5.5 Substitution

Substitution Method for Integration

Integration by substitution reverses the chain rule. Where the chain rule lets you differentiate composite functions, substitution lets you anti-differentiate them. It works by swapping out a messy piece of the integrand for a single new variable, turning a hard integral into a simpler one.

Substitution for Indefinite Integrals

The core idea: if you spot a composite function $f(g(x))$ multiplied by (or closely related to) $g'(x)$, you can substitute $u = g(x)$ and rewrite everything in terms of $u$.

Here's the process, step by step:

1. Choose $u$. Pick the "inner function" or the piece of the integrand whose derivative also appears (up to a constant). For example, in $\int \sin(3x+1)\, dx$, let $u = 3x + 1$.

2. Find $du$. Differentiate your choice: $du = 3\,dx$, so $dx = \frac{1}{3}\,du$.

3. Rewrite the integral in terms of $u$. Replace every $x$-expression and $dx$: $\int \sin(3x+1)\, dx = \int \sin(u) \cdot \frac{1}{3}\, du$

4. Integrate with respect to $u$ using formulas you already know: $\frac{1}{3}\int \sin(u)\, du = -\frac{1}{3}\cos(u) + C$

5. Substitute back. Replace $u$ with the original expression: $-\frac{1}{3}\cos(3x+1) + C$

A common mistake is forgetting to convert every part of the integrand to $u$. If any $x$ is left over after substitution, something went wrong. Either your choice of $u$ needs adjusting, or the leftover $x$ can be solved for using your substitution equation.

Substitution in Definite Integrals

The process is the same, with one key difference: you convert the limits of integration to $u$-values so you never have to substitute back.

1. Choose $u$ and find $du$ exactly as before.

2. Transform the limits. If $u = 3x + 1$ and the original limits are $x = 0$ to $x = 2$:

   • Lower limit: $u(0) = 3(0) + 1 = 1$
   • Upper limit: $u(2) = 3(2) + 1 = 7$

3. Rewrite and evaluate: $\int_0^2 \sin(3x+1)\, dx = \frac{1}{3}\int_1^7 \sin(u)\, du = \frac{1}{3}\Big[-\cos(u)\Big]_1^7 = \frac{1}{3}\big(-\cos 7 + \cos 1\big)$

Because you already plugged in $u$-limits, the answer is a number. No back-substitution needed.

Recognizing When to Use Substitution

Substitution works best when part of the integrand is a function of a simpler expression, and the derivative of that expression is also present (possibly off by a constant factor). Here are the most common patterns:

• Powers of linear terms: $\int (2x-3)^5\, dx$. Let $u = 2x - 3$, so $du = 2\,dx$. The integral becomes $\frac{1}{2}\int u^5\, du$.
• Trig functions of linear terms: $\int \cos(4x+2)\, dx$. Let $u = 4x + 2$, so $du = 4\,dx$. You get $\frac{1}{4}\int \cos(u)\, du$.
• Exponentials of linear terms: $\int e^{-5x+1}\, dx$. Let $u = -5x + 1$, so $du = -5\,dx$. This gives $-\frac{1}{5}\int e^u\, du$.
• Roots of linear terms: $\int \sqrt{x+7}\, dx$. Let $u = x + 7$, so $du = dx$. Rewrite as $\int u^{1/2}\, du$ and use the power rule.
• Non-linear inner functions: $\int 2x\cos(x^2)\, dx$. Let $u = x^2$, so $du = 2x\,dx$. The $2x$ already sitting in the integrand is exactly $du$, giving $\int \cos(u)\, du$.

That last example shows the real power of substitution beyond simple linear terms. The key question to ask yourself: Is there a function and its derivative both present in the integrand? If yes, substitution is almost certainly the right move.

Beyond Basic Substitution

Substitution is your first tool for integrals that aren't straightforward. Later in calculus, you'll encounter integration by parts (for products of unrelated functions like $x\,e^x$) and trigonometric substitution (for expressions like $\sqrt{a^2 - x^2}$). Substitution also shows up frequently when solving differential equations. Building strong pattern recognition here pays off throughout the rest of the course.