Substitution for a hidden inner variable, integrals from easy to advanced lesson

What is the exact value of $\int_0^{\sqrt{\pi/2}} 2x\cos(x^2)\,dx$ ? The function $\cos(x^2)$ on its own has no antiderivative built from elementary functions, but here it is multiplied by $2x$ , the derivative of the expression $x^2$ inside the cosine. This page shows how that pairing lets a new variable $u = x^2$ absorb the whole factor $2x\,dx$ , how the bounds move with the variable, why the change of variable is legitimate on this interval, and why the integral equals $1$ .

Key terms

Inner function

Definition In a composition $f(g(x))$ , the function $g$ is the inner function and $f$ is the outer function. The value $g(x)$ is computed first and then used as the input of $f$ . For substitution the inner function is the natural candidate for the new variable $u = g(x)$ , and the integrand is ready for the substitution rule when the remaining factor is $g'(x)$ or a constant multiple of it.

In plain words Look for an expression that sits inside another function, such as the input of a cosine, the base of a power or the exponent of $e$ . Differentiate that expression and look for the result among the other factors.

Example In $2x\cos(x^2)$ the inner function is $x^2$ , the outer function is cosine, and the derivative $2x$ of the inner function is exactly the other factor. In $x\cos(x^2)$ the factor $x$ is half of that derivative, which is still enough, because $x\,dx = \tfrac12\,du$ . In $\cos(x^2)$ alone the derivative of the inside is missing, and this function has no elementary antiderivative at all.

Chain rule in reverse

Definition Let $g$ be differentiable on an interval $I$ , and let $F$ be an antiderivative of $f$ on an interval that contains every value $g(x)$ with $x$ in $I$ . The chain rule gives

\begin{aligned} \frac{d}{dx}F(g(x)) &= F'(g(x))\,g'(x) \\[0.4em] &= f(g(x))\,g'(x), \end{aligned}

so $F(g(x))$ is an antiderivative of $f(g(x))\,g'(x)$ on $I$ , and

\begin{aligned} &\int f(g(x))\,g'(x)\,dx \\[0.4em] &\quad = F(g(x)) + C. \end{aligned}

In plain words Every chain rule derivative has the shape of an outer derivative, evaluated at the inside, times the derivative of the inside. An integrand with that shape can be integrated by recognizing which function was differentiated.

Example Since

\frac{d}{dx}\sin(x^2) = \cos(x^2)\cdot 2x,

it follows that $\int 2x\cos(x^2)\,dx = \sin(x^2) + C$ on the whole real line. In the same way $\int \cos x\,e^{\sin x}\,dx = e^{\sin x} + C$ . Differentiating the answer and getting the integrand back is the check that the reversal was done correctly.

Differential du

Definition For $u = g(x)$ with $g$ differentiable, the differential of $u$ is $du = g'(x)\,dx$ . In a substitution the whole product $g'(x)\,dx$ is replaced by $du$ , and a constant multiple is handled the same way, $c\,g'(x)\,dx = c\,du$ . The notation records the linear part of the exact change of $u$ ,

\begin{aligned} \Delta u &= g(x + \Delta x) - g(x) \\[0.4em] &= g'(x)\,\Delta x + \varepsilon\,\Delta x, \end{aligned}

where $\varepsilon \to 0$ as $\Delta x \to 0$ .

In plain words When $g$ is increasing, a thin strip of width $\Delta x$ at position $x$ corresponds to a strip on the $u$ -axis whose width is close to $g'(x)\,\Delta x$ . The factor $g'(x)$ in the integrand measures this stretching, so it is absorbed into $du$ rather than dropped.

Example For $u = x^2$ the differential is $du = 2x\,dx$ , and the exact change is

\begin{aligned} \Delta u &= (x + \Delta x)^2 - x^2 \\[0.4em] &= 2x\,\Delta x + (\Delta x)^2. \end{aligned}

At $x = 1$ with $\Delta x = 0.1$ the exact change is $0.21$ and the linear part $2x\,\Delta x$ is $0.2$ . The extra term $(\Delta x)^2$ shrinks faster than $\Delta x$ , so it disappears in the limit of thin strips. In the integral the group $2x\,dx$ becomes $du$ and $\cos(x^2)$ becomes $\cos u$ , so $2x\cos(x^2)\,dx$ becomes $\cos u\,du$ . Replacing only $x^2$ by $u$ and leaving $2x\,dx$ in place would mix two variables in one integral.

