What are the key techniques for solving integration problems in Pure Mathematics 2?

In Pure Mathematics 2, particularly under Integration 2, key techniques include integration by parts, substitution, and partial fractions. These methods help in solving complex integrals that cannot be integrated directly. Understanding when and how to apply each technique is crucial for solving integration problems effectively in the Edexcel International A Levels curriculum.

How do you apply integration by parts in Integration 2?

Integration by parts is a technique derived from the product rule of differentiation. It is used when integrating the product of two functions. The formula is ∫u dv = uv - ∫v du, where you choose u and dv from the integrand. This method is particularly useful in Pure Mathematics 2 for integrals involving polynomials and exponential or trigonometric functions.

What is the role of substitution in solving integrals in Pure Mathematics 2?

Substitution is a powerful method used to simplify integrals by changing variables. In Integration 2, it is often used when an integral contains a composite function. By substituting a part of the integrand with a new variable, the integral becomes easier to solve. This technique is essential for tackling complex integrals in the Edexcel International A Levels Mathematics curriculum.

How are partial fractions used in Integration 2?

Partial fractions are used to decompose rational functions into simpler fractions that are easier to integrate. This technique is particularly useful in Integration 2 when dealing with integrals of rational functions. By expressing the integrand as a sum of simpler fractions, you can apply basic integration techniques to solve the integral, a skill emphasized in the Edexcel International A Levels.

What are common mistakes to avoid when integrating in Pure Mathematics 2?

Common mistakes in Integration 2 include incorrect application of integration techniques, such as choosing the wrong function for u and dv in integration by parts, or incorrect substitution. Another frequent error is neglecting the constant of integration. Understanding the properties of integrals and practicing various problems can help avoid these mistakes, which is crucial for success in the Edexcel International A Levels Mathematics exams.

Why is "Writing $\int \sin x \, dx = \cos x$ (missing minus)" a common mistake in Integration 2?

Why it happens: Confusion with the derivative pair: ()' = , but = -. How to avoid it: Memory rule: 'integrating flips the sign'. So x \, dx = - x + C. Conversely x \, dx = x + C (no sign flip). Source: WMA12/01 examiner reports — flagged annually

Why is "Omitting $+ C$ on an indefinite integral" a common mistake in Integration 2?

Why it happens: Habit slip — students write the antiderivative and move on. How to avoid it: EVERY indefinite integral has a + C. Examiners deduct 1 mark in any 'find the integral' question without it. Definite integrals (with limits) do NOT need + C.

Why is "Substituting $u = g(x)$ but forgetting to replace $dx$ with $du / g'(x)$" a common mistake in Integration 2?

Why it happens: Treating the substitution as 'just renaming the variable'. How to avoid it: Always do TWO things: (1) replace g(x) with u, AND (2) replace g'(x) \, dx with du. Differentiate the substitution: du / dx = g'(x), so du = g'(x) \, dx.

Why is "Choosing $u = e^{x}$ in $\int x e^{x} \, dx$" a common mistake in Integration 2?

Why it happens: Random choice without strategy. How to avoid it: LIATE rule: pick u in the order Logarithmic, Inverse trig, Algebraic, Trig, Exponential — whichever appears EARLIEST in this list. In x e^x dx, x is algebraic (A) and e^x is exponential (E), so pick u = x (A before E).

Why is "Computing $\int (\text{lower} - \text{upper}) \, dx$ and reporting a negative area" a common mistake in Integration 2?

Why it happens: Subtracting the curves in the wrong order. How to avoid it: Test the order by evaluating both curves at an interior x-value: whichever is bigger is the upper. Always subtract LOWER FROM UPPER. If you compute a negative answer, take the absolute value (and explain why) — but the cleaner approach is to do the subtraction in the right order from the start.

Pure Mathematics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Integration 2

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Integration 2 Study Notes — Edexcel IAL Mathematics P2 (WMA12/01, 2018 spec — 2026 onwards)

Integration of $\sin x$ , $\cos x$ , $e^{x}$ , $1/x$ . Integration by substitution (simple cases). Introductory integration by parts. Evaluation of definite integrals and area between two curves.

