What are the key techniques for solving integration problems in Integration 3 of Pure Mathematics 3?

In Integration 3 of Pure Mathematics 3, key techniques include integration by parts, substitution, and partial fractions. These methods are essential for solving complex integrals that cannot be addressed by basic integration rules. Understanding when and how to apply each technique is crucial for success in the Edexcel International A Levels curriculum.

How does integration by parts work in the context of Integration 3?

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 Integration 3 for handling integrals involving logarithmic, exponential, and trigonometric functions.

What is the role of substitution in solving integrals in Pure Mathematics 3?

Substitution is a method used to simplify integrals by changing variables. In Integration 3, it helps transform a complex integral into a simpler form that is easier to evaluate. This technique is especially useful when dealing with integrals involving composite functions or when the integrand suggests a substitution that simplifies the expression.

How are partial fractions used in Integration 3 to solve integrals?

Partial fractions decomposition is a technique used to break down rational functions into simpler fractions that are easier to integrate. In Integration 3, this method is applied to integrals where the integrand is a rational function, allowing students to integrate each simpler fraction individually. This approach is essential for tackling complex rational expressions in the Edexcel International A Levels syllabus.

What are some common mistakes to avoid when solving integration problems in Pure Mathematics 3?

Common mistakes in Integration 3 include incorrect application of integration techniques, such as choosing the wrong function for u in integration by parts, or failing to properly simplify expressions before integrating. Additionally, students often overlook the importance of adding the constant of integration. Careful attention to detail and practice with a variety of problems can help avoid these errors.

Why is "$\int e^{3x}\,dx = e^{3x} + C$ (missing the $\dfrac{1}{3}$)" a common mistake in Integration 3?

Why it happens: Treating the chain rule effect as zero. How to avoid it: Mental check: differentiate the answer. If you get 3 e^3x but should have e^3x, you need 1/3 at the front. ALWAYS verify by differentiation. Source: WMA13/01 examiner reports — flagged annually

Why is "$\int \dfrac{1}{x}\,dx = \ln x + C$ (without modulus)" a common mistake in Integration 3?

Why it happens: Treating x as defined for all x. How to avoid it: only takes positive arguments. The full antiderivative is |x| + C. Mark schemes require the modulus.

Why is "Forgetting to replace $dx$ when substituting" a common mistake in Integration 3?

Why it happens: Only substituting for the x-terms inside the integrand. How to avoid it: When you set u = g(x), IMMEDIATELY write du = g'(x)\,dx and solve for dx. Substitute BOTH the x-expression AND dx before integrating.

Why is "Computing a definite integral via substitution but using old $x$-limits" a common mistake in Integration 3?

Why it happens: Forgetting that the variable has changed. How to avoid it: Either (i) change the limits to u-values and stay in u, OR (ii) integrate in u, back-substitute to x, then apply original x-limits. NEVER mix. Source: WMA13/01 mark schemes — strict on limit-changes

Why is "Going round in circles with IBP (e.g. $\int e^{x}\sin x\,dx$)" a common mistake in Integration 3?

Why it happens: Both functions stay the 'same kind' after differentiation/integration. How to avoid it: For e^x x\,dx, apply IBP TWICE — the integral reappears, treat the equation algebraically: I = e^x x - e^x x = e^x x - e^x x - I, so I = 1/2 e^x( x - x) + C.

Pure Mathematics 3Edexcel International A Levels Mathematics (XMA01-YMA01)

Integration 3

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Integration 3 Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

Integration of $e^{ax + b}$ , $\dfrac{1}{ax + b}$ , $\sin(ax + b)$ , $\cos(ax + b)$ — the chain-rule reverse for linear inner functions. Trig integrals via double-angle identities (e.g. $\int \sin^{2}x\,dx$ ). Substitution method and iterated integration by parts $\int u\,dv = uv - \int v\,du$ — the LIATE heuristic for choosing $u$ .

