Grade 12 · Pure Mathematics 2 · Cambridge A-Level 9709 · Age 17–18
Welcome to Further Integration
Further integration unlocks the computation of much harder integrals needed for finding volumes of revolution, areas between curves, and solving differential equations. Building on basic integration, you will master four powerful techniques: substitution, integration by parts, partial fractions, and standard trigonometric forms. These methods appear throughout Cambridge A-Level 9709 Paper 2 and Paper 3.
Integrate by substitution — recognise and reverse the chain rule
Use integration by parts once for products like x·eˣ or x·cos(x)
Use integration by parts twice for expressions like x²·eˣ
Integrate rational functions using partial fractions
Integrate 1/(ax+b) to obtain a logarithmic result
Integrate standard trig forms including sin²(x) using double angle identities
Integrate e^(ax+b) correctly with the 1/a factor
Evaluate definite integrals using all four methods
Compute areas under and between curves using advanced integration
Choose the correct integration method given any integrand
Topics in This Module
Integration by Substitution
Reverse the chain rule by introducing a new variable u = g(x)
Integration by Parts
Handle products using ∫u dv = uv − ∫v du; use LIATE to choose u
Using Partial Fractions
Decompose rational functions then integrate each term as ln|...|
Standard Trig Integrals
Use double angle identities to integrate sin²(x) and cos²(x)
Definite Integrals
Apply all methods to definite integrals; change limits with substitution
Applications
Area between curves, volumes of revolution setup, differential equations
Learn 1 — Integration by Substitution
Integration by substitution is the reverse of the chain rule. When an integrand has the form f(g(x))·g'(x), let u = g(x) so the integral transforms into a simpler form in u.
Method Steps
Step 1: Identify a suitable substitution u = g(x) — usually the inner function of a composite.
Step 2: Differentiate to find du/dx, then express dx = du/g'(x).
Step 3: Substitute both the function and dx into the integral so everything is in terms of u.
Step 4: Integrate with respect to u.
Step 5: Back-substitute to return to x (for indefinite integrals).
For definite integrals: change the limits using u = g(x) at the original limits — do NOT back-substitute.
Example A: ∫ x(x²+1)⁵ dx
Let u = x²+1 → du/dx = 2x → x dx = du/2
∫ x(x²+1)⁵ dx = ∫ u⁵ · (du/2) = (1/2)·(u⁶/6) + C = u⁶/12 + C
Back-substitute: (x²+1)⁶/12 + C
Example B: ∫ sin(3x+2) dx
Let u = 3x+2 → du = 3 dx → dx = du/3
∫ sin(u) · du/3 = −cos(u)/3 + C = −cos(3x+2)/3 + C
Example C: ∫ x·e^(x²) dx
Let u = x² → du = 2x dx → x dx = du/2
∫ e^u · du/2 = e^u/2 + C = e^(x²)/2 + C
Definite Integrals with Substitution — Change the Limits!
Example: ∫₀¹ 2x·e^(x²) dx. Let u = x²: when x=0, u=0; when x=1, u=1.
∫₀¹ e^u du = [e^u]₀¹ = e¹ − e⁰ = e − 1 ≈ 1.718
When to Use Substitution
Look for an integrand of the form f(g(x))·g'(x). The derivative of the inner function must appear (or nearly appear) as a factor. Common signals: composite functions, exponentials with a function in the exponent, trig of a polynomial, rational functions where numerator is derivative of denominator.
Warning: Always transform dx as well as the function. Forgetting to replace dx is the most common error. If du = g'(x)dx, then dx = du/g'(x).
Learn 2 — Integration by Parts
Integration by parts handles integrals of products where substitution does not apply. It is derived directly from the product rule of differentiation.
∫ u · (dv/dx) dx = uv − ∫ v · (du/dx) dx
LIATE Rule — Choosing u
Choose u as the first type in the list that appears in your integrand: Logarithm → Inverse trig → Algebraic (polynomial) → Trigonometric → Exponential
The remaining factor becomes dv/dx. The goal is for ∫v du to be simpler than the original.
