Integration by substitution (u-substitution) — AP Calculus BC
1. What is U-Substitution? ★☆☆☆☆ ⏱ 2 min
Integration by substitution (commonly called u-substitution) is the core technique for integrating composite functions, and it is the inverse of the chain rule for differentiation. It rewrites complex integrands into a form that matches the standard library of basic antiderivatives for power, trigonometric, exponential, and logarithmic functions.
This topic makes up 17–20% of your total AP Calculus BC exam score as part of Unit 6, and it is a prerequisite for all advanced integration techniques covered in BC, including integration by parts, trigonometric substitution, and partial fractions.
2. Indefinite U-Substitution for Composite Functions ★★☆☆☆ ⏱ 3 min
Indefinite u-substitution is used to find the general antiderivative (with constant of integration) of a composite function. It follows directly from reversing the chain rule:
\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)
Integrating both sides gives the core identity:
\int f'(g(x)) g'(x) dx = f(g(x)) + C
To apply substitution, set $u = g(x)$, the inner function of the composite. By the chain rule, $\frac{du}{dx} = g'(x)$, which rearranges to $du = g'(x) dx$. Substituting simplifies the integral to $\int f'(u) du = f(u) + C$, which you can integrate with basic rules, then substitute back $u = g(x)$ to get the final result in terms of $x$. The key requirement is that all $x$-terms must be replaced with $u$-terms after substitution.
3. Definite U-Substitution and Changing Limits ★★☆☆☆ ⏱ 3 min
For definite integrals, the most efficient approach (and the one expected on the AP exam) is to change the limits of integration to match the new variable $u$, eliminating the need to substitute back to $x$ after integration. For an integral over $x = a$ to $x = b$, if $u = g(x)$, the lower limit for $u$ is $g(a)$ and the upper limit for $u$ is $g(b)$. The formula becomes:
\int_{x=a}^{x=b} f'(g(x)) g'(x) dx = \int_{u=g(a)}^{u=g(b)} f'(u) du
This method reduces arithmetic error by removing the extra substitution step required if you first find the indefinite antiderivative in terms of $x$. While you can technically solve for the indefinite antiderivative, substitute back, then evaluate at the original $x$ limits, this adds unnecessary work and room for error.
4. U-Substitution with a Missing Constant Factor ★★★☆☆ ⏱ 3 min
A very common scenario in u-substitution problems is when $du$ does not exactly match the remaining $x$-terms in the integrand, but only differs by a constant multiple. For example, if we have $\int x \cos(x^2) dx$, setting $u = x^2$ gives $du = 2x dx$, but the integrand only has $x dx$. In this case, we can rearrange the $du$ equation to solve for the missing term: $x dx = \frac{du}{2}$. This works only when the difference is a constant factor—you can always pull constant factors out of an integral, so this adjustment is valid.
Common Pitfalls
Why: Students mix the two methods for evaluating definite u-sub, confusing the changed-limits approach with the substitute-back approach.
Why: Students rush through substitution and forget to account for the extra constant factor.
Why: Students do not remember the reverse chain rule structure, which targets the inner function for substitution.
Why: Students remember the antiderivative of $\frac{1}{x}$ but forget the rule carries over to substituted variables.
Why: Students rush to integrate and do not check that all $x$-terms are rewritten.