Standard Integrals

Every derivative you learnt in P3 becomes an integral when you read it backwards. This note builds the table of standard integrals for Paper 3: exponentials, $\dfrac{1}{x}$ and the logarithm, sine, cosine, $\sec^2$ and the inverse tangent, each with a linear expression $ax + b$ inside. Every harder technique (identities, partial fractions, parts, substitution) ends by reducing an integral to one of these, so they must be automatic. Almost every P3 paper uses them, usually inside a definite integral with an exact answer in terms of $e$, $\ln$ or $\pi$.

Integration as reverse differentiation

To integrate $f(x)$ is to find a function $F(x)$ whose derivative is $f(x)$. Such an $F$ is an antiderivative, and because any constant differentiates to zero, there is a whole family of them:

$$\int f(x)\, dx = F(x) + c, \qquad \text{where } F'(x) = f(x)$$

The $c$ is the arbitrary constant of integration. It must appear in every indefinite integral. In a definite integral it cancels, so it is left out:

$$\int_a^b f(x)\, dx = \Big[F(x)\Big]_a^b = F(b) - F(a)$$

This means every result in this note can be checked by differentiating. If you are ever unsure of a sign or a factor, differentiate your answer and see whether you get back the integrand. It takes ten seconds and catches most errors.

The standard integrals

Each line of this table is a derivative read from right to left.

Derivative you know	Integral it gives
$\dfrac{d}{dx}\left(\dfrac{x^{n+1}}{n+1}\right) = x^n$	$\displaystyle\int x^n\, dx = \frac{x^{n+1}}{n+1} + c,\ n \ne -1$
$\dfrac{d}{dx}(e^x) = e^x$	$\displaystyle\int e^x\, dx = e^x + c$
$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$	$\displaystyle\int \frac{1}{x}\, dx = \ln\lvert x\rvert + c$
$\dfrac{d}{dx}(\cos x) = -\sin x$	$\displaystyle\int \sin x\, dx = -\cos x + c$
$\dfrac{d}{dx}(\sin x) = \cos x$	$\displaystyle\int \cos x\, dx = \sin x + c$
$\dfrac{d}{dx}(\tan x) = \sec^2 x$	$\displaystyle\int \sec^2 x\, dx = \tan x + c$
$\dfrac{d}{dx}\left(\tan^{-1} x\right) = \dfrac{1}{1 + x^2}$	$\displaystyle\int \frac{1}{1 + x^2}\, dx = \tan^{-1} x + c$

Two points need care.

The power rule fails at $n = -1$, because it would divide by zero. The gap is filled by the logarithm: $\displaystyle\int x^{-1}\, dx = \ln\lvert x\rvert + c$. Only a power of exactly $-1$ gives a logarithm. $\displaystyle\int x^{-2}\, dx = -x^{-1} + c$, with no logarithm in sight.

Why the modulus? $\ln x$ only exists for $x > 0$, but $\dfrac{1}{x}$ is perfectly happy for $x < 0$. For negative $x$, $\dfrac{d}{dx}\ln(-x) = \dfrac{-1}{-x} = \dfrac{1}{x}$, so $\ln(-x)$ is an antiderivative there. The single expression $\ln\lvert x\rvert$ covers both sides. In a definite integral over positive values you can write ordinary brackets; in an indefinite integral, keep the modulus.

All trigonometry is in radians. $\dfrac{d}{dx}\sin x = \cos x$ is only true when $x$ is in radians, so the same is true of every trigonometric integral. Set your calculator to radians before evaluating any definite integral numerically.

A linear expression inside

The chain rule says that differentiating $F(ax + b)$ produces an extra factor of $a$:

$$\frac{d}{dx}F(ax + b) = a\,F'(ax + b)$$

Reading that backwards, integrating $f(ax + b)$ gives $F(ax + b)$ divided by $a$. That one idea extends the whole table.

The rule "divide by the coefficient of $x$" only works when the inside is linear. $\displaystyle\int e^{x^2}\, dx$ is not $\dfrac{e^{x^2}}{2x}$: differentiate that and you do not get $e^{x^2}$ back. Non-linear insides need the methods in Integrating f'(x)/f(x) and related forms or substitution.

