∫ Integration by Substitution
A technique for solving complex integrals
🎯 When Do We Use Substitution?
Substitution is appropriate when we have a composite function – a function inside a function.
Clear indicators: when you see something like:
- \((2x+3)^5\) – an expression raised to a power
- \(\sqrt{x^2+1}\) – square root of an expression
- \(e^{3x}\) – exponential with an expression
- \(\sin(2x)\) – trigonometric function of an expression
💡 The Core Idea
Substitution is the "reverse chain rule".
If in differentiation we used the chain rule:
\([f(g(x))]' = f'(g(x)) \cdot g'(x)\)
Then in integration we do the reverse:
\(\int f'(g(x)) \cdot g'(x) \, dx = f(g(x)) + C\)
📝 Method Steps
Step 1: Choose a substitution
Define \(u = g(x)\) (usually the "inner" expression)
Step 2: Compute du
Differentiate: \(du = g'(x) \, dx\)
Or: \(dx = \frac{du}{g'(x)}\)
Step 3: Substitute into the integral
Replace all x and dx with u and du
Step 4: Solve
Evaluate the new integral (which is usually simpler)
Step 5: Back-substitute for x
Replace \(u = g(x)\) back
✏️ Example 1: Power of a Linear Expression
Evaluate: \(\int (2x+3)^5 \, dx\)
Solution:
Step 1: Choose substitution
\(u = 2x + 3\)
Step 2: Compute du
\(\frac{du}{dx} = 2\)
\(du = 2 \, dx\)
\(dx = \frac{du}{2}\)
Step 3: Substitute
\(\int (2x+3)^5 \, dx = \int u^5 \cdot \frac{du}{2}\)
\(= \frac{1}{2} \int u^5 \, du\)
Step 4: Solve
\(= \frac{1}{2} \cdot \frac{u^6}{6} + C = \frac{u^6}{12} + C\)
Step 5: Back-substitute for x
\(= \frac{(2x+3)^6}{12} + C\)
Answer: \(\frac{(2x+3)^6}{12} + C\)
⚡ Quick Formula for Linear Expressions
When a linear expression \((ax + b)\) appears inside a function:
\(\int f(ax+b) \, dx = \frac{1}{a} \cdot F(ax+b) + C\)
where F is the antiderivative of f
💡 In words: divide by the coefficient of x!
Examples:
| \(\int (3x+1)^4 \, dx = \frac{(3x+1)^5}{5 \cdot 3} + C = \frac{(3x+1)^5}{15} + C\) |
| \(\int e^{5x} \, dx = \frac{e^{5x}}{5} + C\) |
| \(\int \cos(2x) \, dx = \frac{\sin(2x)}{2} + C\) |
| \(\int \frac{1}{4x-1} \, dx = \frac{\ln|4x-1|}{4} + C\) |
✏️ Example 2: Root of an Expression
Evaluate: \(\int x\sqrt{x^2+1} \, dx\)
Solution:
Step 1: Choose substitution (what is under the root)
\(u = x^2 + 1\)
Step 2: Compute du
\(du = 2x \, dx\)
\(x \, dx = \frac{du}{2}\)
Step 3: Substitute
\(\int x\sqrt{x^2+1} \, dx = \int \sqrt{u} \cdot \frac{du}{2}\)
\(= \frac{1}{2} \int u^{\frac{1}{2}} \, du\)
Step 4: Solve
\(= \frac{1}{2} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{1}{3} u^{\frac{3}{2}} + C\)
Step 5: Back-substitute for x
\(= \frac{1}{3} (x^2+1)^{\frac{3}{2}} + C = \frac{1}{3}\sqrt{(x^2+1)^3} + C\)
Answer: \(\frac{1}{3}(x^2+1)^{\frac{3}{2}} + C\)
✏️ Example 3: Exponential Function
Evaluate: \(\int x \cdot e^{x^2} \, dx\)
Solution:
Step 1: Substitution
\(u = x^2\)
Step 2: du
\(du = 2x \, dx \implies x \, dx = \frac{du}{2}\)
Steps 3+4: Substitute and solve
\(\int x \cdot e^{x^2} \, dx = \int e^u \cdot \frac{du}{2} = \frac{1}{2} e^u + C\)
Step 5: Back to x
\(= \frac{1}{2} e^{x^2} + C\)
Answer: \(\frac{1}{2} e^{x^2} + C\)
✏️ Example 4: Logarithm
Evaluate: \(\int \frac{\ln x}{x} \, dx\)
Solution:
Step 1: Substitution
\(u = \ln x\)
Step 2: du
\(du = \frac{1}{x} \, dx\)
Steps 3+4: Substitute and solve
\(\int \frac{\ln x}{x} \, dx = \int u \, du = \frac{u^2}{2} + C\)
Step 5: Back to x
\(= \frac{(\ln x)^2}{2} + C\)
Answer: \(\frac{(\ln x)^2}{2} + C\)
✏️ Example 5: Definite Integral with Substitution
Evaluate: \(\int_0^1 x(x^2+1)^3 \, dx\)
Solution:
Step 1: Substitution
\(u = x^2 + 1\)
\(du = 2x \, dx \implies x \, dx = \frac{du}{2}\)
Step 2: Convert the limits!
