\)
Back-substitute to get \(x\)
Then evaluate with original limits

❌ Common Mistakes to Avoid

Mistake 1: Forgetting to substitute ALL \(x\)'s (must have only \(u\)'s and \(du\))
Mistake 2: Not adjusting constants when \(du\) doesn't match exactly
Mistake 3: Forgetting to back-substitute (answer must be in terms of original variable!)
Mistake 4: For definite integrals: forgetting to change limits OR back-substituting when you changed limits
Mistake 5: Choosing \(u\) poorly (if derivative doesn't appear, wrong choice!)
Mistake 6: Forgetting +C for indefinite integrals
Mistake 7: Trying to use substitution when it's not needed (simple power rule case)
Mistake 8: Sign errors when adjusting constants
Mistake 9: For definite integrals: using wrong limits after substitution
Mistake 10: Not verifying answer by differentiating

📝 Practice Problems

Set A: Basic Substitution

\(\int 6x(3x^2 - 1)^4 \, dx\)
\(\int x^2\sqrt{x^3 + 1} \, dx\)
\(\int e^{5x} \, dx\)

Answers:

\(\frac{(3x^2 - 1)^5}{5} + C\) (let \(u = 3x^2 - 1\))
\(\frac{2(x^3 + 1)^{3/2}}{9} + C\) (let \(u = x^3 + 1\))
\(\frac{e^{5x}}{5} + C\) (let \(u = 5x\))

Set B: Intermediate

\(\int \frac{x^2}{x^3 - 5} \, dx\)
\(\int \sin x \cos^4 x \, dx\)
\(\int_1^3 \frac{2x}{x^2 + 1} \, dx\)

Answers:

\(\frac{1}{3}\ln|x^3 - 5| + C\)
\(-\frac{\cos^5 x}{5} + C\)
\(\ln 10 - \ln 2 = \ln 5\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Clear substitution shown: Write \(u =\) and \(du =\)
Proper setup: Show the integral in terms of \(u\)
For definite integrals: Either change limits OR back-substitute (be consistent!)
Final answer in correct variable: Must be in terms of original variable (unless definite with changed limits)
+C for indefinite integrals: Always include!
Show all steps: Even if straightforward

⚡ Ultimate Quick Reference

U-SUBSTITUTION CHECKLIST

The 6-Step Process:

Choose \(u\) = inside function
Find \(du = u'\,dx\)
Substitute (all \(x\)'s → \(u\)'s)
Integrate \(\int f(u)\,du\)
Back-substitute \(u = g(x)\)
Add +C (indefinite only)

For Definite Integrals:

Option A: Change limits to \(u\)-values, then evaluate (no back-sub needed)
Option B: Keep original limits, back-substitute, then evaluate

Master U-Substitution! U-substitution is the reverse Chain Rule and the most important integration technique! The key is recognizing when a function and its derivative both appear in the integrand. Choose \(u\) as the "inside" function (composite part), find \(du = u'\,dx\), substitute everything to eliminate all \(x\)'s, integrate the simpler expression in \(u\), then back-substitute to get the answer in \(x\). Common patterns: \(\int f'(x)[f(x)]^n\,dx\), \(\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C\), and \(\int f'(x)e^{f(x)}\,dx = e^{f(x)} + C\). For definite integrals, either change the limits to \(u\)-values (recommended—no back-substitution needed!) or keep original limits and back-substitute before evaluating. Always check: derivative of your answer should give back the integrand. Don't forget +C for indefinite integrals! Practice recognizing substitution opportunities—it's tested on EVERY AP® Calculus exam! 🎯✨