Integration Revision Notes — 9709 Paper 3

The standard integrals

Syllabus requirement · §3.5

“

extend the idea of 'reverse differentiation' to include the integration of $\mathrm{e}^{ax+b}$ , 1/(ax+b), sin(ax+b), cos(ax+b), sec²(ax+b) and 1/(x² + a²). Including examples such as 1/(2 + 3x²).

”

Integration undoes differentiation: $\displaystyle\int f(x)\,\mathrm{d}x$ asks "what function has $f(x)$ as its derivative?" In Paper 1 the answer was always a power of $x$ . Paper 3 adds the functions whose derivatives you learnt in the Differentiation note, and each result below is one of those derivatives read backwards.

Start with the chain rule on a function of $ax+b$ . Differentiating the outside function and then multiplying by the derivative of the inside, $a$ , gives

\frac{\mathrm{d}}{\mathrm{d}x}\,\mathrm{e}^{ax+b} = a\,\mathrm{e}^{ax+b}

So differentiating $\mathrm{e}^{ax+b}$ produces an extra factor of $a$ . To get back to plain $\mathrm{e}^{ax+b}$ we must start from $\tfrac1a\mathrm{e}^{ax+b}$ :

\frac{\mathrm{d}}{\mathrm{d}x}\left(\tfrac1a\mathrm{e}^{ax+b}\right) = \tfrac1a\cdot a\,\mathrm{e}^{ax+b} = \mathrm{e}^{ax+b} \quad\Longrightarrow\quad \int \mathrm{e}^{ax+b}\,\mathrm{d}x = \tfrac1a\mathrm{e}^{ax+b} + c

The same happens with every function of $ax+b$ : integrate the outside function as if the bracket were just $x$ , then divide by $a$ , the number in front of $x$ .

Derivative you already know	Read backwards
$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(ax+b) = \dfrac{a}{ax+b}$	$\displaystyle\int\frac{1}{ax+b}\,\mathrm{d}x = \tfrac1a\ln\lvert ax+b\rvert + c$
$\dfrac{\mathrm{d}}{\mathrm{d}x}\cos(ax+b) = -a\sin(ax+b)$	$\displaystyle\int\sin(ax+b)\,\mathrm{d}x = -\tfrac1a\cos(ax+b) + c$
$\dfrac{\mathrm{d}}{\mathrm{d}x}\sin(ax+b) = a\cos(ax+b)$	$\displaystyle\int\cos(ax+b)\,\mathrm{d}x = \tfrac1a\sin(ax+b) + c$
$\dfrac{\mathrm{d}}{\mathrm{d}x}\tan(ax+b) = a\sec^2(ax+b)$	$\displaystyle\int\sec^2(ax+b)\,\mathrm{d}x = \tfrac1a\tan(ax+b) + c$

The minus sign in the $\sin$ row comes from the minus in the derivative of $\cos$ . All trig integrals assume the angle is in radians, just as the trig derivatives do.

The six standard forms

\int \mathrm{e}^{ax+b}\,\mathrm{d}x = \tfrac{1}{a}\mathrm{e}^{ax+b} + c

Divide by the number in front of x

\int \frac{1}{ax+b}\,\mathrm{d}x = \tfrac{1}{a}\ln|ax+b| + c

A linear bottom gives a log

\int \sin(ax+b)\,\mathrm{d}x = -\tfrac{1}{a}\cos(ax+b) + c

Note the minus sign

\int \cos(ax+b)\,\mathrm{d}x = \tfrac{1}{a}\sin(ax+b) + c

\int \sec^2(ax+b)\,\mathrm{d}x = \tfrac{1}{a}\tan(ax+b) + c

\int \frac{1}{x^2+a^2}\,\mathrm{d}x = \tfrac{1}{a}\tan^{-1}\!\left(\tfrac{x}{a}\right) + c

Here a is the square root of the number added to x²

What the formula list gives you

The formula list in the exam gives these results for plain $x$ (for example $\int\sin x\,\mathrm{d}x = -\cos x$ ), and it gives the $\tan^{-1}$ result in full. It does not give the $ax+b$ versions: dividing by $a$ is your job.

The list of derivatives on the formula list can be read backwards too. Three that turn up inside Paper 3 questions:

\int\sec x\tan x\,\mathrm{d}x = \sec x \qquad \int\mathrm{cosec}^2x\,\mathrm{d}x = -\cot x \qquad \int\mathrm{cosec}\,x\cot x\,\mathrm{d}x = -\mathrm{cosec}\,x

Each comes from a derivative you know: for example $\dfrac{\mathrm{d}}{\mathrm{d}x}\sec x = \sec x\tan x$ . With $ax+b$ inside, divide by $a$ as before.