The inverse tangent integral

The last line deserves its own derivation, because the $\dfrac{1}{a}$ arrives in a different way. Write $x^2 + a^2 = a^2\left(1 + \left(\tfrac{x}{a}\right)^2\right)$. Then

$$\frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right) = \frac{1}{1 + \left(\frac{x}{a}\right)^2}\cdot\frac{1}{a} = \frac{a^2}{a^2 + x^2}\cdot\frac{1}{a} = \frac{a}{x^2 + a^2}$$

So $\dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)$ differentiates to $\dfrac{1}{x^2 + a^2}$. Notice that the constant is $a^2$, so you must square-root it to find $a$: for $\dfrac{1}{x^2 + 9}$, $a = 3$; for $\dfrac{1}{x^2 + 5}$, $a = \sqrt{5}$.

When $x^2$ has a coefficient, as in $\dfrac{1}{2 + 3x^2}$, take that coefficient out first so that the denominator is exactly $x^2 + a^2$:

$$\frac{1}{2 + 3x^2} = \frac{1}{3}\cdot\frac{1}{x^2 + \frac{2}{3}}, \qquad a = \sqrt{\tfrac{2}{3}}$$

The alternative is to see $p + qx^2$ as $p + (\sqrt{q}\,x)^2$, a linear inside $\sqrt{q}\,x$, and divide by its coefficient $\sqrt{q}$. Both give the same answer.

Preparing the integrand

Many integrands are not in the table as written but become standard after a line of algebra. Before integrating, rewrite the integrand as a sum of standard terms.

• Split a fraction over a single-term denominator. $\dfrac{x^2 + 3}{x} = x + \dfrac{3}{x}$, which integrates to $\dfrac{x^2}{2} + 3\ln\lvert x\rvert + c$.
• Expand brackets. $\left(e^x + 2\right)^2 = e^{2x} + 4e^x + 4$.
• Divide by an exponential term by term. $\dfrac{e^{2x} + 1}{e^x} = e^x + e^{-x}$.
• Use index laws. $\dfrac{1}{\sqrt{2x + 1}} = (2x + 1)^{-1/2}$ and $\dfrac{3}{e^{2x}} = 3e^{-2x}$.
• Use identities for $\sin^2 x$, $\cos^2 x$ and $\tan^2 x$; these have their own note, Integrating trigonometric functions.

You cannot split a fraction whose denominator is a sum: $\dfrac{1}{x + x^2} \ne \dfrac{1}{x} + \dfrac{1}{x^2}$. That case needs partial fractions.

Exact answers from definite integrals

P3 questions almost always say "find the exact value" or "show that the integral equals ...". You will need the laws of logarithms to tidy up:

$$\ln a + \ln b = \ln ab, \qquad \ln a - \ln b = \ln\frac{a}{b}, \qquad k\ln a = \ln a^k, \qquad \ln 1 = 0, \qquad e^{\ln a} = a$$

and the exact values of the trigonometric functions at $0$, $\tfrac{\pi}{6}$, $\tfrac{\pi}{4}$, $\tfrac{\pi}{3}$ and $\tfrac{\pi}{2}$, plus $\tan^{-1} 1 = \tfrac{\pi}{4}$, $\tan^{-1}\sqrt{3} = \tfrac{\pi}{3}$ and $\tan^{-1}\tfrac{1}{\sqrt{3}} = \tfrac{\pi}{6}$.

Areas

The area between a curve $y = f(x)$, the $x$-axis and the lines $x = a$, $x = b$ is $\displaystyle\int_a^b f(x)\, dx$ when the curve is above the axis. If the curve is below the axis the integral is negative, and the area is its modulus. If the curve crosses the axis between $a$ and $b$, split the integral at the crossing point. Area between two curves is $\displaystyle\int_a^b (\text{upper} - \text{lower})\, dx$. These ideas are from P1 (see Areas and definite integrals); in P3 they are combined with the new functions.

Worked examples

The curves $y = e^x$ and $y = 3 - 2e^{-x}$ cross at $x = 0$ and $x = \ln 2 \approx 0.693$. The thin region between them has area $3\ln 2 - 2$.

The region bounded by $y = 4 - e^{2x}$ and the axes, between $x = 0$ and $x = \ln 2$, has area $4\ln 2 - \tfrac{3}{2}$. Beyond $x = \ln 2$ the curve drops below the axis.

