5.10 Indefinite integrals and substitution by inspection

Integration reverses differentiation. Once the derivatives of $\sin x$, $\cos x$, $e^x$ and $\ln x$ and the chain rule are known, each of them can be read backwards to give a new integral. The result is a short list of standard integrals, two rules that extend it to many more functions, and a habit of checking every answer by differentiating.

These integrals are the working tools of the rest of calculus. They give exact areas under curves of exponential and trigonometric functions, displacements from velocities, and total quantities, such as the volume of water that has flowed out of a tank, from a rate of change.

Integration as the reverse of differentiation

An antiderivative of a function $f$ is any function $F$ with $F'(x) = f(x)$. Adding a constant to $F$ does not change its derivative, so every function that has one antiderivative has infinitely many, and they differ only by a constant. The indefinite integral collects all of them in one expression:

The function being integrated, $f(x)$, is the integrand, and $C$ is the constant of integration. The $dx$ shows that the variable of integration is $x$.

Because integration undoes differentiation, every derivative result gives an integral. For example, $\frac{d}{dx}(\sin x) = \cos x$ means that $\int \cos x\,dx = \sin x + C$. This also gives a reliable check: differentiating the answer to an integral must give back the integrand exactly.

Standard integrals

Powers of x with rational exponents

The power rule for derivatives, $\frac{d}{dx}(x^{n+1}) = (n + 1)x^n$, holds for every rational power. Dividing by $n + 1$ gives the rule for integrating a power.

The rule applies to roots and reciprocals once they are written as powers: $\sqrt{x} = x^{\frac{1}{2}}$, $\sqrt[3]{x^2} = x^{\frac{2}{3}}$, $\frac{1}{x^3} = x^{-3}$ and $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$. Constant multiples and sums are integrated term by term, just as they are differentiated term by term:

Dividing by a fractional power is the step that causes most errors. Dividing by $\frac{3}{2}$ is the same as multiplying by $\frac{2}{3}$, so $\int x^{\frac{1}{2}}\,dx = \frac{2}{3}x^{\frac{3}{2}} + C$.

There is no product rule or quotient rule for integration. A product or quotient must first be rewritten as a sum of powers, by expanding brackets or by dividing each term of the numerator by the denominator.

The integral of 1/x

The power rule leaves out exactly one power, $x^{-1}$. Its integral comes from the derivative of the natural logarithm: $\frac{d}{dx}(\ln x) = \frac{1}{x}$ for $x > 0$. But $\frac{1}{x}$ is also defined for negative $x$, where $\ln x$ is not. For $x < 0$, the function $\ln(-x)$ is defined, and by the chain rule

Both cases are covered by the absolute value $|x|$, which equals $x$ when $x > 0$ and $-x$ when $x < 0$.

The graph of $y = \ln|x|$ is the graph of $y = \ln x$ together with its reflection in the $y$-axis. On the left branch the curve falls as $x$ increases, so its gradient is negative, exactly as $\frac{1}{x}$ is negative there. The vertical asymptote at $x = 0$ splits the domain in two: an integral of $\frac{1}{x}$ never crosses $x = 0$.

Exponential and trigonometric functions

The exponential function is its own derivative, so it is also its own integral. The derivatives $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$ give the two trigonometric integrals. The second one carries a minus sign: since $\frac{d}{dx}(-\cos x) = \sin x$, the integral of $\sin x$ is $-\cos x$.

The four functions $\sin x$, $\cos x$, $-\sin x$ and $-\cos x$ form a cycle. Differentiating moves one step clockwise round the cycle and integrating moves one step anticlockwise, so four integrations (or four differentiations) bring a function back to itself.

These results hold only when $x$ is measured in radians, because the derivatives of $\sin x$ and $\cos x$ are only this simple in radians. All the standard integrals are collected in the table.

Finding a function from its derivative

An indefinite integral is a family of functions. Their graphs are vertical translations of one another, because changing $C$ moves the whole graph up or down without changing its gradient at any $x$. One extra piece of information, a boundary condition such as a point on the curve or the value of the function at a given $x$, picks out a single member of the family. This is set out for polynomials in 5.5; the method is the same for every integral.