Three standard forms in one integral

Find $\displaystyle\int\left(6\mathrm{e}^{3x+1} - 4\sin 2x + \frac{5}{2x-1}\right)\mathrm{d}x$ .

Show full working

1
Integrate term by term. The first term is $6\times\mathrm{e}^{ax+b}$ with $a=3$ , so divide by $3$ : $\int 6\mathrm{e}^{3x+1}\,\mathrm{d}x = 6\times\tfrac13\mathrm{e}^{3x+1} = 2\mathrm{e}^{3x+1}$
A constant multiple just waits outside; only the e part is integrated.
2
The second term is $-4\times\sin(ax+b)$ with $a=2$ . The integral of $\sin 2x$ is $-\tfrac12\cos 2x$ : $\int -4\sin 2x\,\mathrm{d}x = -4\times\left(-\tfrac12\cos 2x\right) = 2\cos 2x$
Two minus signs meet here, one from the question and one from integrating sin. They give a plus. Losing one of them is the usual slip.
3
The third term is $5\times\dfrac{1}{ax+b}$ with $a=2$ , so divide by $2$ : $\int\frac{5}{2x-1}\,\mathrm{d}x = 5\times\tfrac12\ln|2x-1| = \tfrac52\ln|2x-1|$
The 2 in front of x is what we divide by. The −1 in the bracket changes nothing.
4
Add the three results and one constant: $2\mathrm{e}^{3x+1} + 2\cos 2x + \tfrac52\ln|2x-1| + c$
One + c covers all three terms, because constants added together are still just a constant.

Answer

$2\mathrm{e}^{3x+1} + 2\cos 2x + \tfrac52\ln|2x-1| + c$

Check any answer by differentiating it: you should get back exactly what you were asked to integrate.

Where the $\tan^{-1}$ result comes from

The Differentiation note gives $\dfrac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}x = \dfrac{1}{1+x^2}$ . Apply it to $\tan^{-1}\!\left(\dfrac xa\right)$ , where the inside $\dfrac xa$ has derivative $\dfrac1a$ :

\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}\!\left(\frac xa\right) = \frac{1}{1+\left(\frac xa\right)^2}\times\frac1a = \frac{1}{a\left(1+\frac{x^2}{a^2}\right)} = \frac{a}{a^2+x^2}

That is $a$ times what we want, so divide by $a$ :

\int\frac{1}{x^2+a^2}\,\mathrm{d}x = \frac1a\tan^{-1}\!\left(\frac xa\right) + c

The result only fits when the $x^2$ has no number in front of it. If it has one, such as $4x^2$ , take that number out of the whole denominator first.

Keep the two fraction shapes apart: a linear bottom such as $\dfrac{1}{2x+3}$ gives a log; $x^2$ plus a positive number such as $\dfrac{1}{x^2+9}$ gives $\tan^{-1}$ .

Getting the denominator into shape first

Find the exact value of $\displaystyle\int_0^{\frac32}\frac{1}{9+4x^2}\,\mathrm{d}x$ .

Show full working

1
The $x^2$ has a $4$ in front of it, so take $4$ out of the whole denominator: $9+4x^2 = 4\left(\tfrac94 + x^2\right)$
The tan⁻¹ result needs x² on its own. Dividing 9 by 4 as well is the step people forget.
2
Move the $\tfrac14$ outside the integral: $\int\frac{1}{9+4x^2}\,\mathrm{d}x = \frac14\int\frac{1}{x^2+\tfrac94}\,\mathrm{d}x$
3
Match with $x^2+a^2$ : $a^2 = \tfrac94$ , so $a = \tfrac32$ .
a is the square root of the number added to x², not the number itself.
4
Apply the result $\dfrac1a\tan^{-1}\!\left(\dfrac xa\right)$ with $a=\tfrac32$ . Here $\dfrac1a = \dfrac23$ and $\dfrac xa = \dfrac{2x}{3}$ : $\frac14\times\frac23\tan^{-1}\!\left(\frac{2x}{3}\right) = \frac16\tan^{-1}\!\left(\frac{2x}{3}\right)$
5
Upper limit $x=\tfrac32$ : $\;\dfrac{2x}{3} = 1$ , and $\tan^{-1}1 = \tfrac14\pi$ , so the value is $\tfrac16\times\tfrac14\pi = \tfrac{1}{24}\pi$ .
tan⁻¹ gives an angle, and in integration that angle is in radians.
6
Lower limit $x=0$ : $\;\tan^{-1}0 = 0$ , so the value is $0$ .
7
Upper minus lower: $\;\tfrac{1}{24}\pi - 0 = \tfrac{1}{24}\pi$ .