At a glance

Power rule: $\int x^{n} dx = \dfrac{x^{n + 1}}{n + 1} + C$ for $n \neq -1$ .
Standard integrals: $\int \sin x \, dx = -\cos x + C$ ; $\int \cos x \, dx = \sin x + C$ .
Exp/log: $\int e^{x} dx = e^{x} + C$ ; $\int (1/x) dx = \ln|x| + C$ .
Substitution: $u = g(x) \Rightarrow du = g'(x) dx$ . Edexcel always tells you which substitution.
Parts: $\int u \, dv = uv - \int v \, du$ — in the formula booklet.
Definite integral: $F(b) - F(a)$ . No $+ C$ .
Area between curves: $\int_{a}^{b} (\text{upper} - \text{lower}) \, dx$ .

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

P2 8.1 — Integrate $\sin kx$ , $\cos kx$ , $e^{kx}$ and $1/x$ (each times a constant) as standard results.
P2 8.2 — Integrate using a given substitution.
P2 8.3 — Use integration by parts in straightforward cases.
P2 8.4 — Evaluate definite integrals; understand their geometrical interpretation as area.
P2 8.5 — Find the area of a region between a curve and the $x$ -axis, or between two curves.

Standard integrals

Reverse of the differentiation table. Sign flips on $\int \sin$ .

Integrand	Antiderivative
$x^{n}$ ( $n \neq -1$ )	$\dfrac{x^{n + 1}}{n + 1}$
$\sin x$	$-\cos x$
$\cos x$	$\sin x$
$e^{x}$	$e^{x}$
$1/x$	$\ln
$a^{x}$	$\dfrac{a^{x}}{\ln a}$

(All antiderivatives need a $+ C$ for indefinite integrals.)

Sign flip alert. $\int \sin x \, dx = -\cos x + C$ , NOT $\cos x$ . Memory: 'integrating $\sin$ flips the sign'. $\int \cos x \, dx = \sin x + C$ has NO sign flip.

Linear chain. When the inner function is linear, divide by its coefficient:

$\int \sin(kx) \, dx = -\dfrac{1}{k} \cos(kx) + C, \qquad \int e^{kx} \, dx = \dfrac{1}{k} e^{kx} + C, \qquad \int \dfrac{1}{ax + b} \, dx = \dfrac{1}{a} \ln|ax + b| + C.$

Worked example. $\displaystyle \int (2 \sin x - 3 \cos x + \dfrac{4}{x} - e^{x}) \, dx = -2 \cos x - 3 \sin x + 4 \ln|x| - e^{x} + C$ .

Linear chain example. $\int \cos 3x \, dx = \tfrac{1}{3} \sin 3x + C$ . Differentiate to check: $\tfrac{1}{3} \cdot 3 \cos 3x = \cos 3x$ . ✓

The area between two curves on [a, b] is ∫ₐᵇ (upper curve − lower curve) dx.

$\int \sin x = -\cos x$ (FLIP).
$\int \cos x = \sin x$ (NO flip).
$\int 1/x = \ln|x|$ — modulus included.
Linear chain: divide by inner coefficient.
Always $+ C$ for indefinite integrals.

Integration by substitution

Reverse of the chain rule. Edexcel tells you the substitution.

Idea. Reverse of the chain rule. If the integrand has the form $f(g(x)) g'(x)$ , set $u = g(x)$ . Then $du = g'(x) dx$ and the integral becomes $\int f(u) \, du$ .

Procedure.

Identify $u$ (always TOLD in P2).
Differentiate: $du = g'(x) dx$ .
Rewrite the integrand in terms of $u$ and $du$ .
Integrate.
Substitute back to get the answer in $x$ .

Worked example. Use $u = x^{2} + 1$ to find $\int 4x (x^{2} + 1)^{3} \, dx$ .

$du = 2x \, dx$ , so $2x \, dx = du$ . The integrand becomes $4x (x^{2} + 1)^{3} \, dx = 2(x^{2} + 1)^{3} \cdot 2x \, dx = 2 u^{3} \, du$ .
$\int 2 u^{3} \, du = \tfrac{1}{2} u^{4} + C$ .
Back-substitute: $\tfrac{1}{2}(x^{2} + 1)^{4} + C$ .

Verify by differentiating. $\dfrac{d}{dx}[\tfrac{1}{2}(x^{2} + 1)^{4}] = \tfrac{1}{2} \cdot 4(x^{2} + 1)^{3} \cdot 2x = 4x (x^{2} + 1)^{3}$ . ✓

Definite-integral substitution. Change the LIMITS too. If $u = g(x)$ and $x$ ranges from $a$ to $b$ , then $u$ ranges from $g(a)$ to $g(b)$ . No need to back-substitute at the end.

