Math Core

The basic rules handle $\int \cos x\,dx$ , but what about $\int x\cos(x^2)\,dx$ ? Integrands like this come from the chain rule, and substitution is the technique that runs the chain rule in reverse. It is the most important integration technique in AP Calculus AB.

The chain rule, backward

The chain rule says $\dfrac{d}{dx} F\big(u(x)\big) = F'\big(u(x)\big)\,u'(x)$ . Reading that as an antiderivative statement,

\int f\big(u(x)\big)\,u'(x)\,dx = F\big(u(x)\big) + C, \quad \text{where } F' = f.

So look for an integrand that contains a function $u(x)$ together with its derivative $u'(x)$ as a factor. Then rename $u(x)$ as $u$ and replace $u'(x)\,dx$ with $du$ :

\int f\big(u(x)\big)\,u'(x)\,dx = \int f(u)\,du.

u-substitution steps

Choose $u$ : usually the "inside" function, such as the expression inside a power, a root, a trig function or an exponent.
Compute $du = u'(x)\,dx$ .
Rewrite the entire integral in terms of $u$ and $du$ . No $x$ may be left over. Adjust constant factors if needed.
Integrate with respect to $u$ .
Substitute back to write the answer in terms of $x$ .

Worked example: A power of an inside function

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

Let $u = x^2 + 3$ . Then $du = 2x\,dx$ , which is exactly the rest of the integrand.

\int 2x\,(x^2 + 3)^5\,dx = \int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2 + 3)^6}{6} + C.

Check by differentiating: $\dfrac{6(x^2 + 3)^5 \cdot 2x}{6} = 2x(x^2 + 3)^5$ .

Adjusting for a constant factor

Often the derivative of $u$ is present except for a constant factor. Since constants can move in and out of an integral, you can fix this by solving for the piece you have.

Worked example: Fixing a missing constant

Find $\displaystyle\int x\cos(x^2)\,dx$ .

Let $u = x^2$ , so $du = 2x\,dx$ . The integrand has $x\,dx$ , not $2x\,dx$ . Solve: $x\,dx = \dfrac{1}{2}\,du$ .

\int x\cos(x^2)\,dx = \int \cos u \cdot \frac{1}{2}\,du = \frac{1}{2}\sin u + C = \frac{1}{2}\sin(x^2) + C.

A common special case is a linear inside function: $\displaystyle\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + C$ . For example, $\displaystyle\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C$ and $\displaystyle\int \cos(5x)\,dx = \frac{1}{5}\sin(5x) + C$ .

Common mistake

You can only adjust for a constant factor. In $\displaystyle\int \cos(x^2)\,dx$ , the needed factor $2x$ is missing entirely, and you can't pull a variable out of the integral to fix it. Substitution doesn't work there. (That integral has no elementary antiderivative at all.)

Substitution in definite integrals

With a definite integral, the limits $a$ and $b$ are values of $x$ . When you switch to $u$ , you can change the limits to the matching values of $u$ and never go back to $x$ .

Worked example: Changing the limits

Evaluate $\displaystyle\int_0^2 \frac{x}{x^2 + 1}\,dx$ .

Let $u = x^2 + 1$ , so $du = 2x\,dx$ and $x\,dx = \dfrac{1}{2}\,du$ . Change the limits:

when $x = 0$ , $u = 1$ ;
when $x = 2$ , $u = 5$ .

\int_0^2 \frac{x}{x^2 + 1}\,dx = \frac{1}{2}\int_1^5 \frac{1}{u}\,du = \frac{1}{2}\Big[\ln\lvert u\rvert\Big]_1^5 = \frac{1}{2}(\ln 5 - \ln 1) = \frac{1}{2}\ln 5.

Tip

If you'd rather not change the limits, you can find the antiderivative in terms of $u$ , substitute back to $x$ , and then use the original limits. What you can't do is mix them: $u$ -limits with an $x$ -antiderivative, or the reverse, gives a wrong answer.

Algebra first: long division and completing the square

Some integrands need to be rearranged before any rule or substitution applies.

Long division. When a rational function's numerator has degree at least as large as its denominator's, divide first.

Worked example: Divide, then integrate

Find $\displaystyle\int \frac{x^2 + 2x + 3}{x + 1}\,dx$ .

Rewrite the numerator: $x^2 + 2x + 3 = (x + 1)^2 + 2 = (x + 1)(x + 1) + 2$ . So

\frac{x^2 + 2x + 3}{x + 1} = x + 1 + \frac{2}{x + 1}.

Integrate each piece (the last one with $u = x + 1$ ):

\int \left(x + 1 + \frac{2}{x + 1}\right) dx = \frac{x^2}{2} + x + 2\ln\lvert x + 1\rvert + C.

Completing the square. A quadratic denominator with no real roots can be rewritten as $(x - h)^2 + k^2$ , which leads to arctangent. When $k = 1$ , the substitution $u = x - h$ gives

\int \frac{1}{(x - h)^2 + 1}\,dx = \arctan(x - h) + C.

For instance, $x^2 + 4x + 5 = (x + 2)^2 + 1$ , so $\displaystyle\int \frac{dx}{x^2 + 4x + 5} = \arctan(x + 2) + C$ .

Practice

Practice 1

Find the antiderivative $F$ of $f(x) = 3x^2(x^3 + 1)^4$ that satisfies $F(0) = 0$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

$\displaystyle\int \cos(5x)\,dx =$

Practice 3

Evaluate $\displaystyle\int_0^{\pi/2} \sin^3 x \cos x\,dx$ .