At a glance

$\int e^{ax + b}\,dx = \dfrac{1}{a} e^{ax + b} + C$ .
$\int \dfrac{1}{ax + b}\,dx = \dfrac{1}{a} \ln|ax + b| + C$ (MODULUS!).
$\int \sin(ax + b)\,dx = -\dfrac{1}{a}\cos(ax + b) + C$ .
$\sin^{2}x = \dfrac{1 - \cos 2x}{2}$ → $\int \sin^{2}x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ .
Substitution: $u = g(x)$ , $du = g'(x)\,dx$ . Change limits for definite integrals.
Integration by parts: $\int u\,dv = uv - \int v\,du$ . Use LIATE for $u$ .
$+C$ always for indefinite; $\ln|\cdot|$ ALWAYS uses modulus.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P3 7.1 — Integrate standard functions including $e^{ax+b}$ , $\dfrac{1}{ax+b}$ and trig with linear inner.
P3 7.2 — Use double-angle identities to integrate $\sin^{2}x$ , $\cos^{2}x$ etc.
P3 7.3 — Use integration by substitution to evaluate integrals.
P3 7.4 — Use integration by parts, including iterated application.

Standard integrals with linear inner

Linear inner → divide by the coefficient of $x$ .

Pattern. Whenever the inner function is LINEAR ( $ax + b$ ), the integral is the same as the basic form times $\dfrac{1}{a}$ .

Integral	Result
$\int e^{ax + b}\,dx$	$\dfrac{1}{a} e^{ax + b} + C$
$\int \dfrac{1}{ax + b}\,dx$	$\dfrac{1}{a} \ln\\|ax + b\\| + C$
$\int \sin(ax + b)\,dx$	$-\dfrac{1}{a} \cos(ax + b) + C$
$\int \cos(ax + b)\,dx$	$\dfrac{1}{a} \sin(ax + b) + C$
$\int \sec^{2}(ax + b)\,dx$	$\dfrac{1}{a} \tan(ax + b) + C$

Why the $\dfrac{1}{a}$ ? Reverse of chain rule. Differentiating $\dfrac{1}{a} e^{ax + b}$ gives $\dfrac{1}{a} \cdot a \cdot e^{ax + b} = e^{ax + b}$ . The $a$ cancels.

The modulus. $\ln$ requires positive argument. To integrate $\dfrac{1}{ax + b}$ for ALL valid $x$ , write $\ln|ax + b|$ .

Worked example. $\int e^{3x - 1}\,dx = \dfrac{1}{3} e^{3x - 1} + C$ .

Worked example. $\int \dfrac{2}{4x - 5}\,dx = 2 \cdot \dfrac{1}{4}\ln|4x - 5| + C = \dfrac{1}{2}\ln|4x - 5| + C$ .

Check by differentiation. Always reasonable: differentiate your answer; should give the integrand.

Linear inner ⇒ divide by $a$ .
$\ln$ takes MODULUS — never write $\ln(ax + b)$ alone.
Always $+ C$ for indefinite.
Verify by differentiation.

Trig integrals via double-angle

Rewrite $\sin^{2}$ and $\cos^{2}$ using $\cos 2x$ .

Half-angle identities (rearranged from double-angle): $\sin^{2}x = \dfrac{1 - \cos 2x}{2}, \qquad \cos^{2}x = \dfrac{1 + \cos 2x}{2}.$

Why use them? Direct integration of $\sin^{2}x$ is impossible — but the rewrite makes it trivial.

Worked example. $\int \sin^{2}x\,dx$ .

$\int \sin^{2}x\,dx = \int \dfrac{1 - \cos 2x}{2}\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ .

Worked example. $\int \cos^{2}x\,dx$ .

$\int \cos^{2}x\,dx = \int \dfrac{1 + \cos 2x}{2}\,dx = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$ .

Higher even powers. $\int \sin^{4}x\,dx$ : use $\sin^{4}x = (\sin^{2}x)^{2} = \left(\dfrac{1 - \cos 2x}{2}\right)^{2}$ , expand and use the half-angle identity again on the resulting $\cos^{2}(2x)$ term.