Worked example. $\int_{0}^{1} 4x(x^{2} + 1)^{3} \, dx$ with $u = x^{2} + 1$ .

When $x = 0$ , $u = 1$ . When $x = 1$ , $u = 2$ .
$\int_{1}^{2} 2 u^{3} \, du = [\tfrac{u^{4}}{2}]_{1}^{2} = 8 - 1/2 = 15/2$ .

Edexcel always tells you the substitution.
Replace $g(x)$ with $u$ AND $g'(x) dx$ with $du$ .
Definite integral: change the limits.
Verify by differentiating.

Integration by parts

$\int u \, dv = uv - \int v \, du$ . Pick $u$ via LIATE.

Formula. From the product rule, $\int u \, dv = uv - \int v \, du$ .

(GIVEN in the IAL formula booklet.)

LIATE rule for choosing $u$ . Pick $u$ to be whichever appears EARLIEST in the list:

Logarithmic
Inverse trigonometric
Algebraic
Trigonometric
Exponential

So in $\int x e^{x} \, dx$ : $x$ is algebraic (A), $e^{x}$ is exponential (E). A before E, so $u = x$ .

Procedure.

Pick $u$ (via LIATE). The remaining factor including $dx$ is $dv$ .
Differentiate $u$ to find $du$ .
Integrate $dv$ to find $v$ .
Apply the formula.
The remaining integral $\int v \, du$ should be EASIER than the original. If not, you picked $u$ wrong.

Worked example. $\int x e^{x} \, dx$ .

$u = x$ , $du = dx$ . $dv = e^{x} \, dx$ , $v = e^{x}$ .
$\int x e^{x} \, dx = x e^{x} - \int e^{x} \, dx = x e^{x} - e^{x} + C = (x - 1) e^{x} + C$ .

Worked example. $\int \ln x \, dx$ .

$u = \ln x$ , $du = (1/x) \, dx$ . $dv = dx$ , $v = x$ .
$\int \ln x \, dx = x \ln x - \int x \cdot (1/x) \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C$ .

When to apply parts twice. Integrals like $\int x^{2} e^{x} \, dx$ need parts TWICE (first with $u = x^{2}$ , reducing to $\int x e^{x}$ , then parts again). P2 questions rarely require this — they're more common in P3.

$\int u \, dv = uv - \int v \, du$ .
Choose $u$ by LIATE.
Each parts step must reduce the difficulty.
$\int \ln x$ uses $u = \ln x$ , $dv = dx$ — classic.

Definite integrals and area

$\int_{a}^{b} f = F(b) - F(a)$ . No $+ C$ . Geometric meaning: signed area.

Fundamental theorem of calculus. If $F$ is an antiderivative of $f$ :

$\int_{a}^{b} f(x) \, dx = F(b) - F(a) \equiv [F(x)]_{a}^{b}.$

No $+ C$ in definite integrals — the constants cancel in $F(b) - F(a)$ .

Geometric interpretation. $\int_{a}^{b} f(x) \, dx$ is the signed area between the curve $y = f(x)$ and the $x$ -axis on $[a, b]$ . Regions where $f > 0$ count positively; regions where $f < 0$ count negatively.

For absolute area (always positive), integrate $|f|$ — typically by splitting at the $x$ -intercepts.

Worked example. $\displaystyle \int_{0}^{\pi/2} (\cos x + 2 \sin x) \, dx$ .

Antiderivative: $\sin x - 2 \cos x$ .
$\sin(\pi/2) - 2 \cos(\pi/2) = 1 - 0 = 1$ .
$\sin 0 - 2 \cos 0 = 0 - 2 = -2$ .
Integral: $1 - (-2) = 3$ .

Properties.

$\int_{a}^{a} f = 0$ .
$\int_{b}^{a} f = -\int_{a}^{b} f$ (reversing limits flips sign).
$\int_{a}^{c} f = \int_{a}^{b} f + \int_{b}^{c} f$ (additivity).
$\int_{a}^{b} (cf + g) = c \int_{a}^{b} f + \int_{a}^{b} g$ (linearity).

$\int_{a}^{b} f = F(b) - F(a)$ , NO $+ C$ .
Signed area: $f > 0$ counts positive, $f < 0$ negative.
Reversing limits flips the sign.
Split at zeros if you need absolute area.

Area between two curves

$\int_{a}^{b}(\text{upper} - \text{lower}) \, dx$ where $a$ and $b$ are intersection points.