U-substitution

Definition Let $g$ be continuously differentiable on $[a, b]$ , and let $f$ be continuous on an interval that contains every value $g(x)$ with $a \le x \le b$ . Then

\begin{aligned} &\int_a^b f(g(x))\,g'(x)\,dx \\[0.4em] &\quad = \int_{g(a)}^{g(b)} f(u)\,du. \end{aligned}

To use it, set $u = g(x)$ , replace $g'(x)\,dx$ by $du$ , and replace the bounds $a$ and $b$ by $g(a)$ and $g(b)$ . The rule holds as stated even when $g(b) < g(a)$ and even when $g$ is not one-to-one. Because $f$ is continuous it has an antiderivative $F$ , and the fundamental theorem, applied to $F(g(x))$ on the left and to $F(u)$ on the right, makes both sides equal to $F(g(b)) - F(g(a))$ .

In plain words Rename the inner expression as a new coordinate. Its derivative, together with $dx$ , becomes the width $du$ of a strip in the new coordinate, and the endpoints move to where the inner function sends them. Both integrals describe the same accumulated total in different coordinates.

Example For the integral of this page take $u = x^2$ , so $du = 2x\,dx$ . The function $x^2$ is continuously differentiable, and cosine is continuous on $[0, \pi/2]$ . The bounds $x = 0$ and $x = \sqrt{\pi/2}$ become $u = 0$ and $u = \pi/2$ , and

\begin{aligned} &\int_0^{\sqrt{\pi/2}} 2x\cos(x^2)\,dx \\[0.4em] &\quad = \int_0^{\pi/2} \cos u\,du \\[0.4em] &\quad = \sin\frac{\pi}{2} - \sin 0 = 1. \end{aligned}

Differentiating $\sin(x^2)$ returns $2x\cos(x^2)$ , which confirms the antiderivative in the original variable. The size is plausible too. On $[0, \pi/2]$ the cosine lies between $0$ and $1$ , so its integral lies between $0$ and $\pi/2 \approx 1.571$ , and the original integrand is nonnegative on its interval because $2x$ and $\cos(x^2)$ both are.

Changing the bounds

Definition The bounds of a definite integral are values of its variable of integration. When $u = g(x)$ and $x$ runs from $a$ to $b$ , the new integral runs from $u = g(a)$ to $u = g(b)$ , in that order even when $g(a) > g(b)$ . The alternative is to find an antiderivative in $u$ , substitute $u = g(x)$ back, and evaluate at the original bounds $a$ and $b$ . Both routes give the same number.

In plain words The new limits are the old endpoints written in the new coordinate. An integrand in $u$ with bounds in $x$ answers a different question.

Example For $u = x^2$ the endpoints $x = 0$ and $x = \sqrt{\pi/2}$ map to $u = 0$ and $u = \pi/2$ . Returning to $x$ instead gives the antiderivative $\sin(x^2)$ and the same value, $\sin(\pi/2) - \sin 0 = 1$ . Keeping the old bounds with the new integrand would give

\begin{aligned} \int_0^{\sqrt{\pi/2}} \cos u\,du &= \sin\bigl(\sqrt{\pi/2}\bigr) \\[0.4em] &\approx 0.950, \end{aligned}

which is wrong, even though it looks reasonable.

One-to-one substitution

Definition A substitution $u = g(x)$ is one-to-one on $[a, b]$ when different points of $[a, b]$ give different values of $u$ . For a continuous $g$ this means that $g$ is strictly increasing or strictly decreasing on $[a, b]$ , so every $u$ between $g(a)$ and $g(b)$ comes from exactly one $x$ , written $x = g^{-1}(u)$ . The substitution rule above does not require this, because the integrand already contains the whole factor $g'(x)\,dx$ . It is required when part of the integrand must be rewritten through $x = g^{-1}(u)$ , since one formula for $x$ in terms of $u$ then has to hold on the whole interval. When $g$ is not one-to-one, split the interval into pieces on which it is.

In plain words If the integrand already has the shape $f(g(x))\,g'(x)$ , substitute without further checks. If you have to solve $u = g(x)$ for $x$ , make sure each value of $u$ comes from only one $x$ .