Practice

Answers

1. $\dfrac{(2x + 3)^6}{6 \times 2} + c = \dfrac{(2x + 3)^6}{12} + c$.

2. $-\tfrac{1}{4}e^{1 - 4x} + 3\ln\lvert x\rvert + c$.

3. $\Big[\tan x - \tfrac{1}{2}\cos 2x\Big]_0^{\pi/4} = \left(1 - \tfrac{1}{2}\cos\tfrac{\pi}{2}\right) - \left(0 - \tfrac{1}{2}\right) = 1 - 0 + \tfrac{1}{2} = \dfrac{3}{2}$.

4. Split: $\dfrac{x^2 + 2}{x} = x + \dfrac{2}{x}$. Then $\Big[\tfrac{1}{2}x^2 + 2\ln x\Big]_1^3 = \left(\tfrac{9}{2} + 2\ln 3\right) - \left(\tfrac{1}{2} + 0\right) = 4 + 2\ln 3 = 4 + \ln 9$.

5. (a) $a = 5$: $\tfrac{1}{5}\tan^{-1}\left(\tfrac{x}{5}\right) + c$. (b) $\dfrac{1}{1 + 9x^2} = \dfrac{1}{9}\cdot\dfrac{1}{x^2 + \frac{1}{9}}$, $a = \tfrac{1}{3}$, so the integral is $\tfrac{1}{9}\cdot 3\tan^{-1}(3x) = \tfrac{1}{3}\tan^{-1}(3x)$. Between the limits: $\tfrac{1}{3}\tan^{-1} 1 - 0 = \dfrac{\pi}{12}$.

6. $\dfrac{1}{5 + 2x^2} = \dfrac{1}{2}\cdot\dfrac{1}{x^2 + \frac{5}{2}}$, $a = \sqrt{\tfrac{5}{2}}$. The integral is $\dfrac{1}{2}\sqrt{\dfrac{2}{5}}\tan^{-1}\left(x\sqrt{\dfrac{2}{5}}\right) + c = \dfrac{1}{\sqrt{10}}\tan^{-1}\left(\dfrac{\sqrt{10}\,x}{5}\right) + c$.

7. Expand and divide: $\dfrac{e^{2x} - 2e^x + 1}{e^x} = e^x - 2 + e^{-x}$. Then $\Big[e^x - 2x - e^{-x}\Big]_0^1 = \left(e - 2 - e^{-1}\right) - (1 - 0 - 1) = e - 2 - \dfrac{1}{e}$.

8. $\Big[-2e^{-x/2}\Big]_0^a = -2e^{-a/2} + 2 = 1$, so $e^{-a/2} = \tfrac{1}{2}$, $-\tfrac{a}{2} = \ln\tfrac{1}{2} = -\ln 2$, and $a = 2\ln 2$ (or $\ln 4$).

9. $V = \pi\displaystyle\int_0^2 y^2\, dx = \pi\int_0^2 \frac{4}{4x + 1}\, dx = \pi\Big[\ln(4x + 1)\Big]_0^2 = \pi(\ln 9 - \ln 1) = \pi\ln 9$ (or $2\pi\ln 3$).

10. $\displaystyle\int_0^k \frac{1}{3 + x^2}\, dx = \Big[\tfrac{1}{\sqrt{3}}\tan^{-1}\tfrac{x}{\sqrt{3}}\Big]_0^k = \tfrac{1}{\sqrt{3}}\tan^{-1}\tfrac{k}{\sqrt{3}}$. Setting this equal to $\dfrac{\pi}{6\sqrt{3}}$ gives $\tan^{-1}\tfrac{k}{\sqrt{3}} = \tfrac{\pi}{6}$, so $\tfrac{k}{\sqrt{3}} = \tan\tfrac{\pi}{6} = \tfrac{1}{\sqrt{3}}$ and $k = 1$. Then $\displaystyle\int_0^1 \frac{1}{x + 1}\, dx = \Big[\ln(x + 1)\Big]_0^1 = \ln 2$, so the required value is $\dfrac{\pi}{6\sqrt{3}} + \ln 2$ (which can be written $\dfrac{\sqrt{3}\,\pi}{18} + \ln 2$).