Tangent and cotangent. $\int \tan^{2}x\,dx$ : use $\tan^{2}x = \sec^{2}x - 1$ : $\int \tan^{2}x\,dx = \tan x - x + C.$

Similarly $\int \cot^{2}x\,dx = -\cot x - x + C$ .

Odd powers. $\int \sin^{3}x\,dx$ : peel off one $\sin x$ , use $\sin^{2}x = 1 - \cos^{2}x$ : $\int \sin^{3}x\,dx = \int (1 - \cos^{2}x)\sin x\,dx \;\;[\text{substitute } u = \cos x] = -\cos x + \dfrac{\cos^{3}x}{3} + C.$

$\sin^{2}x = (1 - \cos 2x)/2$ , $\cos^{2}x = (1 + \cos 2x)/2$ .
Direct application gives $\dfrac{x}{2} \mp \dfrac{\sin 2x}{4} + C$ .
$\int \tan^{2}x\,dx = \tan x - x + C$ .
Odd powers: substitute the conjugate trig function.

Integration by substitution

$u = g(x)$ , $du = g'(x)\,dx$ . Change LIMITS too.

When to substitute. When the integrand looks like $f(g(x)) \cdot g'(x)$ — there's an inner function and (some multiple of) its derivative.

Algorithm.

Pick $u = g(x)$ — the 'inner' function.
Compute $du = g'(x)\,dx$ , isolate $dx$ .
Substitute throughout, including $dx$ .
Integrate in $u$ .
(Indefinite) Back-substitute $u = g(x)$ in the final answer. (Definite) Change the $x$ -limits to $u$ -limits BEFORE integrating, then evaluate.

Worked example. $\int x\sqrt{x^{2} + 1}\,dx$ .

$u = x^{2} + 1 \Rightarrow du = 2x\,dx \Rightarrow x\,dx = \dfrac{1}{2}du$ .
$\int x \sqrt{x^{2} + 1}\,dx = \int \sqrt{u} \cdot \dfrac{1}{2}\,du = \dfrac{1}{2} \cdot \dfrac{u^{3/2}}{3/2} + C = \dfrac{1}{3} u^{3/2} + C$ .
Back-substitute: $\dfrac{1}{3}(x^{2} + 1)^{3/2} + C$ .

Definite integral with change of limits.

$\displaystyle\int_{0}^{\pi/2}\sin^{3}x \cos x\,dx$ with $u = \sin x$ :

$du = \cos x\,dx$ .
New limits: $x = 0 \Rightarrow u = 0$ ; $x = \pi/2 \Rightarrow u = 1$ .
$\displaystyle\int_{0}^{1} u^{3}\,du = \left[\dfrac{u^{4}}{4}\right]_{0}^{1} = \dfrac{1}{4}$ .

Choosing $u$ . Look for:

A function inside another function (composite). Pick $u$ = inner.
A bracketed expression to some power. Pick $u$ = bracket.
An expression whose derivative appears nearby. Pick $u$ = that expression.

Worked example. $\int \dfrac{x}{x^{2} + 1}\,dx$ .

$u = x^{2} + 1 \Rightarrow du = 2x\,dx \Rightarrow x\,dx = \dfrac{1}{2}du$ .
$\int \dfrac{1}{u} \cdot \dfrac{1}{2}\,du = \dfrac{1}{2}\ln|u| + C = \dfrac{1}{2}\ln(x^{2} + 1) + C$ (no modulus needed since $x^{2} + 1 > 0$ ).

$u = g(x)$ , $du = g'(x)\,dx$ .
Substitute $u$ AND $dx$ — both, every time.
Definite: change LIMITS, stay in $u$ .
Indefinite: integrate in $u$ then back-substitute.

Integration by parts

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

Formula. $\int u\,dv = uv - \int v\,du.$

Derived from the product rule. If the integrand is a PRODUCT of two functions, this technique can untangle it.

LIATE rule. Choose $u$ in this priority order:

L — Logarithm ( $\ln x$ ).
I — Inverse trig ( $\arcsin x$ , etc.).
A — Algebraic (polynomial, $x^{n}$ ).
T — Trig ( $\sin x$ , $\cos x$ ).
E — Exponential ( $e^{x}$ ).

