Calculus I · Adding Up the Infinite

Substitution: the chain rule, run backward

Most antiderivatives aren’t in the table — but many can be transformed into ones that are, by the single most useful integration technique, and it is precisely the chain rule reversed. Recall the chain rule: $\tfrac{d}{dx}F(g(x)) = F'(g(x))\,g'(x)$. Read that equation from right to left and it becomes an antiderivative rule:

$$\int F'(g(x))\,g'(x)\,dx = F(g(x)) + C.$$

In practice you spot an “inside function” $g(x)$ whose derivative $g'(x)$ is also sitting in the integrand (up to a constant factor), and you rename: let $u = g(x)$, so that $du = g'(x)\,dx$. The integral collapses into one in $u$ that you can do from the table.

Example. $\displaystyle\int 2x(x^2+1)^5\,dx$. The inside is $g(x) = x^2+1$, and its derivative $2x$ is right there. Let $u = x^2+1$, $du = 2x\,dx$: $$\int 2x(x^2+1)^5\,dx = \int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C.$$ Differentiate the answer (chain rule!) and you get $2x(x^2+1)^5$ back — substitution and the chain rule are genuinely inverse moves.

Example. $\displaystyle\int x\,e^{x^2}\,dx$. Inside is $x^2$, derivative $2x$ — present up to the factor of $2$. Let $u = x^2$, $du = 2x\,dx$, so $x\,dx = \tfrac12\,du$: $$\int x\,e^{x^2}\,dx = \int e^u \cdot \tfrac12\,du = \tfrac12 e^u + C = \tfrac12 e^{x^2} + C.$$

Example — the logarithm pattern. $\displaystyle\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx$. The inside is $\cos x$, and its derivative $-\sin x$ is sitting on top (up to the minus sign). Let $u = \cos x$, $du = -\sin x\,dx$: $$\int \frac{\sin x}{\cos x}\,dx = \int \frac{-1}{u}\,du = -\ln|u| + C = -\ln|\cos x| + C.$$ Whenever the top of a fraction is the derivative of the bottom, substitution turns the whole thing into a logarithm — a pattern worth knowing on sight: $\displaystyle\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)| + C$.

This is the mirror image of Chapter 4. There, the product rule and chain rule were tools for taking derivatives; here, reversed, they become tools for finding antiderivatives. (The product rule’s reverse — integration by parts — is the other great technique, and it arrives in Chapter 11 as a payoff of the Fundamental Theorem.) For now, substitution applies cleanly to indefinite integrals; using it on definite integrals, where the limits themselves transform, is a Chapter 11 refinement.

Toolkit: Substitution

1. Find an inside function $u = g(x)$ whose derivative $g'(x)$ appears in the integrand (up to a constant).
2. Compute $du = g'(x)\,dx$ and rewrite the entire integral in terms of $u$ — no $x$ may remain.
3. Integrate in $u$ using the table, then substitute $x$ back.
4. Always check by differentiating your answer. Substitution is the chain rule reversed, so the chain rule is your proofreader.

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