Answer

$\dfrac{1}{24}\pi$

Why there is a modulus in $\ln|ax+b|$

$\ln$ only accepts positive numbers, but $\dfrac{1}{ax+b}$ is still a sensible function where $ax+b$ is negative. The modulus makes the answer work in both cases.

In a definite integral where $ax+b$ stays positive, you can drop the modulus. If $ax+b$ is zero somewhere between the limits, the integral has no value at all, because the graph shoots off to infinity there.

When the top is as big as the bottom: divide first

None of these results fits $\dfrac{x^2+3}{x^2+1}$ , because the top has the same degree (the same highest power of $x$ ) as the bottom. A fraction like this is called improper, and you must divide before you can integrate. Often you can do it by writing the top as "the bottom plus what is left":

\frac{x^2+3}{x^2+1} = \frac{\left(x^2+1\right) + 2}{x^2+1} = 1 + \frac{2}{x^2+1}

Now both pieces are standard:

\int\frac{x^2+3}{x^2+1}\,\mathrm{d}x = \int 1\,\mathrm{d}x + 2\int\frac{1}{x^2+1}\,\mathrm{d}x = x + 2\tan^{-1}x + c

This check (is the top's degree lower than the bottom's?) comes before every method in this note. When the split is not obvious, use polynomial division as in the Algebra note.

Dividing first when the split is not obvious

Find the exact value of $\displaystyle\int_0^2\frac{x^2+2x+3}{x+1}\,\mathrm{d}x$ .

Show full working

1
The top has degree $2$ and the bottom degree $1$ , so divide first. Divide the leading term $x^2$ by $x$ : the first term of the quotient is $x$ .
Polynomial division works one term at a time, exactly as in the Algebra note.
2
Multiply back and subtract: $x(x+1) = x^2 + x$ , and $\left(x^2+2x+3\right) - \left(x^2+x\right) = x + 3$ .
3
Divide the new leading term $x$ by $x$ : the next term of the quotient is $1$ . Multiply back and subtract: $1(x+1) = x+1$ , and $(x+3) - (x+1) = 2$ .
The remainder 2 has lower degree than x + 1, so the division stops here.
4
So the quotient is $x+1$ and the remainder is $2$ : $\frac{x^2+2x+3}{x+1} = x + 1 + \frac{2}{x+1}$
Check: (x + 1)(x + 1) + 2 = x² + 2x + 1 + 2 = x² + 2x + 3.
5
Integrate each term: $\int\left(x + 1 + \frac{2}{x+1}\right)\mathrm{d}x = \tfrac12x^2 + x + 2\ln(x+1)$
x + 1 is positive between 0 and 2, so no modulus is needed.
6
Upper limit $x=2$ : $\;\tfrac12(4) + 2 + 2\ln 3 = 4 + 2\ln 3$ .
7
Lower limit $x=0$ : $\;0 + 0 + 2\ln 1 = 0$ .
8
Upper minus lower: $\;4 + 2\ln 3$ .

Answer

$4 + 2\ln 3$

Divide, then a tan⁻¹

9709/33 O/N 2024 Q98 marks

(a) Find the quotient and remainder when $x^4 + 16$ is divided by $x^2 + 4$ . [3]
(b) Hence show that $\displaystyle\int_2^{2\sqrt3}\frac{x^4+16}{x^2+4}\,\mathrm{d}x = \tfrac43\left(\pi + 4\right)$ . [5]

