Topic 6.9 Notes – Integrating Using Substitution

1. What Substitution Is and Why It Works

Think back to the Chain Rule:

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

When you integrate, you’re undoing that pattern. So if you see something that looks like

$$\int f'(g(x)) \cdot g'(x)\,dx$$

you can reverse the process.

Substitution means:

• Let $u = g(x)$ (the inner function).
• Then $du = g'(x)\,dx$.
• Rewrite the integral entirely in terms of $u$.

You’re basically renaming the inside to clean up the expression.

A quick example

$$\int 6x(3x^2+5)^4\,dx$$

• Let $u = 3x^2+5$
• Then $du = 6x\,dx$

The integral becomes:

$$\int u^4\,du = \frac{u^5}{5} + C$$

Back-substitute:

$$\frac{(3x^2+5)^5}{5} + C$$

That worked because we had a function and its derivative multiplied together.

If it looks like Chain Rule in reverse, substitution is usually the move.

2. The Substitution Process

Indefinite Integrals

When there are no bounds:

1. Choose $u$
   Pick the inside of parentheses, radicals, trig, exponentials, denominators, etc.
2. Differentiate
   Find $du = g'(x)\,dx$.
3. Rewrite everything in terms of $u$
   No $x$'s should remain.
4. Integrate
5. Back-substitute
6. Add +C

If an $x$ is still floating around, you’re not done rewriting.

Definite Integrals

For definite integrals, you must deal with bounds correctly.

Suppose:

$$\int_{0}^{2} x e^{x^2}\,dx$$

Let $u = x^2$, so $du = 2x\,dx$.

We only have $x\,dx$, so rewrite:

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

Now change the bounds:

• When $x=0$, $u=0$
• When $x=2$, $u=4$

So the integral becomes:

$$\frac{1}{2}\int_{0}^{4} e^u\,du$$

Now integrate and evaluate. No need to substitute back.

If you switch to $u$, your limits must also switch to $u$.
Leaving x-bounds with a u-integral is a guaranteed point loss on an FRQ.

3. When Substitution Is the Right Tool

Here’s what usually signals substitution:

A. Composite function with its derivative

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

Let $u=5x$. Then $du=5dx$.
Adjust by multiplying/dividing by 5.

B. Power of an expression

$$\int (x^3 - 2)^7 \cdot 3x^2\,dx$$

The outer power and inner derivative are both present.

C. Rational form where numerator is derivative of denominator

$$\int \frac{4x}{x^2+3}\,dx$$

Let $u = x^2+3$. This becomes:

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

These show up a lot on AP multiple choice.

D. Exponentials with linear inside

$$\int e^{7x-1}\,dx$$

Let $u=7x-1$. Adjust by $1/7$.

Students often forget that small constant factor.

4. Algebra That Makes Substitution Work

Sometimes you’re close but not quite there.

Common adjustments:

• Factor constants out
• Multiply and divide to create the missing derivative
• Rewrite radicals as powers
• Rewrite fractions with negative exponents

Example:

$$\int \frac{x}{\sqrt{x^2+9}}\,dx$$

Rewrite as:

$$\int \frac{x}{(x^2+9)^{1/2}}\,dx$$

Let $u=x^2+9$, $du=2x\,dx$.
You’re off by a factor of 2, so adjust with $1/2$.

Small algebra tweaks often unlock the substitution.

5. Common Mistakes That Cost Points

• Leaving an $x$ in the integral after switching to $u$.
• Forgetting to change bounds on definite integrals.
• Forgetting to back-substitute on indefinite integrals.
• Choosing a messy $u$ that makes things worse.
• Dropping constant factors when adjusting for $du$.

On AP FRQs, they look closely at setup. Even if algebra slips later, a correct substitution setup usually earns credit.