6.9 Integrating Using Substitution

U-substitution, also called change of variables, is the integration technique that reverses the chain rule. You pick an inner function $u=g(x)$, replace its derivative factor with $du$, integrate in terms of $u$, and then switch back to $x$ or update the limits for a definite integral. For AP Calculus, make every part of the integral use either $u$-values or $x$-values before evaluating.

Why This Matters for the AP Calculus Exam

Substitution shows up across Unit 6, which carries one of the larger weightings on both the AB and BC exams. It is the most common antiderivative technique you will reach for, and you need to recognize when an integrand is a factor of a chain-rule derivative without being told to substitute. On both multiple-choice and free-response questions, fast and accurate substitution lets you evaluate integrals that come up inside accumulation problems, area and volume setups, differential equations, and motion problems later in the course.

When you evaluate a definite integral by hand, present a correct antiderivative and use clear notation. Include the constant of integration with indefinite integrals, watch your parentheses, and avoid stringing together equal signs between expressions that are not actually equal. This keeps your work readable and easy to follow.

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Key Takeaways

• U-substitution reverses the chain rule: look for an integrand in the form $f(g(x)) \cdot g'(x)$.
• Choose $u = g(x)$ so that $g'(x)$ (or a constant multiple of it) already appears in the integrand.
• Find $du = g'(x)\,dx$, then replace every $x$-piece so the integral is fully in terms of $u$.
• For a definite integral, either change the limits to $u$-values or back-substitute to $x$ before using the original limits. Do not mix the two.
• After integrating in $u$, back-substitute to $x$ for indefinite integrals and add $+C$.
• If the derivative factor is off by only a constant, factor that constant out front; if it is off by a variable factor, substitution alone will not work.

How U-Substitution Works

Substitution replaces part of the integrand with a new variable to turn a messy integral into one you recognize. You usually choose the new variable based on an inner function whose derivative is already present in the integral.

Recall the Chain Rule: $\frac{d}{dx}(f(g(x)))=f'(g(x))\cdot g'(x)$ Since antidifferentiation reverses differentiation, u-substitution works as the reverse of the chain rule. You identify an inner expression whose derivative also shows up in the integrand, which is exactly the pattern the chain rule produces.

Problem Solving

The step-by-step process:

1. Identify the inner function. Look for an expression, often inside a root, a trig function, an exponent, or a denominator, whose derivative also appears in the integrand.
2. Choose the new variable. Set $u$ equal to that inner function.
3. Differentiate. Find $\frac{du}{dx}$ and solve for $du$. Don't forget the chain rule here.
4. Rewrite the integral. Substitute $u$ and $du$ so the whole integral is in terms of $u$. For a definite integral, you can change the limits of integration to $u$-values now.
5. Simplify. Make sure no $x$ remains. The integral should now match a basic antiderivative form.
6. Evaluate. Integrate with respect to $u$.
7. Back-substitute. Replace $u$ with the original expression in $x$ and add $+C$ for indefinite integrals. If you already changed the limits, just evaluate at the new $u$-bounds.

This technique takes practice, but once you internalize the pattern it becomes one of your fastest tools.

U-Substitution Practice Problems

1) Basic Substitution with an Indefinite Integral

Evaluate: $\int 2x\cos(x^2)\,dx$

Identify the inner function. You can integrate $2x$ and $\cos(x)$ on their own, but $\cos(x^2)$ is harder because of the inner function $x^2$. That inner function is what you want to replace.

Choose the new variable. $u = x^2$

Differentiate. $\frac{du}{dx} = 2x \quad\Rightarrow\quad du = 2x\,dx$ Notice that $2x\,dx$ already appears in the integral.

Rewrite the integral. $\int \cos(x^2)\,2x\,dx = \int \cos(u)\,du$

Evaluate. $\int \cos(u)\,du = \sin(u) + C$

Back-substitute. $\int 2x\cos(x^2)\,dx = \sin(x^2) + C$

2) Substitution with a Definite Integral

Evaluate: $\int_{1}^{2}\frac{2x}{(x^2+1)^2}\,dx$

There are two valid approaches with definite integrals:

1. Change the limits of integration as you substitute, or
2. Back-substitute to $x$ before evaluating with the original limits.

