Integration by Substitution

u-Substitution

By the end of this lesson, you will be able to use u-substitution to evaluate indefinite and definite integrals of composite functions.

The Pattern

$u$-substitution is the chain rule in reverse. The chain rule says:

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

So: $\int F'(g(x)) \cdot g'(x), dx = F(g(x)) + C$

When you see an integral of the form $\int f(g(x)) \cdot g'(x), dx$, let $u = g(x)$.

The Substitution Procedure

1. Choose $u = g(x)$ (usually the "inside" function)
2. Compute $du = g'(x), dx$
3. Rewrite the integral entirely in terms of $u$ (no $x$ left)
4. Integrate with respect to $u$
5. Back-substitute $u = g(x)$

Examples

Example 1: $\displaystyle\int 2x(x^2+1)^5, dx$

Let $u = x^2+1$; $du = 2x, dx$

$= \displaystyle\int u^5, du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C$

Example 2: $\displaystyle\int \cos(3x), dx$

Let $u = 3x$; $du = 3, dx$, so $dx = du/3$

$= \displaystyle\int \cos u, \frac{du}{3} = \frac{\sin u}{3} + C = \frac{\sin(3x)}{3} + C$

Example 3: $\displaystyle\int \frac{\ln x}{x}, dx$

Let $u = \ln x$; $du = \frac{1}{x}, dx$

$= \displaystyle\int u, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$

Definite Integrals with u-Substitution

Method 1: change the limits of integration.

When $u = g(x)$, change limits: $x = a \to u = g(a)$, $x = b \to u = g(b)$.

Example: $\displaystyle\int_0^1 2xe^{x^2}, dx$

$u = x^2$; $du = 2x, dx$; limits: $x=0\to u=0$; $x=1\to u=1$

$= \displaystyle\int_0^1 e^u, du = [e^u]_0^1 = e - 1$

Method 2: integrate, back-substitute, then evaluate at original limits. Either works.

Recognizing When to Substitute

The integral contains $f(g(x)) \cdot g'(x)$. Common patterns:

• Inside function and its derivative both appear
• A "chunk" that's repeated
• Something inside parentheses with the derivative of the inside also present