Substitution

Every differentiation rule runs backwards into an integration technique. The chain rule runs backwards into substitution, which is the one technique AB needs and the foundation of every technique BC adds.

The chain rule says

$$\frac{d}{dx}F\big(u(x)\big) = F'\big(u(x)\big)\cdot u'(x)$$

Read right to left: anything of the form $f(u)\cdot u'$ integrates to $F(u)$. The whole method is recognising that shape.

The shape to look for

An integrand that contains a function and its own derivative, up to a constant factor.

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

The inside is $x^2$; its derivative is $2x$; and $2x$ is sitting right there. Substitute $u = x^2$, so $du = 2x\,dx$, and the integral becomes $\int\cos u\,du = \sin u = \sin(x^2)$.

Both directions checked. The second line is the honest test of any integration: differentiate your answer and see whether you get the integrand back. That takes seconds and catches nearly every mistake, which makes skipping it inexcusable on an exam.

Working the bookkeeping

$$\int x\,e^{x^2}\,dx$$

The derivative of $x^2$ is $2x$ and only $x$ is present — off by a constant, which is always fixable.

Let $u = x^2$, so $du = 2x\,dx$, hence $x\,dx = \tfrac12\,du$:

$$\int x e^{x^2}dx = \tfrac12\int e^u\,du = \tfrac12 e^u = \tfrac12 e^{x^2}$$

Definite integrals: change the limits

With definite integrals there are two routes, and one is reliably safer.

Substitute the limits too. When $u = x^2$ and $x$ runs from $0$ to $2$, then $u$ runs from $0$ to $4$:

$$\int_0^2 x e^{x^2}dx = \tfrac12\int_0^4 e^u\,du = \tfrac12\left(e^4 - 1\right)$$

Or convert back to $x$ first, then use the original limits. Both are correct; the first is less error-prone, because forgetting to convert back is a standard way to evaluate the antiderivative at the wrong numbers.

Choosing $u$

When more than one candidate presents itself, these heuristics hold up:

• The inside of a composition. In $\cos(x^2)$, take $u = x^2$.
• Whatever is under a root, inside a denominator, or in an exponent.
• The thing whose derivative is also present, which is the real criterion — the others are shortcuts to it.

A quick test before committing: compute $du$ and see whether the rest of the integrand is exactly $du$ up to a constant. If a stray $x$ survives, that choice of $u$ is wrong. Trying the other candidate takes fifteen seconds.

Those three are all the same move. In the first, $u = \ln x$ gives $du = \tfrac{dx}{x}$, which is exactly what remains. In the second, $\tan x = \tfrac{\sin x}{\cos x}$ and $u = \cos x$ gives $du = -\sin x\,dx$. In the third, $u = x^2 + 1$ gives $du = 2x\,dx$, off by the usual constant.

Once the pattern is visible, these stop being separate problems.

Practice