Example On $[0, \sqrt{\pi/2}]$ every $x$ is nonnegative, so $x^2$ is strictly increasing and the substitution is one-to-one. Its derivative $2x$ is zero at the endpoint $x = 0$ , which does no harm, because the rule replaces the product $2x\,dx$ by $du$ and never divides by $2x$ . On $[-1, 1]$ the same substitution is not one-to-one, yet the rule still applies directly and gives

\begin{aligned} &\int_{-1}^{1} 2x\cos(x^2)\,dx \\[0.4em] &\quad = \int_1^1 \cos u\,du = 0, \end{aligned}

which is correct, because the integrand is an odd function.

Check yourself

Evaluate the integral

\int_0^1 x(x^2+1)^3\,dx.

Show the answer

The inner function is $x^2 + 1$ , with derivative $2x$ , and the integrand contains half of it. With $u = x^2 + 1$ we have $du = 2x\,dx$ , so $x\,dx = \tfrac12\,du$ . The bounds $x = 0$ and $x = 1$ become $u = 1$ and $u = 2$ , and

\begin{aligned} &\int_0^1 x(x^2+1)^3\,dx \\[0.4em] &\quad = \frac12\int_1^2 u^3\,du \\[0.4em] &\quad = \frac12\cdot\frac{2^4 - 1^4}{4} = \frac{15}{8}. \end{aligned}

As a check, the derivative of $\tfrac18(x^2+1)^4$ is $x(x^2+1)^3$ , and its values at $1$ and $0$ are $2$ and $\tfrac18$ , whose difference is $\tfrac{15}{8}$ .

Evaluate the integral below. A classmate gets $-\tfrac13$ . Where can the sign go wrong?

\int_0^{\pi/2} \sin x\cos^2 x\,dx

Show the answer

Take $u = \cos x$ , so $du = -\sin x\,dx$ and $\sin x\,dx = -du$ . The bounds $x = 0$ and $x = \pi/2$ become $u = 1$ and $u = 0$ , in decreasing order, which the rule allows. Then

\begin{aligned} &\int_0^{\pi/2} \sin x\cos^2 x\,dx \\[0.4em] &\quad = \int_1^0 u^2\,(-du) \\[0.4em] &\quad = \int_0^1 u^2\,du = \frac13. \end{aligned}

There are two sign changes, the minus in $du = -\sin x\,dx$ and the reversed bounds, and they cancel. Dropping either one alone gives $-\tfrac13$ . That answer cannot be right, because $\sin x\cos^2 x \ge 0$ on $[0, \pi/2]$ , so the integral is positive.

A student evaluates the integral below by setting $u = x^2$ , writing $x = \sqrt{u}$ and $dx = du/(2\sqrt{u})$ . Both bounds become $u = 1$ , so the student answers $0$ . What went wrong, and what is the correct value?

\int_{-1}^{1} x^2\,dx

Show the answer

The integrand $x^2\,dx$ does not contain the factor $2x\,dx$ , so the student had to solve for $x$ . The formula $x = \sqrt{u}$ holds only for $x \ge 0$ . On $[-1, 1]$ the substitution $u = x^2$ is not one-to-one, since every $u$ in $(0, 1]$ comes from two values of $x$ , and on $[-1, 0]$ the correct formula is $x = -\sqrt{u}$ , with $dx = -du/(2\sqrt{u})$ .

Split the interval at $0$ . On $[0, 1]$ the bounds become $0$ and $1$ , and on $[-1, 0]$ they become $1$ and $0$ , so

\begin{aligned} \int_0^1 x^2\,dx &= \int_0^1 \frac{\sqrt{u}}{2}\,du = \frac13, \\[0.4em] \int_{-1}^0 x^2\,dx &= \int_1^0 \Bigl(-\frac{\sqrt{u}}{2}\Bigr)\,du \\[0.4em] &= \int_0^1 \frac{\sqrt{u}}{2}\,du = \frac13. \end{aligned}

Here $u/(2\sqrt{u}) = \sqrt{u}/2$ for $u > 0$ . The total is $\tfrac23$ , which agrees with the power rule, $\bigl[x^3/3\bigr]_{-1}^{1} = \tfrac13 + \tfrac13$ . The answer $0$ was impossible from the start, since $x^2$ is positive except at $x = 0$ .

Substitution for a hidden inner variable

Key terms

Inner function

Chain rule in reverse

Differential du

U-substitution

Changing the bounds

One-to-one substitution

Check yourself

How this lesson connects

Builds on

Used later in