Integration by Substitution

The Method (u-Substitution)

The substitution method reverses the chain rule:

$\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du = F(u) + C$

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

Steps for Indefinite Integrals

1. Choose $u = g(x)$ (usually the "inside" function)
2. Compute $du = g'(x) \, dx$
3. Rewrite the integral in terms of $u$
4. Integrate with respect to $u$
5. Substitute back $x$ for $u$

Steps for Definite Integrals

Two methods:

Method 1: Change limits

1. When $x = a$, $u = g(a)$
2. When $x = b$, $u = g(b)$
3. Evaluate using new limits (no back-substitution needed)

Method 2: Substitute back

1. Find antiderivative in terms of $u$
2. Substitute back to $x$
3. Evaluate at original limits

Common Substitutions

Integrand contains	Try $u =$
$(ax + b)^n$	$ax + b$
$\sin(ax)$ or $\cos(ax)$	$ax$
$e^{ax}$	$ax$
$\sqrt{a^2 - x^2}$	$a \sin(\theta)$
$\ln(x)$	$\ln(x)$
$x^{n-1}$ inside $f(x^n)$	$x^n$

Examples

Example 1: Basic substitution

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

Let $u = 2x + 1$, $du = 2dx$, $dx = \frac{du}{2}$

$= \frac{1}{2} \int u^5 \, du = \frac{1}{2} \cdot \frac{u^6}{6} + C = \frac{(2x + 1)^6}{12} + C$

Example 2: Trigonometric

$\int \sin^3(x) \cos(x) \, dx$

Let $u = \sin(x)$, $du = \cos(x) \, dx$

$= \int u^3 \, du = \frac{u^4}{4} + C = \frac{\sin^4(x)}{4} + C$

Example 3: Exponential

$\int x \cdot e^{x^2} \, dx$

Let $u = x^2$, $du = 2x \, dx$, $x \, dx = \frac{du}{2}$

$= \frac{1}{2} \int e^u \, du = \frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C$

Example 4: Logarithmic

$\int \frac{1}{x \ln(x)} \, dx$

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

$= \int \frac{1}{u} \, du = \ln|u| + C = \ln|\ln(x)| + C$

Example 5: Definite integral (changing limits)

$\int_0^1 x\sqrt{1 - x^2} \, dx$

Let $u = 1 - x^2$, $du = -2x \, dx$ When $x = 0$, $u = 1$ When $x = 1$, $u = 0$

$= -\frac{1}{2} \int_1^0 \sqrt{u} \, du = \frac{1}{2} \int_0^1 \sqrt{u} \, du$

$= \frac{1}{2} \cdot \left[\frac{u^{3/2}}{3/2}\right]_0^1 = \frac{1}{2} \cdot \frac{2}{3} \cdot 1 = \frac{1}{3}$

Example 6: Adjusting for missing constants

$\int \sin(5x) \, dx$

Let $u = 5x$, $du = 5dx$, $dx = \frac{du}{5}$

$= \frac{1}{5} \int \sin(u) \, du = -\frac{1}{5}\cos(u) + C = -\frac{1}{5}\cos(5x) + C$

Example 7: Rational with substitution

$\int \frac{x}{x^2 + 1} \, dx$

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

$= \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln(x^2 + 1) + C$

Note: Absolute value not needed since $x^2 + 1 > 0$.

Example 8: Trigonometric (tan/sec)

$\int \tan(x) \, dx = \int \frac{\sin(x)}{\cos(x)} \, dx$

Let $u = \cos(x)$, $du = -\sin(x) \, dx$

$= -\int \frac{1}{u} \, du = -\ln|u| + C = -\ln|\cos(x)| + C = \ln|\sec(x)| + C$