The curves all have the gradient function $\frac{dy}{dx} = \cos 2x$, because the chain rule gives $\frac{d}{dx}\left(\frac{1}{2}\sin 2x\right) = \frac{1}{2} \times 2\cos 2x = \cos 2x$. Substituting the point $\left(\frac{\pi}{4}, \frac{5}{2}\right)$ into $y = \frac{1}{2}\sin 2x + C$ gives $\frac{1}{2} + C = \frac{5}{2}$, so $C = 2$.

Composites with a linear function

Why the integral is divided by a

Many integrands are a standard function of a linear expression $ax + b$, such as $\cos(3x + 1)$, $e^{4 - 2x}$ or $\frac{1}{2x - 5}$. Each is a composite function whose inner function is $ax + b$. Differentiating a composite of this kind with the chain rule multiplies by the derivative of the inner function, which is the constant $a$:

To reverse this, integrate the outer function as usual and then divide by $a$. If $F$ is an antiderivative of $f$, then

For example, $\frac{d}{dx}\sin(3x + 1) = 3\cos(3x + 1)$, which is three times too large, so $\int \cos(3x + 1)\,dx = \frac{1}{3}\sin(3x + 1) + C$.

Expanding $(3x + 2)^5$ and integrating term by term gives the same family of functions, but it takes far longer and produces a different-looking constant term. Two correct answers to an indefinite integral may look different and still both be right, as long as they differ only by a constant.

Negative and fractional powers of a linear expression

Roots and reciprocals of a linear expression are written as powers in the same way as roots and reciprocals of $x$. The only exception is the power $-1$, which gives a logarithm.

Equivalent forms of a logarithmic answer

Integrals of reciprocals can be written in more than one correct form. Dividing first gives $\int \frac{1}{3x}\,dx = \frac{1}{3}\int \frac{1}{x}\,dx = \frac{1}{3}\ln|x| + C$, while the linear rule with $a = 3$ gives $\frac{1}{3}\ln|3x| + C$. Both are right, because

and $\frac{1}{3}\ln 3$ is a constant that is absorbed into $C$. When an answer has to be given in a particular form, such as $\ln(k(2x + 1)^3)$, the constant is often written as $\ln k$ so that the laws of logarithms can combine the terms.

The linear rule only works for linear inner functions

Dividing by the derivative of the inner function is valid only when that derivative is a constant. For $\int (x^2 + 1)^3\,dx$, the inner function $x^2 + 1$ has derivative $2x$, which is not constant, and $\frac{(x^2 + 1)^4}{4 \times 2x}$ is wrong: differentiating it with the quotient rule does not give back $(x^2 + 1)^3$. Such an integrand must be expanded first, or, if a factor of $2x$ is present, integrated by inspection (the reverse chain rule).

Rewriting with trigonometric identities

Some trigonometric integrands are not on the list but can be rewritten as one that is. The most common are $\sin^2 x$ and $\cos^2 x$. The double angle identity $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ (3.6) rearranges to

Each right-hand side is a constant plus a cosine of the linear function $2x$, which can be integrated. Similarly, $\sin x \cos x = \frac{1}{2}\sin 2x$.

Integration by inspection: the reverse chain rule

Recognising a chain rule derivative

The chain rule differentiates a composite function $F(g(x))$ as

where $f = F'$. Reading this backwards gives an integral:

An integrand of this shape is the product of a composite function $f(g(x))$ and the derivative $g'(x)$ of its inner function. For example, in $\int 2x(x^2 + 1)^4\,dx$ the inner function is $g(x) = x^2 + 1$ and the factor $2x$ is exactly $g'(x)$.

The word "inspection" means that the answer is found by looking at the integrand and recognising its structure, not by a fixed algorithm. The method works in four steps.

1. Identify the inner function $g(x)$, usually the expression inside a bracket, a root, an exponent or the denominator.

2. Check that the rest of the integrand is a constant multiple of $g'(x)$. If it is not, inspection does not work.

3. Integrate the outer function as if $g(x)$ were a single variable, and write the guess in terms of $g(x)$.

4. Differentiate the guess and compare it with the integrand. Multiply the guess by whatever constant makes them agree.

Adjusting the constant factor

The factor in front of the composite does not have to match $g'(x)$ exactly. It is enough that it is a constant multiple of $g'(x)$, because a constant can be moved outside the integral and corrected for afterwards.