Show full working

1
(a) Divide the leading terms: $x^4 \div x^2 = x^2$ . Multiply back and subtract: $x^2\left(x^2+4\right) = x^4 + 4x^2$ , and $\left(x^4 + 16\right) - \left(x^4 + 4x^2\right) = -4x^2 + 16$ .
Write the missing x³, x² and x terms of x⁴ + 16 as zeros if you set it out as long division.
2
Divide again: $-4x^2 \div x^2 = -4$ . Multiply back and subtract: $-4\left(x^2+4\right) = -4x^2 - 16$ , and $\left(-4x^2 + 16\right) - \left(-4x^2 - 16\right) = 32$ .
3
So the quotient is $x^2 - 4$ and the remainder is $32$ : $\frac{x^4+16}{x^2+4} = x^2 - 4 + \frac{32}{x^2+4}$
4
(b) Integrate the quotient: $\displaystyle\int\left(x^2 - 4\right)\mathrm{d}x = \tfrac13x^3 - 4x$ .
5
Integrate the remainder term with the $\tan^{-1}$ result, $a^2 = 4$ so $a = 2$ : $\int\frac{32}{x^2+4}\,\mathrm{d}x = 32\times\tfrac12\tan^{-1}\!\left(\tfrac x2\right) = 16\tan^{-1}\!\left(\tfrac x2\right)$
6
So the antiderivative is $\;\tfrac13x^3 - 4x + 16\tan^{-1}\!\left(\tfrac x2\right)$ .
7
Upper limit $x = 2\sqrt3$ : $x^3 = 8\times3\sqrt3 = 24\sqrt3$ , and $\tfrac x2 = \sqrt3$ with $\tan^{-1}\sqrt3 = \tfrac13\pi$ : $\tfrac13(24\sqrt3) - 8\sqrt3 + 16\times\tfrac13\pi = 8\sqrt3 - 8\sqrt3 + \tfrac{16}{3}\pi = \tfrac{16}{3}\pi$
(2√3)³ = 2³ × (√3)³ = 8 × 3√3.
8
Lower limit $x = 2$ : $\tfrac x2 = 1$ and $\tan^{-1}1 = \tfrac14\pi$ : $\tfrac83 - 8 + 16\times\tfrac14\pi = -\tfrac{16}{3} + 4\pi$
9
Upper minus lower: $\tfrac{16}{3}\pi - \left(-\tfrac{16}{3} + 4\pi\right) = \tfrac{16}{3}\pi - 4\pi + \tfrac{16}{3} = \tfrac43\pi + \tfrac{16}{3} = \tfrac43\left(\pi + 4\right) \;\blacksquare$
16π/3 − 12π/3 = 4π/3. Then take out the common factor 4/3.

Answer

(a) quotient $x^2 - 4$ , remainder $32$ (b) shown

Your turn

Match each integrand to its standard form. Watch the number in front of x inside each bracket, and anything that must be taken out or divided first.