Pick $u$ = whichever appears EARLIER in this list. The other becomes $dv$ .

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

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

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

LIATE: $\ln x$ (logarithm) before $x$ (algebraic). $u = \ln x$ , $dv = x\,dx$ .
$du = \dfrac{1}{x}\,dx$ , $v = \dfrac{x^{2}}{2}$ .
$\int x\ln x\,dx = \dfrac{x^{2}}{2}\ln x - \int \dfrac{x^{2}}{2} \cdot \dfrac{1}{x}\,dx = \dfrac{x^{2}}{2}\ln x - \dfrac{x^{2}}{4} + C$ .

Iterated IBP. For $\int x^{2}\sin x\,dx$ : apply IBP, then IBP again on the resulting integral.

Step 1: $u = x^{2}$ , $dv = \sin x\,dx \Rightarrow$ $\int x^{2}\sin x\,dx = -x^{2}\cos x + \int 2x\cos x\,dx$ .
Step 2: $u = 2x$ , $dv = \cos x\,dx \Rightarrow \int 2x\cos x\,dx = 2x\sin x - \int 2\sin x\,dx = 2x\sin x + 2\cos x + C$ .
Combine: $-x^{2}\cos x + 2x\sin x + 2\cos x + C$ .

Circular IBP. $\int e^{x}\sin x\,dx$ : both $e^{x}$ and $\sin x$ keep their form after differentiation/integration. After two IBPs the integral REAPPEARS — solve algebraically: $I = e^{x}\sin x - \int e^{x}\cos x\,dx, \;\; I' = e^{x}\cos x + \int e^{x}\sin x\,dx = e^{x}\cos x + I.$ Substitute: $I = e^{x}\sin x - (e^{x}\cos x + I) = e^{x}\sin x - e^{x}\cos x - I$ , so $2I = e^{x}(\sin x - \cos x)$ , giving $I = \dfrac{e^{x}(\sin x - \cos x)}{2} + C$ .

$\int \ln x\,dx$ . Sometimes called the 'trick' — but it's just IBP with $u = \ln x$ , $dv = dx$ :

$du = \dfrac{1}{x}\,dx$ , $v = x$ .
$\int \ln x\,dx = x\ln x - \int 1\,dx = x\ln x - x + C$ .

$\int u\,dv = uv - \int v\,du$ .
LIATE: pick $u$ in order L > I > A > T > E.
Iterated for polynomial × trig (or × exponential).
Circular IBP for trig × exponential — solve algebraically.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\int e^{ax + b}\,dx = \dfrac{1}{a} e^{ax + b} + C$ .
$\int \dfrac{1}{ax + b}\,dx = \dfrac{1}{a}\ln|ax + b| + C$ (always modulus).
$\int \sin(ax + b)\,dx = -\dfrac{1}{a}\cos(ax + b) + C$ .
$\int \sec^{2}(ax + b)\,dx = \dfrac{1}{a}\tan(ax + b) + C$ .
$\sin^{2}x = \dfrac{1 - \cos 2x}{2}$ , $\cos^{2}x = \dfrac{1 + \cos 2x}{2}$ .
$\int u\,dv = uv - \int v\,du$ — LIATE.
$\int \ln x\,dx = x\ln x - x + C$ .

How it’s examined

Integration 3 is a $10$ - $15$ -mark question on WMA13/01 every sitting. Typical setups: standard integrals + a definite integral ( $4$ - $5$ marks); $\int \sin^{2}x\,dx$ or related ( $3$ - $4$ marks); use substitution to find a definite integral, with change-of-limits required ( $6$ marks); use integration by parts ( $5$ - $7$ marks); iterated IBP ( $6$ - $8$ marks). Examiner reports flag: missing modulus in $\ln|\cdot|$ , missing $1/a$ factor, missing $+ C$ , wrong limit-changes in substitution. Always verify by differentiating your answer when time permits.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 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 3

Step-by-step solutions to past-paper-style questions on integration 3, written exactly the way a tutor would explain them at the board.