Procedure.

Find intersection points. Set $y_{1} = y_{2}$ and solve for $x$ . These give the integration limits $a$ and $b$ .
Determine which curve is on top. Pick an interior $x$ -value and evaluate both. The bigger one is the UPPER curve.
Integrate the difference. $A = \displaystyle \int_{a}^{b}(\text{upper} - \text{lower}) \, dx$ .

Worked example. Area between $y = x^{2}$ and $y = 4x - x^{2}$ .

Set equal: $x^{2} = 4x - x^{2}$ , so $2 x^{2} - 4x = 0$ , giving $x = 0, 2$ .
At $x = 1$ : $y_{1} = 1$ , $y_{2} = 3$ . So $y = 4x - x^{2}$ is upper.
$A = \int_{0}^{2}[(4x - x^{2}) - x^{2}] \, dx = \int_{0}^{2}(4x - 2 x^{2}) \, dx = [2 x^{2} - \tfrac{2}{3} x^{3}]_{0}^{2} = 8 - \tfrac{16}{3} = \tfrac{8}{3}$ .

Multiple regions. If two curves cross more than twice within the region of interest, split the integral into pieces — in each piece, the same curve is on top.

Vertical strips vs horizontal strips. Standard approach is vertical strips: integrate w.r.t. $x$ . For some regions it's easier to use horizontal strips (integrate w.r.t. $y$ , swap roles of upper/lower). In P2 you can stick with vertical strips for all questions.

Sign safety. If you compute a negative answer, you subtracted in the wrong order. Take the absolute value (and note this in working).

Intersections give the limits.
Check which curve is on top at an interior point.
Subtract LOWER from UPPER.
Split if the curves cross within the region.

How it’s examined

Integration 2 appears in WMA12/01 P2 every sitting (Jan & Jun), typically as Q5-Q14 — worth $10$ - $15$ marks total. Typical questions: integrate a sum of standard functions (4 marks); definite integral with trig and exponentials (4-5 marks); integration by given substitution (5-6 marks); integration by parts on $x e^{x}$ or $\ln x$ (5 marks); area between two curves with quadratic intersections (6-8 marks). Mark schemes award the antiderivative explicitly (often 2-3 A1 marks per term), then the application of limits, then the final value. A* students should aim for full marks here — every integral is verifiable by differentiation.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA12/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Integration 2

Step-by-step solutions to past-paper-style questions on integration 2, written exactly the way a tutor would explain them at the board.

