Integration by substitution

The chain rule, backwards

The chain rule (lesson 2.3.3) differentiates a function inside a function. Take the slope of the outer function, then multiply by the slope of the inner one. So an integrand of that shape, an outer function $f$ applied to $g(x)$ and multiplied by $g'(x)$, is a derivative you can recognise.

Take $(x^2 + 1)^4$. The inner function $x^2 + 1$ has slope $2x$, so the chain rule gives

Read backwards, $\int 8x(x^2 + 1)^3\,dx = (x^2 + 1)^4 + C$. Dividing by 4 gives $\int 2x(x^2 + 1)^3\,dx = \frac{(x^2 + 1)^4}{4} + C$.

Renaming the inside

Substitution does this reading backwards for you. Call the inner function $u = x^2 + 1$.

Its slope is $\frac{du}{dx} = 2x$, written $du = 2x\,dx$. A small step $dx$ in $x$ moves $u$ by $2x$ times as much: the stretch factor of lesson 2.3.3.

Now each piece of the integrand has a $u$ version: $(x^2 + 1)^3$ is $u^3$, and $2x\,dx$ is $du$. The integral becomes one in $u$ alone:

A missing constant

In $\int x\,e^{x^2}\,dx$ the inner function is the exponent, $u = x^2$. Its derivative is $2x$, but the integrand has only $x$. A missing constant can be fixed: $du = 2x\,dx$ gives $x\,dx = \frac{1}{2}du$.

1. 1
   $$\begin{gathered} u = x^2, \quad du = 2x\,dx \\ x\,dx = \tfrac{1}{2}\,du \end{gathered}$$

   Why?
2. 2
   $$\int x\,e^{x^2}\,dx = \frac{1}{2}\int e^u\,du$$

   Why?
3. 3
   $$\frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C$$

   Why?

Check with the chain rule: $\frac{d}{dx}\,\frac{1}{2}e^{x^2} = \frac{1}{2}e^{x^2} \cdot 2x = x\,e^{x^2}$.

Only a constant can be fixed this way. In $\int e^{x^2}\,dx$ there is no $x$ in front to turn into $du$, and substitution fails (see the last section).

Choosing $u$

Look for a function sitting inside something else: inside a power, an exponent, a root, a log or the bottom of a fraction. Then check that its derivative appears as a factor, up to a constant. Each line of the table leaves out the $+ C$.

The first row has a linear inner function, $u = 3x + 1$. Its derivative is the constant 3, so it is always there: $dx = \frac{1}{3}du$, and the answer gains a factor $\frac{1}{3}$. This is the rule $\int e^{kx}\,dx = \frac{e^{kx}}{k} + C$ of lesson 2.5.2, now for any outer function.

In the second row $\ln(x^2 + 1)$ needs no absolute value (lesson 2.5.2), because $x^2 + 1$ is always positive. In the third, the log itself is the inner function, and $\frac{1}{x}$ is its derivative.

The last two rows have inner functions that fall, with negative slope. The minus sign in $du$ comes through to the answer. Lesson 2.6.3 meets $x\,e^{-x^2}$ again, under the bell curve.

Definite integrals: change the limits

A definite integral can be finished in two ways, and they give the same number.

1. Change the limits. Turn each limit into a value of $u$ and finish in $u$, without going back to $x$.
2. Switch back. Find the antiderivative in $x$ first, then use the original limits, as in lesson 2.5.3.

Take $\int_0^1 \frac{2x}{x^2 + 1}\,dx$ with $u = x^2 + 1$ and $du = 2x\,dx$. At $x = 0$, $u = 1$; at $x = 1$, $u = 2$. So

Switching back gives the same: $\big[\ln(x^2 + 1)\big]_0^1 = \ln 2 - \ln 1 = \ln 2$.

One area, two curves

Substitution does not change an area; it redraws it. The two integrals above measure regions under different curves, over different intervals, and the regions have the same area.

A strip of width $dx$ becomes a strip of width $du = 2x\,dx$, so it is $2x$ times as wide. Its height $\frac{2x}{x^2 + 1}$ becomes $\frac{1}{u}$, which is $2x$ times lower. The two factors cancel, and every strip keeps its area.

Midpoint sums (lesson 2.5.1) with equal strips show it in numbers. The two curves have different shapes, so the sums differ at first, but both settle on $\ln 2 \approx 0.693147$.

When the new limits run backwards

In $\int_0^1 x(1 - x^2)^3\,dx$, take $u = 1 - x^2$. Then $du = -2x\,dx$, so $x\,dx = -\frac{1}{2}du$. At $x = 0$, $u = 1$, and at $x = 1$, $u = 0$: the new integral runs from 1 down to 0.

Swapping the limits flips the sign (lesson 2.5.3), which cancels the minus sign from $du$.

Switching back instead

For $\int_1^e \frac{\ln x}{x}\,dx$, the table gives the antiderivative $\frac{(\ln x)^2}{2}$. With the original limits, $\frac{(\ln e)^2}{2} - \frac{(\ln 1)^2}{2} = \frac{1}{2} - 0 = \frac{1}{2}$. Changing the limits instead gives $\int_0^1 u\,du = \frac{1}{2}$.

When substitution does not help

Substitution needs the derivative of the inside to be in the integral. Here are three cases where it is not.

• The missing factor is not a constant. $\int e^{x^2}\,dx$ has nothing to pair with $du = 2x\,dx$. Forcing $u = x^2$ gives $\int \frac{e^u}{2\sqrt{u}}\,du$, which is no easier. In fact $e^{x^2}$ has no antiderivative made of familiar functions; SymPy needs a new one, erfi.
• There is a simpler route. $\int (x^2 + 1)^2\,dx$ lacks the $2x$, but multiplying out (lesson 2.5.2) gives $\frac{x^5}{5} + \frac{2x^3}{3} + x + C$.
• A product of two kinds of function, such as $x\,e^x$. Neither factor is the derivative of an inside. Lesson 2.6.2 handles these with integration by parts.

Dividing by a missing $2x$ does not work either. The derivative of $\frac{e^{x^2}}{2x}$ is $\left(1 - \frac{1}{2x^2}\right)e^{x^2}$, not $e^{x^2}$. Lesson 2.6.3 meets the close relative $e^{-x^2}$, whose area is found another way.

Case study 2.6.5 uses a substitution to turn part of an option price into an area under the bell curve.

Common mistakes

1. Convert $dx$ as well. $\int (3x + 1)^5\,dx$ is $\frac{(3x + 1)^6}{18} + C$, not $\frac{(3x + 1)^6}{6} + C$, whose derivative is $3(3x + 1)^5$.
2. Leave no $x$ behind. An integral with both $u$ and $x$ in it cannot be done in $u$.
3. Only a constant can be fixed. $\frac{e^{x^2}}{2x}$ is not an antiderivative of $e^{x^2}$.
4. Keep the minus sign from a falling inner function: $\int x\,e^{-x^2}\,dx = -\frac{1}{2}e^{-x^2} + C$.
5. Change the limits with the variable. $\int_0^1 2x(x^2 + 1)^3\,dx = \int_1^2 u^3\,du = \frac{15}{4}$. Keeping the old limits gives $\frac{1}{4}$.
6. Keep the new limits in order. When $u$ falls, the new lower limit is the larger number. Swap them only if you also flip the sign.
7. Choose one way to finish. After changing the limits, finish in $u$; do not also put $x$ back.