1Integrate $e^{3x + 2}$ and $\dfrac{1}{4x - 1}$ (4 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q3• standard integrals, linear inner
Question
Find (a) $\int e^{3x + 2}\,dx$ and (b) $\int \dfrac{1}{4x - 1}\,dx$ .
Step-by-step solution
1. Step 1
  (a) Standard form: $\int e^{ax + b}\,dx = \dfrac{1}{a} e^{ax + b} + C$ . With $a = 3$ :
  $\int e^{3x + 2}\,dx = \dfrac{1}{3} e^{3x + 2} + C$
2. Step 2
  (b) Standard form: $\int \dfrac{1}{ax + b}\,dx = \dfrac{1}{a}\ln|ax + b| + C$ . With $a = 4$ :
  $\int \dfrac{1}{4x - 1}\,dx = \dfrac{1}{4}\ln|4x - 1| + C$
Answer
(a) $\dfrac{1}{3} e^{3x + 2} + C$ . (b) $\dfrac{1}{4}\ln|4x - 1| + C$ .
Examiner tip
(a) M1 form. A1 final. (b) M1 form with modulus. A1 final. The MODULUS in $\ln|ax + b|$ matters: $\ln$ requires positive argument, so the modulus extends the formula to negative values. Mark schemes always require the modulus.
2Integrate $\sin^{2}x$ using double-angle (5 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q4• trig integral, double-angle
Question
Find $\int \sin^{2}x\,dx$ .
Step-by-step solution
1. Step 1
  Use $\cos 2x = 1 - 2\sin^{2}x \Rightarrow \sin^{2}x = \dfrac{1 - \cos 2x}{2}$ .
2. Step 2
  Substitute:
  $\int \sin^{2}x\,dx = \int \dfrac{1 - \cos 2x}{2}\,dx = \dfrac{1}{2}\int (1 - \cos 2x)\,dx$
3. Step 3
  Integrate term-by-term:
  $= \dfrac{1}{2}\left(x - \dfrac{\sin 2x}{2}\right) + C = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$
Answer
$\int \sin^{2}x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ .
Examiner tip
M1 for use of double-angle identity. A1 for the correct form $\dfrac{1 - \cos 2x}{2}$ . M1 for integrating. A1 for the $\dfrac{x}{2}$ term. A1 for the $-\dfrac{\sin 2x}{4}$ term and $+ C$ . Analogous: $\int \cos^{2}x\,dx = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$ via $\cos 2x = 2\cos^{2}x - 1$ .
3Integration by substitution: $\int x \sqrt{x^{2} + 1}\,dx$ (5 marks, P3)
Extended• Adapted from WMA13/01 January 2023 Q5• substitution
Question
Use the substitution $u = x^{2} + 1$ to find $\int x \sqrt{x^{2} + 1}\,dx$ .
Step-by-step solution
1. Step 1
  $u = x^{2} + 1 \Rightarrow \dfrac{du}{dx} = 2x \Rightarrow x\,dx = \dfrac{1}{2}du$ .
2. Step 2
  Substitute:
  $\int x \sqrt{x^{2} + 1}\,dx = \int \sqrt{u} \cdot \dfrac{1}{2}du = \dfrac{1}{2} \int u^{1/2}\,du$
3. Step 3
  Integrate:
  $= \dfrac{1}{2} \cdot \dfrac{u^{3/2}}{3/2} + C = \dfrac{1}{3} u^{3/2} + C$
4. Step 4
  Back-substitute $u = x^{2} + 1$ :
  $= \dfrac{1}{3}(x^{2} + 1)^{3/2} + C$
Answer
$\int x \sqrt{x^{2} + 1}\,dx = \dfrac{1}{3}(x^{2} + 1)^{3/2} + C$ .
Examiner tip
B1 statement of $du$ . M1 substitute and rewrite in $u$ . A1 simplified integral in $u$ . M1 integrate $u^{1/2}$ . A1 final form in $x$ . Always remember to back-substitute. Common error: stopping at $\dfrac{1}{3} u^{3/2} + C$ .