1Standard integrals (4 marks, P2)
Core• Adapted from WMA12/01 January 2024 Q11• standard integrals
Question
Find $\displaystyle\int \left(2 \sin x - 3 \cos x + \dfrac{4}{x} - e^{x}\right) dx$ .
Step-by-step solution
1. Step 1
  Integrate term by term using standard results:
  $\int \sin x \, dx = -\cos x, \quad \int \cos x \, dx = \sin x, \quad \int \dfrac{1}{x} \, dx = \ln|x|, \quad \int e^{x} \, dx = e^{x}$
2. Step 2
  Apply to each term:
  $= 2(-\cos x) - 3(\sin x) + 4 \ln|x| - e^{x} + C$
3. Step 3
  Tidy:
  $-2 \cos x - 3 \sin x + 4 \ln|x| - e^{x} + C$
Answer
$-2 \cos x - 3 \sin x + 4 \ln|x| - e^{x} + C$ .
Examiner tip
B1 for each of the four standard integrals. The $-\cos$ from integrating $\sin$ is the biggest sign-error site; another is forgetting $|x|$ inside the $\ln$ (Edexcel mark schemes accept $\ln x$ without the modulus when context makes $x > 0$ clear, but $\ln|x|$ is always safe). Don't forget $+ C$ for an indefinite integral — that loses 1 mark.
2Definite trig integral (4 marks, P2)
Core• Adapted from WMA12/01 June 2024 Q10• definite integral, trig
Question
Evaluate $\displaystyle\int_{0}^{\pi/2} (\cos x + 2 \sin x) \, dx$ .
Step-by-step solution
1. Step 1
  Antiderivative: $\int (\cos x + 2 \sin x) \, dx = \sin x - 2 \cos x + C$ .
2. Step 2
  Evaluate at upper limit $\pi/2$ : $\sin(\pi/2) - 2 \cos(\pi/2) = 1 - 0 = 1$ .
3. Step 3
  Evaluate at lower limit $0$ : $\sin 0 - 2 \cos 0 = 0 - 2 = -2$ .
4. Step 4
  Subtract:
  $1 - (-2) = 3$
Answer
$3$ .
Examiner tip
M1 for both antiderivatives correct. M1 for substituting the limits. A1 for $1$ . A1 for $-(-2) = +2$ and final answer $3$ . The sign on the lower-limit substitution is where most students lose a mark: $-(-2)$ becomes $+2$ , not $-2$ .
3Integration by substitution (5 marks, P2)
Extended• Adapted from WMA12/01 January 2023 Q11• substitution
Question
Use the substitution $u = x^{2} + 1$ to find $\displaystyle\int 4x (x^{2} + 1)^{3} \, dx$ .
Step-by-step solution
1. Step 1
  Let $u = x^{2} + 1$ . Then $\dfrac{du}{dx} = 2x$ , so $du = 2x \, dx$ , i.e. $2x \, dx = du$ .
2. Step 2
  Rewrite the integrand: $4x (x^{2} + 1)^{3} \, dx = 2 (x^{2} + 1)^{3} \cdot 2x \, dx = 2 u^{3} \, du$ .
3. Step 3
  Integrate:
  $\int 2 u^{3} \, du = \dfrac{2 u^{4}}{4} + C = \dfrac{u^{4}}{2} + C$
4. Step 4
  Substitute back:
  $= \dfrac{(x^{2} + 1)^{4}}{2} + C$
Answer
$\dfrac{(x^{2} + 1)^{4}}{2} + C$ .
Examiner tip
M1 for the substitution and finding $du$ in terms of $dx$ . M1 for rewriting the integrand fully in $u$ . A1 for the integral in $u$ . A1 for substituting back. A1 for $+ C$ . Verification: differentiate $\tfrac{1}{2}(x^{2} + 1)^{4}$ — chain rule gives $\tfrac{1}{2} \cdot 4 (x^{2} + 1)^{3} \cdot 2x = 4 x (x^{2} + 1)^{3}$ . ✓
4Integration by parts on $x e^{x}$ (5 marks, P2)
Extended• Adapted from WMA12/01 June 2023 Q12• integration by parts
Question
Find $\displaystyle\int x e^{x} \, dx$ .
Step-by-step solution
1. Step 1
  Integration by parts: $\int u \, dv = uv - \int v \, du$ . Choose $u = x$ (because it simplifies on differentiation) and $dv = e^{x} \, dx$ .
2. Step 2
  Then $du = dx$ and $v = e^{x}$ .
3. Step 3
  Apply:
  $\int x e^{x} \, dx = x e^{x} - \int e^{x} \, dx$
4. Step 4
  Compute the remaining integral:
  $x e^{x} - e^{x} + C = (x - 1) e^{x} + C$
Answer
$(x - 1) e^{x} + C$ .
Examiner tip
M1 for the parts structure with correct choice of $u$ and $dv$ . A1 for $u, du, v, dv$ stated. M1 for substituting into the formula. A1 for the second integral. A1 for the final answer factored. Choosing $u = x$ is critical — choosing $u = e^{x}$ instead generates an integral that's MORE complicated than the original. Use the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) to pick $u$ .
5Area between two curves (7 marks, P2)
Extended• Adapted from WMA12/01 January 2024 Q14• area, between curves
Question
The curves $y = x^{2}$ and $y = 4x - x^{2}$ enclose a finite region. Find its area.
Step-by-step solution
1. Step 1
  Find intersection points. Set $x^{2} = 4x - x^{2}$ :
  $2 x^{2} - 4x = 0 \;\;\Rightarrow\;\; 2x(x - 2) = 0$
2. Step 2
  Intersection at $x = 0$ and $x = 2$ .
3. Step 3
  Determine which curve is on top. At $x = 1$ : $y_{1} = 1$ , $y_{2} = 4 - 1 = 3$ . So $y = 4x - x^{2}$ is the upper curve on $[0, 2]$ .
4. Step 4
  Area = $\displaystyle\int_{0}^{2} [(4x - x^{2}) - x^{2}] \, dx = \int_{0}^{2} (4x - 2 x^{2}) \, dx$ .
5. Step 5
  Antiderivative:
  $2 x^{2} - \dfrac{2 x^{3}}{3}$
6. Step 6
  Evaluate at $x = 2$ : $2(4) - \dfrac{2(8)}{3} = 8 - \dfrac{16}{3} = \dfrac{24 - 16}{3} = \dfrac{8}{3}$ .
7. Step 7
  At $x = 0$ : $0$ . So area $= \dfrac{8}{3}$ .
Answer
Area $= \dfrac{8}{3}$ square units.
Examiner tip
M1 for setting the curves equal. A1 for $x = 0, 2$ . M1 for the difference (upper minus lower). M1 for the integration. A1 for the antiderivative. M1 for substituting both limits. A1 for $8/3$ . Critical step: subtract LOWER from UPPER to get a positive area — students who subtract the wrong way get $-8/3$ and either lose a mark or have to recover with 'taking absolute value'.