Example A: ∫ x·eˣ dx
LIATE: x is Algebraic, eˣ is Exponential → let u = x, dv/dx = eˣ
du/dx = 1, v = eˣ
∫ x·eˣ dx = x·eˣ − ∫ eˣ·1 dx = x·eˣ − eˣ + C = eˣ(x−1) + C
Example B: ∫ x·cos(x) dx
u = x (Algebraic), dv/dx = cos(x) (Trig)
du/dx = 1, v = sin(x)
∫ x·cos(x) dx = x·sin(x) − ∫ sin(x) dx = x·sin(x) + cos(x) + C
Example C: ∫ ln(x) dx
Write as ∫ ln(x)·1 dx. LIATE: Logarithm comes first → u = ln(x), dv/dx = 1
du/dx = 1/x, v = x
∫ ln(x) dx = x·ln(x) − ∫ x·(1/x) dx = x·ln(x) − ∫ 1 dx = x·ln(x) − x + C
Repeated Integration by Parts: ∫ x²·eˣ dx
First application: u = x², dv = eˣ dx → du = 2x dx, v = eˣ
= x²·eˣ − ∫ 2x·eˣ dx Second application on ∫ 2x·eˣ dx: u = 2x, dv = eˣ dx → du = 2 dx, v = eˣ
= x²·eˣ − [2x·eˣ − ∫ 2eˣ dx] = x²·eˣ − 2x·eˣ + 2eˣ + C = eˣ(x²−2x+2) + C
Always write out u, dv, du, v clearly in a box before starting. Rushing this step causes the most errors in exam conditions.
Learn 3 — Integration by Parts (Advanced)
Cyclic Integration: ∫ eˣ·cos(x) dx
Sometimes applying parts twice returns you to the original integral. You can then solve algebraically.
Let I = ∫ eˣ·cos(x) dx 1st application: u = eˣ, dv = cos(x)dx → du = eˣ dx, v = sin(x)
I = eˣ·sin(x) − ∫ eˣ·sin(x) dx
2nd application on ∫ eˣ·sin(x) dx: u = eˣ, dv = sin(x)dx → du = eˣ dx, v = −cos(x)
∫ eˣ·sin(x) dx = −eˣ·cos(x) + ∫ eˣ·cos(x) dx = −eˣ·cos(x) + I
Substituting back: I = eˣ·sin(x) − (−eˣ·cos(x) + I)
I = eˣ·sin(x) + eˣ·cos(x) − I
2I = eˣ(sin(x) + cos(x)) I = eˣ(sin(x) + cos(x))/2 + C
In cyclic integration, keep the same choice of which factor is u throughout both applications, or the integral will cancel to 0 instead of giving you 2I.
∫ x·ln(x) dx
LIATE: u = ln(x) (Logarithm), dv = x dx (Algebraic)
du = 1/x dx, v = x²/2
∫ x·ln(x) dx = (x²/2)·ln(x) − ∫ (x²/2)·(1/x) dx
= (x²/2)·ln(x) − ∫ x/2 dx
= (x²/2)·ln(x) − x²/4 + C = x²(2ln(x)−1)/4 + C
Tabular Method for Repeated Parts
When integrating by parts multiple times with polynomials, the tabular method speeds up the process.
To find ∫ x³·eˣ dx using tabular method:
Differentiate (u)
Signs
Integrate (dv)
x³
+
eˣ
3x²
−
eˣ
6x
+
eˣ
6
−
eˣ
0
eˣ
Multiply diagonals with alternating signs: ∫ x³·eˣ dx = eˣ(x³ − 3x² + 6x − 6) + C
The tabular method only works when one factor differentiates to zero eventually (i.e. it's a polynomial). For cyclic integrals like eˣ·cos(x), use the algebraic method.
Learn 4 — Integrating Using Partial Fractions
When a rational function (polynomial over polynomial) has a factorable denominator, decompose it into partial fractions first, then integrate each term using the logarithm rule.