4Integration by parts: $\int x e^{x}\,dx$ (4 marks, P3)
Core• Adapted from WMA13/01 June 2024 Q6• integration by parts
Question
Find $\int x e^{x}\,dx$ using integration by parts.
Step-by-step solution
1. Step 1
  Choose $u = x$ (algebraic — A in LIATE), $dv = e^{x}\,dx$ . Then $du = dx$ and $v = e^{x}$ .
2. Step 2
  Apply $\int u\,dv = uv - \int v\,du$ :
  $\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx$
3. Step 3
  Integrate:
  $= x e^{x} - e^{x} + C = (x - 1) e^{x} + C$
Answer
$\int x e^{x}\,dx = (x - 1) e^{x} + C$ .
Examiner tip
M1 state IBP formula with $u, dv$ . A1 derive $du, v$ . M1 apply. A1 final form. LIATE order (Logs, Inverse trig, Algebraic, Trig, Exponential) tells you which to pick as $u$ : 'L' comes before 'A' which comes before 'E', so for $\int x e^{x}\,dx$ pick the algebraic $x$ as $u$ .
5Iterated integration by parts: $\int x^{2} \sin x\,dx$ (6 marks, P3)
Extended• Adapted from WMA13/01 January 2024 Q8• integration by parts, iterated
Question
Find $\int x^{2} \sin x\,dx$ .
Step-by-step solution
1. Step 1
  First IBP: $u = x^{2}$ , $dv = \sin x\,dx$ . Then $du = 2x\,dx$ , $v = -\cos x$ .
  $\int x^{2} \sin x\,dx = -x^{2}\cos x + \int 2x \cos x\,dx$
2. Step 2
  Second IBP on $\int 2x \cos x\,dx$ : $u = 2x$ , $dv = \cos x\,dx$ . Then $du = 2\,dx$ , $v = \sin x$ .
  $\int 2x \cos x\,dx = 2x \sin x - \int 2\sin x\,dx = 2x \sin x + 2\cos x + C$
3. Step 3
  Combine:
  $\int x^{2}\sin x\,dx = -x^{2}\cos x + 2x \sin x + 2\cos x + C$
Answer
$-x^{2}\cos x + 2x\sin x + 2\cos x + C$ (or $(2 - x^{2})\cos x + 2x\sin x + C$ ).
Examiner tip
M1 first IBP. A1 result of first step. M1 second IBP. A1 result of second step. A1 final answer combined. A1 $+ C$ . Iterated IBP: each time, the polynomial degree drops by $1$ . After enough iterations the polynomial is gone.
6Definite integral via substitution (6 marks, P3)
Extended• Adapted from WMA13/01 June 2023 Q7• substitution, definite integral
Question
Use $u = \sin x$ to evaluate $\displaystyle\int_{0}^{\pi/2} \sin^{3}x \cos x\,dx$ .
Step-by-step solution
1. Step 1
  $u = \sin x \Rightarrow du = \cos x\,dx$ . Change limits: $x = 0 \Rightarrow u = 0$ ; $x = \pi/2 \Rightarrow u = 1$ .
2. Step 2
  Rewrite the integral:
  $\int_{0}^{\pi/2} \sin^{3}x \cos x\,dx = \int_{0}^{1} u^{3}\,du$
3. Step 3
  Evaluate:
  $= \left[\dfrac{u^{4}}{4}\right]_{0}^{1} = \dfrac{1}{4} - 0 = \dfrac{1}{4}$
Answer
$\dfrac{1}{4}$ .
Examiner tip
B1 statement of $du$ . M1 change of limits (KEY — Edexcel mark schemes are strict). A1 integral in $u$ . M1 evaluate. A1 final value. Change-of-limits saves the back-substitution step but you MUST state the new limits explicitly.