Key Formulae — Integration 2

The formulae you need to memorise for integration 2 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Power rule (integration)
$\int x^{n} \, dx = \dfrac{x^{n + 1}}{n + 1} + C \quad (n \neq -1)$
$n$
any real number except $-1$
When to use
Integrating polynomials, surds and reciprocal powers (other than $1/x$ ). NOT in the IAL formula booklet — memorise.
Example
$\int x^{3} \, dx = \dfrac{x^{4}}{4} + C$ .
Integrals of $\sin$ and $\cos$
$\int \sin x \, dx = -\cos x + C, \qquad \int \cos x \, dx = \sin x + C$
$x$
angle in radians
When to use
Integrating trig functions (radians only). NOT in the formula booklet.
Example
$\int 3 \sin 2x \, dx = -\dfrac{3}{2} \cos 2x + C$ (linear chain).
Integral of $e^{x}$
$\int e^{x} \, dx = e^{x} + C$
When to use
Standard integral. NOT in the formula booklet.
Example
$\int e^{3x} \, dx = \dfrac{1}{3} e^{3x} + C$ (linear chain).
Integral of $1/x$
$\int \dfrac{1}{x} \, dx = \ln|x| + C$
$x$
$x \neq 0$
When to use
The 'missing' case of the power rule ( $n = -1$ ). The modulus extends the formula to negative $x$ . NOT in the formula booklet.
Example
$\int \dfrac{3}{2x + 1} \, dx = \dfrac{3}{2} \ln|2x + 1| + C$ (linear chain).
Integration by parts
$\int u \, dv = uv - \int v \, du$
$u$
function chosen to differentiate
$dv$
the remaining differential, integrated to give $v$
When to use
Integrals of products $\int u(x) v'(x) \, dx$ where one factor simplifies on differentiation (LIATE rule). GIVEN in the IAL formula booklet (Pure Mathematics).
Example
$\int x e^{x} \, dx = (x - 1) e^{x} + C$ with $u = x$ , $dv = e^{x} \, dx$ .
Integration by substitution
$\int f(g(x)) g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x)$
$g(x)$
inner function (substituted as $u$ )
$g'(x) dx$
becomes $du$
When to use
When the integrand contains a function and its derivative (up to a constant). Edexcel P2 questions always state which substitution to use. NOT directly in the formula booklet (but follows from chain rule).
Example
$u = x^{2} + 1$ in $\int 4x(x^{2} + 1)^{3} \, dx$ .
Area between two curves
$A = \int_{a}^{b} [\text{upper}(x) - \text{lower}(x)] \, dx$
$a, b$
intersection $x$ -values
$\text{upper}, \text{lower}$
the two curves, with upper $\geq$ lower on $[a, b]$
When to use
Finding the enclosed area between two curves. NOT in the formula booklet.
Example
$y = x^{2}$ and $y = 4x - x^{2}$ between $x = 0$ and $x = 2$ : area $= 8/3$ .

Key Definitions and Keywords — Integration 2

Definitions to memorise and the exact keywords mark schemes credit for integration 2 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Indefinite integral
Examiner keyword
An antiderivative plus an arbitrary constant of integration $C$ . $\int f(x) \, dx = F(x) + C$ where $F'(x) = f(x)$ .
Definite integral
Examiner keyword
$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$ . No constant of integration. Represents net signed area between the curve and the $x$ -axis from $x = a$ to $x = b$ .
Substitution (in integration)
Examiner keyword
A change of variable that converts an integral into a simpler form. Introduce a new variable $u = g(x)$ , differentiate to find $du$ , and re-express everything in terms of $u$ .
Integration by parts
Examiner keyword
A technique based on the product rule for differentiation: $\int u \, dv = uv - \int v \, du$ . Used when the integrand is a product whose differentiation/integration is simpler than the original.
Area between curves
The region enclosed by two curves on an interval. Compute as $\int (\text{upper} - \text{lower}) \, dx$ between the curves' intersection points.

