Integration by substitution

Substitution is a change of variable, not a new integration rule — it is a way of packaging the chain rule so an unfamiliar integral becomes a familiar one. The formal statement:

where: $g(x)$ is the "inner function," $t = g(x)$ is the new variable, $g'(x) = \dfrac{dt}{dx}$ is its derivative (which must be present, up to a constant multiple, elsewhere in the integrand), and $F(t)$ is any antiderivative of $f(t)$.

Recognizing when to substitute. Scan the integrand for an inner expression that looks awkward standing alone — a bracket raised to a power, a square root, an exponent, an angle multiplied by a constant — and check whether its derivative (up to a constant factor) is sitting nearby. If it is, that expression is almost always the right choice for $t$. Four patterns recur constantly:

• $\int (ax+b)^n\,dx$: put $t = ax + b$.
• $\int \dfrac{f'(x)}{f(x)}\,dx$: put $t = f(x)$; this always gives $\ln|f(x)| + C$.
• $\int f'(x)\,e^{f(x)}\,dx$: put $t = f(x)$; gives $e^{f(x)} + C$.
• $\int [f(x)]^n\,f'(x)\,dx$ (including $\int \dfrac{f'(x)}{2\sqrt{f(x)}}\,dx$): put $t = f(x)$.

A second family is trigonometric substitution, where $x$ itself is replaced (e.g. $x = a\sin\theta$) to tame expressions like $\sqrt{a^2 - x^2}$, turning them into $a\cos\theta$. This is used heavily for the standard square-root forms later in the chapter, but the mechanics — replace, integrate, substitute back — stay identical.

Two habits matter throughout: always substitute back to the original variable before writing a final answer, and always check for a missing constant factor between $g'(x)$ and the differential term actually present in the integrand.

Evaluate $\displaystyle\int 2x\,(x^2+3)^5\,dx$.

Step 1. Identify the inner function. Here $g(x) = x^2 + 3$, and its derivative $g'(x) = 2x$ is exactly the other factor present in the integrand — a strong signal to substitute with $t = g(x)$.

Step 2. Substitute. Let $t = x^2 + 3$, so $\dfrac{dt}{dx} = 2x$, giving $dt = 2x\,dx$. The integral becomes purely in $t$:

Step 3. Integrate in $t$ and substitute back. $\displaystyle\int t^5\,dt = \dfrac{t^6}{6} + C$, and replacing $t = x^2+3$ gives

Step 4. Verify by differentiating: $\dfrac{d}{dx}\left[\dfrac{(x^2+3)^6}{6}\right] = \dfrac{6(x^2+3)^5 \cdot 2x}{6} = 2x(x^2+3)^5$, which matches the original integrand exactly.

Final answer: $\dfrac{(x^2+3)^6}{6} + C$.

Competitive stretch

Evaluate $\displaystyle\int \tan x\,dx$.

Step 1. Rewrite $\tan x = \dfrac{\sin x}{\cos x}$ and notice the numerator $\sin x$ is (up to sign) the derivative of the denominator, since $\dfrac{d}{dx}(\cos x) = -\sin x$.

Step 2. Substitute $t = \cos x$, so $dt = -\sin x\,dx$, i.e. $\sin x\,dx = -dt$. This turns the integral into $\displaystyle\int \dfrac{-dt}{t} = -\ln|t| + C$.

Step 3. Substitute back $t = \cos x$ and simplify using $-\ln|\cos x| = \ln\left|\dfrac{1}{\cos x}\right|$.

Final answer: $\displaystyle\int \tan x\,dx = -\ln|\cos x| + C = \ln|\sec x| + C$, one of the standard integrals worth memorizing directly, though deriving it once shows substitution recognizing the $\int \frac{f'(x)}{f(x)}dx$ pattern in disguise.

• Substituting $t = g(x)$ but forgetting to also replace $dx$ correctly — every part of the original integral, including the differential, must be rewritten in terms of $t$.
• Leaving a stray $x$ behind after substitution — a valid substitution must leave the integral entirely in terms of $t$ (or a constant that can be pulled outside the integral sign).
• Missing a constant factor: treating $\int x(x^2+3)^5\,dx$ (missing the coefficient $2$) as if it matched the worked example directly. Since $dt = 2x\,dx$, here $x\,dx = \tfrac12 dt$, so a compensating factor of $\tfrac12$ is required.
• Forgetting to substitute back to $x$ at the end, leaving the final answer stranded in terms of $t$ — a definite/indefinite integral in $x$ must be answered in $x$.

Ask the tutor: not sure which substitution to try for a tricky integrand — describe it to the AI tutor and ask for a hint on spotting the derivative pairing.