The Substitution Rule (also called u-substitution) is the "reverse chain rule." It allows us to integrate composite functions by changing variables. If you see a function and its derivative inside an integral, substitution is the way to go.
🎯 In this section you will learn
- The Substitution Rule formula: $\int f(g(x)) g'(x) dx = \int f(u) du$ with $u = g(x)$
- How to choose the right $u$ for substitution
- Applying substitution to indefinite integrals
- Applying substitution to definite integrals (changing the limits)
📌 The Substitution Rule
$$ \int f(g(x)) \, g'(x) \, dx = \int f(u) \, du $$
Let $u = g(x)$, then $du = g'(x) dx$
💡 How to Choose $u$
Look for a function $g(x)$ whose derivative $g'(x)$ also appears in the integral (up to a constant factor). Often, $u$ is the "inside" function of a composition.
Find $\displaystyle \int 2x \cos(x^2) \, dx$.
①
Choose $u$
Let $u = x^2$, then $du = 2x dx$
②
Substitute
$\int 2x \cos(x^2) dx = \int \cos u \, du$
③
Integrate
$\int \cos u \, du = \sin u + C = \sin(x^2) + C$
Find $\displaystyle \int x e^{x^2} \, dx$.
- Let $u = x^2$, then $du = 2x dx$ → $x dx = \frac{1}{2} du$
- $\int x e^{x^2} dx = \int e^u \cdot \frac{1}{2} du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C$
Find $\displaystyle \int (3x+1)^5 \, dx$.
- Let $u = 3x+1$, then $du = 3 dx$ → $dx = \frac{1}{3} du$
- $\int (3x+1)^5 dx = \int u^5 \cdot \frac{1}{3} du = \frac{1}{3} \cdot \frac{u^6}{6} + C = \frac{1}{18} (3x+1)^6 + C$
Find $\displaystyle \int \frac{\ln x}{x} \, dx$.
- Let $u = \ln x$, then $du = \frac{1}{x} dx$
- $\int \frac{\ln x}{x} dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$
📌 Substitution for Definite Integrals
When using substitution on a definite integral, you have two options:
- Find the antiderivative in terms of $x$, then evaluate at the original limits.
- Change the limits: When you substitute $u = g(x)$, change the limits from $x = a$ to $x = b$ into $u = g(a)$ and $u = g(b)$.
Find $\displaystyle \int_0^2 2x e^{x^2} \, dx$.
①
Choose $u$
$u = x^2$, $du = 2x dx$
②
Change limits
When $x = 0$, $u = 0$; when $x = 2$, $u = 4$
③
Substitute and integrate
$\int_0^2 2x e^{x^2} dx = \int_0^4 e^u du = [e^u]_0^4 = e^4 - 1$
⚠️ Common Mistakes
- Forgetting $du$: When you substitute, you must replace $dx$ with $\frac{du}{g'(x)}$.
- Not changing limits: For definite integrals, if you don't change the limits, you must convert back to $x$ before evaluating.
- Choosing the wrong $u$: If your $u$ doesn't simplify the integral, try a different choice.
🔍 Key Takeaways
- Substitution Rule: $\int f(g(x)) g'(x) dx = \int f(u) du$ with $u = g(x)$.
- Choose $u$ as the "inside" function whose derivative also appears.
- For indefinite integrals, substitute back to $x$ at the end (don't forget $+C$).
- For definite integrals, either convert back to $x$ or change the limits of integration.
← Back to Module Page