1.5 Substitution

Big idea

The list of basic antiderivatives is short, and most integrands are not on it. Substitution is the first and most important technique for getting from an integrand that is not on the list to one that is. It is the chain rule run backwards: the chain rule turns a composition into a product of an outer derivative and an inner derivative, so an integrand that already looks like such a product can be collapsed back into a composition.

The pattern to hunt for is a function nested inside another, together with that inner function’s derivative sitting as a factor. Name the inner function $u$, and its derivative times $dx$ becomes $du$. Once every $x$ in the integrand has been replaced, you are left with an integral in $u$ that should be one you recognize. Integrate it, then undo the naming.

Nothing about substitution is guesswork once you see what is being matched. The whole method is one equation, $\int f(g(x))g'(x)\,dx = \int f(u)\,du$, read in whichever direction is convenient. What takes practice is spotting the $g'(x)$ factor when it has been scaled by a constant, split across terms, or hidden inside a rewritten expression.

Definite integrals get one extra decision. The limits $a$ and $b$ are values of $x$, and after the change of variable the integral is in $u$, so those numbers no longer describe the right interval. There are two legitimate ways out, and they are not equally convenient; one of them is worth making a habit.

Decoder

If $u = g(x)$ is differentiable and $f$ is continuous on the range of $g$, then $\int f\big(g(x)\big)g'(x)\,dx = \int f(u)\,du$, and the result is converted back by replacing $u$ with $g(x)$.

Everything hinges on $g'(x)\,dx$ being present. That product is what $du$ means, and if it is absent the substitution does not simplify anything. You may scale it into place: if the integrand carries $x\,dx$ and $du = 2x\,dx$, then $x\,dx = \tfrac{1}{2}\,du$, and the constant $\tfrac{1}{2}$ moves outside the integral freely.

What you may not do is invent the missing factor. Multiplying inside the integral by $x$ and outside by $1/x$ is illegal, because a factor of $x$ is not a constant and does not pass through the integral sign. If the required derivative is genuinely absent, this substitution is the wrong tool for that integrand.

The classic mistake is stopping halfway. An indefinite integral must come back in terms of $x$, so an answer containing $u$ is unfinished. The mirror image of that error shows up in definite integrals: changing the variable without changing the limits leaves an integral in $u$ evaluated at values of $x$, which produces a wrong number silently.

Definitions and results

The substitution rule. For $u = g(x)$ with $g$ differentiable and $f$ continuous on the range of $g$,

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

where $F$ is any antiderivative of $f$.

The differential. Writing $u = g(x)$ gives $du = g'(x)\,dx$. This is a bookkeeping device for rewriting the integrand, and every $dx$ in the integral has to be consumed by it.

Choosing $u$. Look for an inner function: what sits under a root, inside a parenthesis raised to a power, in an exponent, or inside a trigonometric function. Then check whether its derivative appears as a factor, up to a constant multiple.

Constants may be adjusted, variables may not. If $du = k\,g'(x)\,dx$ for a constant $k$, divide by $k$ and continue. There is no corresponding move for a variable factor, since only constants can be moved across the integral sign.

Definite integrals by changing the limits. If $u = g(x)$, then

$$ \int_a^b f\big(g(x)\big)g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du $$

The new limits are the $u$-values at the old endpoints.

Definite integrals by converting back. The alternative is to work out the indefinite integral, return to $x$, and then evaluate at $a$ and $b$. Both routes are correct. Changing the limits is the better habit: it finishes in $u$, so there is no back-substitution step to get wrong, and the new limits act as a reminder that the variable has changed.

When the substitution leaves an extra $x$. If, after setting $u = g(x)$, a lone $x$ survives, solve $u = g(x)$ for $x$ and substitute that too. This rescues integrands like $x\sqrt{x+1}$ that would otherwise look hopeless.

Recognizing the logarithmic pattern. Any integrand of the form $g'(x)/g(x)$ is a substitution away from $\int du/u$, so it integrates to $\ln|g(x)| + C$.

Worked examples

An exact match

Find $\displaystyle\int 2x\sqrt{x^2+1}\,dx$.

The inner function is $x^2+1$, and its derivative $2x$ is sitting right there. Set $u = x^2+1$, so $du = 2x\,dx$ and the integral becomes

$$ \int u^{1/2}\,du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}\big(x^2+1\big)^{3/2} + C $$

Check by differentiating: $\tfrac{2}{3}\cdot\tfrac{3}{2}(x^2+1)^{1/2}\cdot 2x = 2x\sqrt{x^2+1}$. That final factor of $2x$ is the chain rule handing back exactly the piece that was consumed by $du$.

Adjusting by a constant

Find $\displaystyle\int x^2\big(x^3-2\big)^4\,dx$.

Set $u = x^3 - 2$, so $du = 3x^2\,dx$. The integrand has $x^2\,dx$, not $3x^2\,dx$, so write $x^2\,dx = \tfrac{1}{3}\,du$:

$$ \int x^2\big(x^3-2\big)^4 dx = \frac{1}{3}\int u^4\,du = \frac{u^5}{15} + C = \frac{\big(x^3-2\big)^5}{15} + C $$

Differentiating gives $\tfrac{5}{15}(x^3-2)^4(3x^2) = x^2(x^3-2)^4$, as required. Expanding the fourth power instead would have meant integrating a degree-twelve polynomial for the same answer.

A definite integral, done both ways

Evaluate $\displaystyle\int_0^2 2x\big(x^2+1\big)^3\,dx$.

Changing the limits. With $u = x^2+1$ and $du = 2x\,dx$, the endpoint $x = 0$ gives $u = 1$ and $x = 2$ gives $u = 5$. So

$$ \int_0^2 2x\big(x^2+1\big)^3 dx = \int_1^5 u^3\,du = \left[\frac{u^4}{4}\right]_1^5 = \frac{625-1}{4} = 156 $$

Converting back. The indefinite integral is $\tfrac{1}{4}(x^2+1)^4 + C$, and evaluating at the original limits gives $\tfrac{625}{4} - \tfrac{1}{4} = 156$. The two agree, as they must.

Prefer the first route. It ends the moment you evaluate in $u$, and the rewritten limits make it impossible to forget which variable you are in. The second route asks you to carry an unsimplified expression back through a substitution before you are allowed to plug in numbers.

When a stray $x$ survives

Find $\displaystyle\int x\sqrt{x+1}\,dx$.

Set $u = x+1$, so $du = dx$. That handles the root, but the factor $x$ is still in $x$-language. Solve the substitution for $x$: since $u = x+1$, we have $x = u-1$. Now everything converts:

$$ \int x\sqrt{x+1}\,dx = \int (u-1)u^{1/2}\,du = \int\left(u^{3/2} - u^{1/2}\right)du = \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C $$

In terms of $x$,

$$ \int x\sqrt{x+1}\,dx = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C $$

Check at a convenient point by differentiating: the derivative is $(x+1)^{3/2} - (x+1)^{1/2} = (x+1)^{1/2}\big[(x+1) - 1\big] = x\sqrt{x+1}$.

Practice

Start with indefinite integrals. Identify the inner function, write $du$, adjust by a constant if needed, integrate, and convert back to $x$.

Then the definite versions. Change the limits as you change the variable, and evaluate without ever returning to $x$.

Quiz

Five items on choosing a substitution, adjusting constants, and evaluating definite integrals with changed limits.