1
Find $\displaystyle\int \left(\mathrm{e}^{4x-1} + \frac{1}{4x-1}\right)\mathrm{d}x$ .
Stuck? Show hint
Two different standard forms added together. Deal with them one at a time.
Show solution
1. 1
  The first term is $\mathrm{e}^{ax+b}$ with $a=4$ , so divide by $4$ : $\int \mathrm{e}^{4x-1}\,\mathrm{d}x = \tfrac14\mathrm{e}^{4x-1}$
2. 2
  The second term is $\dfrac{1}{ax+b}$ , also with $a=4$ : $\int \frac{1}{4x-1}\,\mathrm{d}x = \tfrac14\ln|4x-1|$
  Same bracket, same a, so the same division by 4 — but this one gives a log.
3. 3
  Add the two results and one constant: $\tfrac14\mathrm{e}^{4x-1} + \tfrac14\ln|4x-1| + c$
Answer
$\tfrac14\mathrm{e}^{4x-1} + \tfrac14\ln|4x-1| + c$
2
Find the exact value of $\displaystyle\int_0^{\frac{1}{12}\pi} \sec^2\!\left(2x+\tfrac{1}{6}\pi\right)\mathrm{d}x$ .
Stuck? Show hint
Divide by the number in front of x, then put each limit into the bracket before taking tan.
Show solution
1. 1
  This is $\sec^2(ax+b)$ with $a=2$ : $\int \sec^2\!\left(2x+\tfrac16\pi\right)\mathrm{d}x = \tfrac12\tan\!\left(2x+\tfrac16\pi\right)$
2. 2
  Upper limit $x=\tfrac{1}{12}\pi$ . The bracket is $2\times\tfrac{1}{12}\pi+\tfrac16\pi = \tfrac16\pi+\tfrac16\pi = \tfrac13\pi$ , and $\tan\tfrac13\pi=\sqrt3$ . Value: $\tfrac12\sqrt3$ .
  Work out the bracket first, then take tan of it.
3. 3
  Lower limit $x=0$ . The bracket is $\tfrac16\pi$ , and $\tan\tfrac16\pi = \dfrac{1}{\sqrt3}$ . Value: $\dfrac{1}{2\sqrt3}$ .
4. 4
  Upper minus lower: $\frac{\sqrt3}{2} - \frac{1}{2\sqrt3}$
5. 5
  Write both over $2\sqrt3$ . Since $\sqrt3\times\sqrt3 = 3$ , $\dfrac{\sqrt3}{2} = \dfrac{3}{2\sqrt3}$ : $\frac{3}{2\sqrt3} - \frac{1}{2\sqrt3} = \frac{2}{2\sqrt3} = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}$
  Multiplying top and bottom by √3 turns 1/√3 into √3/3, a form with no root on the bottom.
Answer
$\dfrac{\sqrt3}{3}$
3
Find $\displaystyle\int \frac{1}{2+3x^2}\,\mathrm{d}x$ .
Stuck? Show hint
The x² has a 3 in front of it. Take 3 out of the whole denominator first.
Show solution
1. 1
  Take $3$ out of the denominator so that $x^2$ stands alone: $2+3x^2 = 3\left(\tfrac23+x^2\right)$
2. 2
  Move the $\tfrac13$ outside: $\frac13\int \frac{1}{x^2+\tfrac23}\,\mathrm{d}x$ Match with $x^2+a^2$ : $a^2=\tfrac23$ , so $a=\sqrt{\tfrac23}$ .
3. 3
  Apply the $\tan^{-1}$ result: $\frac13\times\frac{1}{\sqrt{2/3}}\tan^{-1}\!\left(\frac{x}{\sqrt{2/3}}\right) + c$
4. 4
  Tidy the number in front: $3\sqrt{\tfrac23} = \sqrt{9\times\tfrac23} = \sqrt6$ , so $\dfrac13\times\dfrac{1}{\sqrt{2/3}} = \dfrac{1}{\sqrt6}$ .
  Moving the 3 inside the square root as 9 lets the fractions cancel.
5. 5
  Tidy the inside of $\tan^{-1}$ : $\dfrac{x}{\sqrt{2/3}} = x\sqrt{\tfrac32} = \dfrac{\sqrt6}{2}x$ . So the answer is $\frac{1}{\sqrt6}\tan^{-1}\!\left(\frac{\sqrt6\,x}{2}\right) + c$
  Any correct equivalent form scores; tidying just makes the answer easier to check.
Answer
$\dfrac{1}{\sqrt6}\tan^{-1}\!\left(\dfrac{\sqrt6\,x}{2}\right) + c$
4
Find the exact value of $\displaystyle\int_0^{\sqrt3}\frac{x^2+2}{x^2+1}\,\mathrm{d}x$ .
Stuck? Show hint
Top and bottom have the same degree. Write the top as (x² + 1) + something.
Show solution
1. 1
  Same degree top and bottom, so split first: $\frac{x^2+2}{x^2+1} = \frac{\left(x^2+1\right)+1}{x^2+1} = 1 + \frac{1}{x^2+1}$
2. 2
  Integrate each piece. The second is the $\tan^{-1}$ result with $a=1$ : $\int\left(1+\frac{1}{x^2+1}\right)\mathrm{d}x = x + \tan^{-1}x$
3. 3
  Upper limit $x=\sqrt3$ : $\;\sqrt3 + \tan^{-1}\sqrt3 = \sqrt3 + \tfrac13\pi$ .
  tan(π/3) = √3, so tan⁻¹√3 = π/3.
4. 4
  Lower limit $x=0$ : $\;0 + \tan^{-1}0 = 0$ .
5. 5
  Upper minus lower: $\;\sqrt3 + \tfrac13\pi$ .
Answer
$\sqrt3 + \tfrac13\pi$

Practise the six standard formsReal past-paper questions · Integration of e^(ax+b), 1/(ax+b), sin(ax+b), cos(ax+b), sec^2(ax+b), 1/(x^2+a^2)

Checking your access…

Can you do all of these?

Integrate e, 1/(ax + b), sin, cos and sec² of ax + b, dividing by a
Take the number in front of x² out before using the tan⁻¹ result
Tell a linear bottom (log) from x² plus a number (tan⁻¹)
Divide first when the top's degree is not lower than the bottom's
Spot kf′(x)/f(x) and find k, including tan x = sin x / cos x
Spot a bracket to a power times the bracket's derivative
Rewrite sin², cos², sin A cos A and tan² with an identity before integrating
Turn a product such as sin 3x cos x into a sum, and use the R cos(x + α) form to reach sec²
Read derivatives backwards: sec x tan x → sec x, cosec²x → −cot x
Split one factor off an odd power such as cos³x
Use the result of an earlier part when the question says “Hence”
Choose u for by parts as the factor that gets simpler; ln x and tan⁻¹x must be u
Use by parts twice when the polynomial is x²
Carry out a given substitution: dx, every x, and the limits
Integrate each kind of partial fraction, including the squared-bracket one
Split an Ax + B top over x² + a² into a log and a tan⁻¹
Check a product for a derivative link before using by parts
Find the limits of an area from the curve, and split at any crossing of the x-axis
Use V = π∫y² dx, keeping the π from the first line
Evaluate an integral to infinity as a limit, using e⁻ᵏˣ → 0 and tan⁻¹x → ½π
Give the answer in the form asked for: exact, single log, or a stated form