Key Formulae — Integration 3

The formulae you need to memorise for integration 3 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Integral of $e^{ax + b}$
$\int e^{ax + b}\,dx = \dfrac{1}{a} e^{ax + b} + C$
$a, b$
constants, $a \neq 0$
When to use
Whenever the exponent is LINEAR in $x$ — divide by the coefficient of $x$ . IN the IAL formula booklet (as standard integral with adjustment).
Example
$\int e^{3x - 1}\,dx = \dfrac{1}{3} e^{3x - 1} + C$ .
Integral of $\dfrac{1}{ax + b}$
$\int \dfrac{1}{ax + b}\,dx = \dfrac{1}{a} \ln|ax + b| + C$
$a, b$
constants, $a \neq 0$
When to use
Inversely-linear functions. The MODULUS is essential — without it, the formula is invalid for $ax + b < 0$ . IN the IAL formula booklet (as standard integral).
Example
$\int \dfrac{1}{4x - 1}\,dx = \dfrac{1}{4}\ln|4x - 1| + C$ .
Trig integrals via double-angle
$\sin^{2}x = \dfrac{1 - \cos 2x}{2}, \qquad \cos^{2}x = \dfrac{1 + \cos 2x}{2}$
$x$
in radians
When to use
Integrating $\sin^{2}x$ or $\cos^{2}x$ — rewrite using double-angle, then integrate. Identities are in the IAL formula booklet (derivable from double-angle).
Example
$\int \sin^{2}x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ .
Integration by substitution
$\int f(g(x)) g'(x)\,dx = \int f(u)\,du \;\; (u = g(x))$
$u$
new variable; pick a sub-expression in the integrand
$du$
= g'(x) dx
When to use
When the integrand contains $f(g(x)) g'(x)$ — set $u = g(x)$ to simplify. For definite integrals, change the LIMITS to $u$ -values. NOT directly in the formula booklet — derive from the chain rule.
Example
$\int 2x(x^{2} + 1)^{5}\,dx$ with $u = x^{2} + 1 \Rightarrow \dfrac{u^{6}}{6} + C$ .
Integration by parts
$\int u \, dv = uv - \int v \, du$
$u, dv$
your choice of how to split the integrand
$du = u'(x) dx, v = \int dv$
When to use
Whenever the integrand is a product of a polynomial and a trig/log/exp. LIATE rule: prioritise $u$ in order Logs > Inverse trig > Algebraic > Trig > Exponential. IN the IAL formula booklet.
Example
$\int x \cos x\,dx$ with $u = x$ , $dv = \cos x\,dx$ gives $x\sin x + \cos x + C$ .

Key Definitions and Keywords — Integration 3

Definitions to memorise and the exact keywords mark schemes credit for integration 3 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Indefinite integral (antiderivative)
Examiner keyword
$\int f(x)\,dx = F(x) + C$ where $F'(x) = f(x)$ . The $+ C$ encodes that the antiderivative is determined only up to an additive constant.
Integration by substitution
Examiner keyword
A change of variable $u = g(x)$ that turns the integrand into a simpler form in $u$ . The reverse of the chain rule for differentiation.
Integration by parts
Examiner keyword
Identity derived from the product rule: $\int u\,dv = uv - \int v\,du$ . Used to integrate products of dissimilar functions.
LIATE heuristic
A rule of thumb for choosing $u$ in $\int u\,dv$ : Logarithm, Inverse trig, Algebraic (polynomial), Trig, Exponential. Pick whichever appears EARLIEST in this list as your $u$ .
Change of limits
Examiner keyword
When doing a substitution $u = g(x)$ in a definite integral, replace the $x$ -limits with the corresponding $u$ -limits. This eliminates the need to back-substitute.