Both give the same answer, so use whichever feels cleaner.

Identify the inner function. The expression $x^2 + 1$ works, since its derivative $2x$ is present in the integrand.

Choose the new variable and differentiate. $u = x^2 + 1$ $du = 2x\,dx$

Method 1: Changing the Limits of Integration

Convert both bounds to $u$-values using $u = x^2 + 1$.

Lower bound: $x = 1 \Rightarrow u = (1)^2 + 1 = 2$ Upper bound: $x = 2 \Rightarrow u = (2)^2 + 1 = 5$

Substitute the new bounds and the $u$-expression: $\int_{2}^{5} \frac{1}{u^2}\,du = \int_{2}^{5} u^{-2}\,du$ $= \left. \frac{-1}{u} \right|_{2}^{5}$

Since everything is already in terms of $u$, evaluate directly: $\left. \frac{-1}{u} \right|_{2}^{5} = \left( \frac{-1}{5} \right) - \left( \frac{-1}{2} \right)$

So: $\int_{1}^{2}\frac{2x}{(x^2+1)^2}\,dx = \frac{3}{10}$

Method 2: Substituting Back

Here you temporarily set the bounds aside, then restore them at the end. Since the rewritten integral is in terms of $u$, leave off the $x$-bounds for now: $\int \frac{1}{u^2}\,du = \int u^{-2}\,du = \frac{-1}{u}$

Replace $u$ with its original expression: $\frac{-1}{u} = \frac{-1}{x^2+1}$

Now the antiderivative is in terms of $x$, so apply the original limits: $\left.\frac{-1}{x^2+1}\right|_1^{2} = \frac{-1}{5} - \left(\frac{-1}{2}\right)$

So: $\int_{1}^{2}\frac{2x}{(x^2+1)^2}\,dx = \frac{3}{10}$

Both methods reach the same value, which is a good way to check yourself.

How to Use This on the AP Calculus Exam

MCQ

Many multiple-choice integrals are designed to test whether you spot the substitution quickly. Scan for an inner function whose derivative (up to a constant) is already in the integrand. If you can match the $f(g(x))\cdot g'(x)$ pattern, substitution is usually the intended path.

Free Response

When a free-response integral requires substitution by hand, write a correct antiderivative and show the substitution clearly. Add $+C$ on indefinite integrals, keep parentheses tidy, and avoid equal-sign chains between unequal expressions. For definite integrals, either change the limits or back-substitute, but be explicit about which one you did.

Common Trap

If your $du$ is off by only a constant factor, you can fix it by multiplying and dividing by that constant. If it is off by a variable factor that you cannot account for, plain substitution will not finish the integral, and you need a different approach.

Common Misconceptions

• Integration is not just differentiation in reverse order. You have to recognize structure and choose a strategy; you can't blindly undo steps.
• Forgetting to change $dx$ to $du$. You must replace the differential, not just swap the inner function. Every $x$, including $dx$, has to become a $u$-expression.
• Leaving the limits in $x$ after changing variables. If you switch to $u$ and keep the original $x$-bounds, you will evaluate at the wrong numbers. Either convert the limits to $u$ or back-substitute first.
• Substituting when the derivative factor isn't there. If $g'(x)$ is missing and can't be created by adjusting a constant, $u$-substitution alone won't work.
• Dropping the constant of integration. Indefinite integrals need $+C$; only definite integrals skip it.
• Mixing the two definite-integral methods. Pick changing the limits or back-substituting, not a blend of both.

Related AP Calculus Guides

• Unit 6 Overview: Integration and Accumulation of Change
• 6.11 Integrating Using Integration by Parts
• 6.1 Integration and Accumulation of Change
• 6.12 Integrating Using Linear Partial Fractions
• 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
• 6.4 The Fundamental Theorem of Calculus and Accumulation Functions