Common Mistakes and Misconceptions — Integration 2

The traps other students keep falling into on integration 2 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Writing $\int \sin x \, dx = \cos x$ (missing minus)
WMA12/01 examiner reports — flagged annually
Why it happens
Confusion with the derivative pair: $(\sin)' = \cos$ , but $\int \sin = -\cos$ .
How to avoid it
Memory rule: 'integrating $\sin$ flips the sign'. So $\int \sin x \, dx = -\cos x + C$ . Conversely $\int \cos x \, dx = \sin x + C$ (no sign flip).
✕Omitting $+ C$ on an indefinite integral
Why it happens
Habit slip — students write the antiderivative and move on.
How to avoid it
EVERY indefinite integral has a $+ C$ . Examiners deduct 1 mark in any 'find the integral' question without it. Definite integrals (with limits) do NOT need $+ C$ .
✕Substituting $u = g(x)$ but forgetting to replace $dx$ with $du / g'(x)$
Why it happens
Treating the substitution as 'just renaming the variable'.
How to avoid it
Always do TWO things: (1) replace $g(x)$ with $u$ , AND (2) replace $g'(x) \, dx$ with $du$ . Differentiate the substitution: $du / dx = g'(x)$ , so $du = g'(x) \, dx$ .
✕Choosing $u = e^{x}$ in $\int x e^{x} \, dx$
Why it happens
Random choice without strategy.
How to avoid it
LIATE rule: pick $u$ in the order Logarithmic, Inverse trig, Algebraic, Trig, Exponential — whichever appears EARLIEST in this list. In $\int x e^{x} dx$ , $x$ is algebraic ( $A$ ) and $e^{x}$ is exponential ( $E$ ), so pick $u = x$ ( $A$ before $E$ ).
✕Computing $\int (\text{lower} - \text{upper}) \, dx$ and reporting a negative area
Why it happens
Subtracting the curves in the wrong order.
How to avoid it
Test the order by evaluating both curves at an interior $x$ -value: whichever is bigger is the upper. Always subtract LOWER FROM UPPER. If you compute a negative answer, take the absolute value (and explain why) — but the cleaner approach is to do the subtraction in the right order from the start.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Integration 2 — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Integration 2

Short Study Notes in the form of Slides

Short Study Notes — Integration 2

Detailed Notes

What you’ll learn

How it’s examined

1Standard integrals (4 marks, P2)

2Definite trig integral (4 marks, P2)

3Integration by substitution (5 marks, P2)

4Integration by parts on xexx e^{x}xex (5 marks, P2)

5Area between two curves (7 marks, P2)

Power rule (integration)

Integrals of sin⁡\sinsin and cos⁡\coscos

Integral of exe^{x}ex

Integral of 1/x1/x1/x

Integration by parts

Integration by substitution

Area between two curves

Indefinite integral

Definite integral

Substitution (in integration)

Integration by parts

Area between curves

✕Writing ∫sin⁡x dx=cos⁡x\int \sin x \, dx = \cos x∫sinxdx=cosx (missing minus)

✕Omitting +C+ C+C on an indefinite integral

✕Substituting u=g(x)u = g(x)u=g(x) but forgetting to replace dxdxdx with du/g′(x)du / g'(x)du/g′(x)

✕Choosing u=exu = e^{x}u=ex in ∫xex dx\int x e^{x} \, dx∫xexdx

✕Computing ∫(lower−upper) dx\int (\text{lower} - \text{upper}) \, dx∫(lower−upper)dx and reporting a negative area

Practice questions

Past paper style quiz

Assess your understanding

Integration 2 — frequently asked questions

4Integration by parts on $x e^{x}$ (5 marks, P2)

Integrals of $\sin$ and $\cos$

Integral of $e^{x}$

Integral of $1/x$

✕Writing $\int \sin x \, dx = \cos x$ (missing minus)

✕Omitting $+ C$ on an indefinite integral

✕Substituting $u = g(x)$ but forgetting to replace $dx$ with $du / g'(x)$

✕Choosing $u = e^{x}$ in $\int x e^{x} \, dx$

✕Computing $\int (\text{lower} - \text{upper}) \, dx$ and reporting a negative area