Common Mistakes and Misconceptions — Integration 3

The traps other students keep falling into on integration 3 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕ $\int e^{3x}\,dx = e^{3x} + C$ (missing the $\dfrac{1}{3}$ )
WMA13/01 examiner reports — flagged annually
Why it happens
Treating the chain rule effect as zero.
How to avoid it
Mental check: differentiate the answer. If you get $3 e^{3x}$ but should have $e^{3x}$ , you need $\dfrac{1}{3}$ at the front. ALWAYS verify by differentiation.
✕ $\int \dfrac{1}{x}\,dx = \ln x + C$ (without modulus)
Why it happens
Treating $\ln x$ as defined for all $x$ .
How to avoid it
$\ln$ only takes positive arguments. The full antiderivative is $\ln|x| + C$ . Mark schemes require the modulus.
✕Forgetting to replace $dx$ when substituting
Why it happens
Only substituting for the $x$ -terms inside the integrand.
How to avoid it
When you set $u = g(x)$ , IMMEDIATELY write $du = g'(x)\,dx$ and solve for $dx$ . Substitute BOTH the $x$ -expression AND $dx$ before integrating.
✕Computing a definite integral via substitution but using old $x$ -limits
WMA13/01 mark schemes — strict on limit-changes
Why it happens
Forgetting that the variable has changed.
How to avoid it
Either (i) change the limits to $u$ -values and stay in $u$ , OR (ii) integrate in $u$ , back-substitute to $x$ , then apply original $x$ -limits. NEVER mix.
✕Going round in circles with IBP (e.g. $\int e^{x}\sin x\,dx$ )
Why it happens
Both functions stay the 'same kind' after differentiation/integration.
How to avoid it
For $\int e^{x}\sin x\,dx$ , apply IBP TWICE — the integral reappears, treat the equation algebraically: $I = e^{x}\sin x - \int e^{x}\cos x = e^{x}\sin x - e^{x}\cos x - I$ , so $I = \dfrac{1}{2} e^{x}(\sin x - \cos x) + C$ .

Practice questions

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

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Integration 3 — frequently asked questions

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

Integration 3

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Integrate e3x+2e^{3x + 2}e3x+2 and 14x−1\dfrac{1}{4x - 1}4x−11​ (4 marks, P3)

2Integrate sin⁡2x\sin^{2}xsin2x using double-angle (5 marks, P3)

3Integration by substitution: ∫xx2+1 dx\int x \sqrt{x^{2} + 1}\,dx∫xx2+1​dx (5 marks, P3)

4Integration by parts: ∫xex dx\int x e^{x}\,dx∫xexdx (4 marks, P3)

5Iterated integration by parts: ∫x2sin⁡x dx\int x^{2} \sin x\,dx∫x2sinxdx (6 marks, P3)

6Definite integral via substitution (6 marks, P3)

Integral of eax+be^{ax + b}eax+b

Integral of 1ax+b\dfrac{1}{ax + b}ax+b1​

Trig integrals via double-angle

Integration by substitution

Integration by parts

Indefinite integral (antiderivative)

Integration by substitution

Integration by parts

LIATE heuristic

Change of limits

✕∫e3x dx=e3x+C\int e^{3x}\,dx = e^{3x} + C∫e3xdx=e3x+C (missing the 13\dfrac{1}{3}31​)

✕∫1x dx=ln⁡x+C\int \dfrac{1}{x}\,dx = \ln x + C∫x1​dx=lnx+C (without modulus)

✕Forgetting to replace dxdxdx when substituting

✕Computing a definite integral via substitution but using old xxx-limits

✕Going round in circles with IBP (e.g. ∫exsin⁡x dx\int e^{x}\sin x\,dx∫exsinxdx)

Practice questions

Past paper style quiz

Video lesson

Integration 3 — frequently asked questions

1Integrate $e^{3x + 2}$ and $\dfrac{1}{4x - 1}$ (4 marks, P3)

2Integrate $\sin^{2}x$ using double-angle (5 marks, P3)

3Integration by substitution: $\int x \sqrt{x^{2} + 1}\,dx$ (5 marks, P3)

4Integration by parts: $\int x e^{x}\,dx$ (4 marks, P3)

5Iterated integration by parts: $\int x^{2} \sin x\,dx$ (6 marks, P3)

Integral of $e^{ax + b}$

Integral of $\dfrac{1}{ax + b}$

✕ $\int e^{3x}\,dx = e^{3x} + C$ (missing the $\dfrac{1}{3}$ )

✕ $\int \dfrac{1}{x}\,dx = \ln x + C$ (without modulus)

✕Forgetting to replace $dx$ when substituting

✕Computing a definite integral via substitution but using old $x$ -limits

✕Going round in circles with IBP (e.g. $\int e^{x}\sin x\,dx$ )