When \(x = 0\): \(u = 0^2 + 1 = 1\)
When \(x = 1\): \(u = 1^2 + 1 = 2\)
Step 3: New integral
\(\int_0^1 x(x^2+1)^3 \, dx = \int_1^2 u^3 \cdot \frac{du}{2} = \frac{1}{2} \int_1^2 u^3 \, du\)
Step 4: Evaluate
\(= \frac{1}{2} \Big[ \frac{u^4}{4} \Big]_1^2\)
\(= \frac{1}{2} \left( \frac{16}{4} - \frac{1}{4} \right)\)
\(= \frac{1}{2} \cdot \frac{15}{4} = \frac{15}{8}\)
Answer: \(\frac{15}{8}\)
⚠️ Important: in a definite integral you must also convert the limits to u!
Then there is no need to back-substitute to x at the end.
🎯 How to Choose u?
| Integral type | What to choose as u |
|---|---|
| \(\int f(ax+b) \, dx\) | \(u = ax + b\) |
| \(\int x \cdot f(x^2) \, dx\) | \(u = x^2\) |
| \(\int \frac{f'(x)}{f(x)} \, dx\) | \(u = f(x)\) → result: \(\ln|u|\) |
| \(\int f'(x) \cdot e^{f(x)} \, dx\) | \(u = f(x)\) |
| \(\int \sin^n x \cos x \, dx\) | \(u = \sin x\) |
💡 Rule of thumb:
Choose u to be the expression whose derivative already appears in the integral (or nearly appears)
📋 Important Integrals (to know!)
| Integral | Result |
|---|---|
| \(\int (ax+b)^n \, dx\) | \(\frac{(ax+b)^{n+1}}{a(n+1)} + C\) |
| \(\int \frac{1}{ax+b} \, dx\) | \(\frac{1}{a}\ln|ax+b| + C\) |
| \(\int e^{ax+b} \, dx\) | \(\frac{1}{a}e^{ax+b} + C\) |
| \(\int \sin(ax+b) \, dx\) | \(-\frac{1}{a}\cos(ax+b) + C\) |
| \(\int \cos(ax+b) \, dx\) | \(\frac{1}{a}\sin(ax+b) + C\) |
| \(\int \frac{f'(x)}{f(x)} \, dx\) | \(\ln|f(x)| + C\) |
💡 Tips for the Exam
1️⃣ Spot the pattern
Look for a function whose derivative already appears in the integral
2️⃣ Don't forget to divide
If \(du = 2x \, dx\), then \(x \, dx = \frac{du}{2}\)
3️⃣ Limits in definite integrals
Convert the limits to u, or back-substitute to x before substituting
4️⃣ Verify by differentiating
You can always differentiate the result to confirm you get the integrand
📝 Summary
Substitution method: \(u = g(x)\), \(du = g'(x) \, dx\)
Linear expression: \(\int f(ax+b) \, dx = \frac{1}{a} F(ax+b) + C\)
The key: identify the inner expression and its derivative