Integrating by Substitution — AP® Calculus AB Notes

The basic rules handle $\cos x$ and $x^5$ , but not $2x\cos(x^2)$ or $x(1 + x^2)^3$ . Those integrands came from the chain rule, and u-substitution is the chain rule run backward.

Spotting the chain rule in an integrand

By the chain rule, $\frac{d}{dx}\sin(x^2)$ $= \cos(x^2) \cdot 2x$ . So

$\int 2x\cos(x^2)\,dx = \sin(x^2) + C.$

The pattern: an inside function $g(x) = x^2$ , an outside function applied to it, and the inside's derivative $g'(x) = 2x$ as a factor. Whenever you see $f(g(x))\,g'(x)$ , the answer is $F(g(x)) + C$ , where $F$ is an antiderivative of $f$ .

The method

Choose $u$ : an inside function whose derivative is also in the integrand (up to a constant factor).
Find $du = g'(x)\,dx$ .
Rewrite the whole integral in $u$ . No $x$ may remain.
Integrate in $u$ , then substitute $u = g(x)$ back.
Check by differentiating.

Worked example 1. Find $\int \frac{x}{x^2 + 4}\,dx$ .

Inside: $u = x^2 + 4$ , so $du = 2x\,dx$ and $x\,dx = \frac{1}{2}du$ .

Rewrite: $\int \frac{1}{u}\cdot\frac{1}{2}\,du$ $= \frac{1}{2}\ln|u| + C$ .

Back to $x$ : $\frac{1}{2}\ln(x^2 + 4) + C$ (no absolute value needed: $x^2 + 4 > 0$ ).

Check: $\frac{d}{dx}\left[\frac{1}{2}\ln(x^2 + 4)\right]$ $= \frac{1}{2}\cdot\frac{2x}{x^2 + 4}$ $= \frac{x}{x^2 + 4}$ .

A missing constant factor is fine: multiply and divide by it, as with the $\frac{1}{2}$ above. A missing variable factor is not: $\int (x^2 + 1)^5\,dx$ has no $x$ to match $du = 2x\,dx$ , so substitution fails (expand the power instead).

Linear inside functions

When the inside is $kx$ or $kx + b$ , $du = k\,dx$ , so the rule is just "divide by $k$ ":

$\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C, \qquad \int \cos(5x)\,dx = \frac{1}{5}\sin(5x) + C.$

Check by differentiating: the chain rule puts the $k$ back.

Definite integrals: change the limits

Left: the graph of y = 2x cos(x squared) from x = 0 to the square root of pi/2, about 1.25, with the region under it shaded; its area is 1. Right: the graph of y = cos u from u = 0 to pi/2, with the region under it shaded; its area is also 1. With u = x squared, the x-limits become the u-limits. — **Figure 1.** Same area, new variable, new limits. LevlPrep original diagram.

In a definite integral, the limits are values of the variable. When the variable changes from $x$ to $u$ , the limits must change too: $x = a$ becomes $u = g(a)$ and $x = b$ becomes $u = g(b)$ . The figure shows the idea: the region under $2x\cos(x^2)$ on $[0, \sqrt{\pi/2}]$ and the region under $\cos u$ on $[0, \frac{\pi}{2}]$ have the same area.

Worked example 2. Evaluate $\int_0^1 x(1 + x^2)^3\,dx$ .

$u = 1 + x^2$ , $du = 2x\,dx$ , so $x\,dx = \frac{1}{2}du$ .

Limits: $x = 0 \to u = 1$ ; $x = 1 \to u = 2$ .

$\frac{1}{2}\int_1^2 u^3\,du$ $= \frac{1}{2}\cdot\frac{u^4}{4}\Big|_1^2$ $= \frac{1}{8}(16 - 1)$ $= \frac{15}{8}$ .

The common wrong answer, $\frac{1}{8}$ , comes from evaluating $\frac{u^4}{8}$ at the old limits 0 and 1.

Write the $u$ -limits on the integral sign as soon as you change variables. Mixing a $u$ -integrand with $x$ -limits is a linkage error the scoring guidelines penalize.

Substitution with tables

The same idea works with functions given only by values. Since $\frac{d}{dx}f(g(x))$ $= f'(g(x))g'(x)$ ,

$\int_1^3 f'(g(x))\,g'(x)\,dx = f(g(3)) - f(g(1)).$

Read $g(3)$ and $g(1)$ from the table first, then $f$ at those inputs.

Choosing u

Integrand	Choose u	Result
$2x\cos(x^2)$	$x^2$	$\sin(x^2) + C$
$\sin^3 x\cos x$	$\sin x$	$\frac{\sin^4 x}{4} + C$
$\frac{\ln x}{x}$	$\ln x$	$\frac{(\ln x)^2}{2} + C$
$g'(x)e^{g(x)}$	$g(x)$	$e^{g(x)} + C$