The patterns to look for

Most integrals that can be found by inspection fit one of a few patterns. Each comes from a standard derivative combined with the chain rule.

The second pattern deserves particular attention. Whenever the numerator of a fraction is a constant multiple of the derivative of the denominator, the integral is a logarithm of the denominator. In $\int \frac{2x}{x^2 + 3}\,dx$ the modulus is not needed, because $x^2 + 3$ is always positive.

More integrals by inspection

The same idea applies whenever a function sits inside another function and the derivative of the inside is also a factor. Recognising the inner function is the skill to practise. Typical places for it are an exponent ($e^{x^2}$), a trigonometric argument ($\cos(x^3)$), the base of a power ($\sin^3 x = (\sin x)^3$, $(\ln x)^2$) and the inside of a root ($\sqrt{x^2 + 9}$).

Finding an unknown constant

Some questions give the answer to an integral in terms of an unknown constant and ask for its value. Differentiating the given answer and comparing it with the integrand is the quickest approach.

Integration by substitution

Inspection can also be written out formally as a substitution. The inner function is given a new name, $u = g(x)$, and the derivative $\frac{du}{dx} = g'(x)$ is used to replace $g'(x)\,dx$ by $du$. The integral then becomes a standard integral in $u$. Both methods give the same answer; substitution takes longer to write but makes each step visible, which helps when the constant factor is awkward. Substitution in general, including integrands where the inner function is not obvious, is HL content in 5.16.

Integrals in context

When a rate of change is known, integrating it gives the quantity itself, up to a constant. A known starting value, such as the amount at time $t = 0$, fixes the constant. The rate of change of a cost with respect to the number of items made is called the marginal cost; integrating it gives the total cost, and the constant of integration is the fixed cost of making zero items.

The rate of leaking in the model falls over time, from $20$ litres per hour at $t = 0$ to about $1.6$ litres per hour when the tank empties. This is realistic for a leak near the bottom of a tank, where the flow slows as the water level, and so the pressure, drops.

Exam-style questions on indefinite integrals

Questions on indefinite integrals usually appear on the non-calculator paper, often as the first part of a longer question that later uses the result for an area (5.11) or a kinematics problem (5.9). The two examples show typical structures.

Self-check

• To do:

  I can integrate $x^n$ for any rational $n \neq -1$, after rewriting roots and reciprocals as powers.

• To do:

  I can state the integrals of $\frac{1}{x}$, $e^x$, $\sin x$ and $\cos x$, including the modulus in $\ln|x|$ and the sign of $\int \sin x\,dx$.

• To do:

  I can integrate $f(ax + b)$ for each standard function by dividing by $a$.

• To do:

  I can rewrite $\sin^2 x$ and $\cos^2 x$ with a double angle identity and integrate them.

• To do:

  I can recognise $k\,g'(x)f(g(x))$ and integrate it by inspection, adjusting a constant factor.

• To do:

  I can integrate a fraction whose numerator is a multiple of the derivative of its denominator.

• To do:

  I can carry out the same integral by substitution, giving the answer in terms of $x$.

• To do:

  I can use a boundary condition to find the constant of integration, including in a context.

• To do:

  I can check any indefinite integral by differentiating my answer.

Find $y$ in terms of $x$.

[6]6 marks

Integrate term by term, adding one to each power and dividing by the new power:

In the form it was given, the answer is $x^4 - 4x\sqrt{x} - \frac{1}{x^2} + C$.

Check by differentiating: $4x^3 - 4 \times \frac{3}{2}x^{\frac{1}{2}} + 2x^{-3} = 4x^3 - 6\sqrt{x} + \frac{2}{x^3}$, which is the integrand.

(b) Expand the numerator, then divide each term by $x^{\frac{1}{2}}$:

Integrate term by term:

Each coefficient comes from dividing by the new power: $1 \div \frac{5}{2} = \frac{2}{5}$, $4 \div \frac{3}{2} = \frac{8}{3}$ and $4 \div \frac{1}{2} = 8$.

(b) Take the constant $\frac{5}{2}$ outside the integral:

Substitute $x = 0$, $y = 7$. Since $e^0 = 1$ and $\sin 0 = 0$,

The curve is $y = 4e^x - 3\sin x + 3$.

(b) Here $a = 2$ and the outer function is $\frac{1}{u}$, which integrates to $\ln|u|$:

(c) Here $a = -2$, so divide by $-2$: $\displaystyle\int e^{4 - 2x}\,dx = -\frac{1}{2}e^{4 - 2x} + C$.

(d) Here $a = 3$ and the outer function is $u^5$, which integrates to $\frac{u^6}{6}$:

Check (d): $\frac{d}{dx}\frac{(3x + 2)^6}{18} = \frac{6(3x + 2)^5 \times 3}{18} = (3x + 2)^5$.

(b) Write the integrand as $6(3 - 2x)^{-2}$. The power is $-2$, not $-1$, so the power rule applies. Raise the power to $-1$, divide by $-1$, then divide by $a = -2$:

(c) The inner function is $\frac{1}{2}x$, so $a = \frac{1}{2}$ and dividing by $a$ means multiplying by $2$:

(b) Write $\tan x = \frac{\sin x}{\cos x}$. The derivative of the denominator is $-\sin x$, so the numerator is $-1$ times it:

(c) $\frac{d}{dx}(e^x + 1) = e^x$, the numerator. Since $e^x + 1 > 0$ for every $x$, no modulus is needed:

(d) $\frac{d}{dx}(x^2 + x + 7) = 2x + 1$, and $6x + 3 = 3(2x + 1)$. The quadratic has discriminant $1 - 28 < 0$ and a positive leading coefficient, so it is always positive:

(b) The inner function is $\sin x$, with derivative $\cos x$, which is present:

(c) Write $\sin^3 x$ as $(\sin x)^3$. The inner function is $\sin x$ and $\cos x$ is its derivative. The outer function $u^3$ integrates to $\frac{u^4}{4}$:

(d) Write the integrand as $\frac{1}{x} \times \ln x$. The inner function is $\ln x$, whose derivative $\frac{1}{x}$ is present, and the outer function is $u$, which integrates to $\frac{u^2}{2}$:

(e) Write the integrand as $x(x^2 + 9)^{-\frac{1}{2}}$. The inner function $x^2 + 9$ has derivative $2x$, and $x$ is half of that. Guess $(x^2 + 9)^{\frac{1}{2}}$:

This is exactly the integrand, so

This must equal the integrand $kx(x^2 + 2)^3$, so $k = 8$.

Replace every part of the integral by its expression in $u$:

Integrate with respect to $u$:

Finally, write the answer in terms of $x$:

When $t = 0$, $V = 200$:

So $V = 200 + 80\ln 4 - 80\ln(t + 4)$, which the laws of logarithms simplify to

(b) The tank is empty when $V = 0$:

The letter $K$ is used for the constant because $C$ already stands for the cost. When $x = 0$, $C = 2000$:

So $C(x) = 50x + 300\sqrt{2x + 9} + 1100$, and

The total cost of making 100 chairs is about USD 10 400 (3 s.f.).

Substitute the point $(0, 5)$: $5 = -3 + C$, so $C = 8$.

The marks are for: recognising the reverse chain rule (or making a substitution), the correct integral, substituting the point, and the final answer.

(b) As $x \to \pm\infty$, $\frac{3}{x^2 + 1} \to 0$, so the horizontal asymptote is $y = 8$.

(c) $f'(x) = 0$ only at $x = 0$, where $f'$ changes from negative to positive, so $f(0) = 5$ is the minimum value. The graph approaches $y = 8$ but never reaches it, since $\frac{3}{x^2 + 1} > 0$. The range is $5 \le f(x) < 8$.

(b) "Hence" means use part (a). Integrate the right-hand side, dividing by $2$ for the $\cos 2x$ term:

(c) From (b), $g(x) = \frac{x}{2} - \frac{1}{4}\sin 2x + C$. At $x = \frac{\pi}{2}$, $\sin 2x = \sin \pi = 0$, so

[4]4 marks

[4]4 marks