Key Integration Results
∫ 1/(x+a) dx = ln|x+a| + C ∫ 1/(ax+b) dx = (1/a)·ln|ax+b| + C
Example: ∫ (3x+7)/((x+1)(x+3)) dx
Step 1 — Decompose:
(3x+7)/((x+1)(x+3)) = A/(x+1) + B/(x+3)
Multiply through: 3x+7 = A(x+3) + B(x+1)
x = −1: 4 = 2A → A = 2
x = −3: −2 = −2B → B = 1
Step 2 — Integrate:
∫ [2/(x+1) + 1/(x+3)] dx = 2·ln|x+1| + ln|x+3| + C
= ln|(x+1)²(x+3)| + C
Before decomposing, check the degree of numerator vs denominator. If degree of numerator ≥ degree of denominator, perform polynomial long division first to get a proper fraction. Only then apply partial fractions.
Key rule: For repeated factors (x+a)², use A/(x+a) + B/(x+a)².
For irreducible quadratic (x²+a²), use (Ax+B)/(x²+a²).
Learn 5 — Standard Trig Integrals and Definite Applications
Standard Trig Results
∫ sin(ax) dx = −(1/a)cos(ax) + C
∫ cos(ax) dx = (1/a)sin(ax) + C
∫ sec²(ax) dx = (1/a)tan(ax) + C
Integrating sin²(x) using Double Angle
We cannot integrate sin²(x) directly. Use the identity:
cos(2x) = 1 − 2sin²(x) → sin²(x) = (1 − cos(2x))/2
Use identity: sin(3x)·cos(3x) = sin(6x)/2
Or use substitution u = sin(3x): du = 3cos(3x)dx
∫₀^(π/6) u · du/3 = [u²/6]₀^(π/6) — at x=π/6, u=sin(π/2)=1; at x=0, u=0
= 1/6 − 0 = 1/6 ≈ 0.167
Area Between Curves
Area between y=f(x) and y=g(x) from a to b (where f ≥ g) = ∫ₐᵇ [f(x) − g(x)] dx.
Use any of the integration techniques to evaluate the resulting integral. Always sketch the curves first to identify which is on top.
Common sign error: ∫ sin(ax) dx = −(1/a)cos(ax) + C — the negative sign comes from the chain rule and the 1/a factor is often forgotten.
Mistake 4 — Not Changing Limits for Definite Substitution
✗ Wrong: ∫₀¹ 2x·e^(x²) dx, let u=x². Integrate ∫₀¹ e^u du (limits unchanged!)
✓ Correct: Change limits: x=0 → u=0; x=1 → u=1. Then ∫₀¹ e^u du = e − 1 ≈ 1.718. (Same here, but when limits differ the mistake matters!)
Mistake 5 — Missing the 1/a Factor in Trig Integrals
✗ Wrong: ∫ cos(3x) dx = sin(3x) + C
✓ Correct: ∫ cos(3x) dx = (1/3)sin(3x) + C — always divide by the coefficient of x inside.
Mistake 6 — Cyclic Integration: Not Collecting Terms
✗ Wrong: Stopping after two applications of parts and claiming the answer is 0 (the integral cancels).
✓ Correct: When I appears on both sides, move it to one side: 2I = eˣ(sin x + cos x), then I = eˣ(sin x + cos x)/2 + C. Always add + C after dividing.
Mistake 7 — Improper Fractions in Partial Fractions
✗ Wrong: Decomposing (x²+3x+2)/(x+1)(x+2) directly as A/(x+1) + B/(x+2) without checking degree.
✓ Correct: The numerator has degree 2 and denominator degree 2 — this is improper! Divide first: (x²+3x+2)/(x²+3x+2) = 1 (in this case), then handle the remainder. Always do polynomial division first when degrees match or numerator degree is higher.
Key Formulas Reference
Integral
Result
Notes
∫ xⁿ dx
xⁿ⁺¹/(n+1) + C
n ≠ −1
∫ 1/x dx
ln|x| + C
x ≠ 0
∫ 1/(ax+b) dx
(1/a)ln|ax+b| + C
a ≠ 0
∫ eˣ dx
eˣ + C
∫ e^(ax+b) dx
(1/a)e^(ax+b) + C
a ≠ 0
∫ sin(ax) dx
−(1/a)cos(ax) + C
Note the minus
∫ cos(ax) dx
(1/a)sin(ax) + C
∫ sec²(ax) dx
(1/a)tan(ax) + C
∫ sin²(x) dx
x/2 − sin(2x)/4 + C
Use cos(2x)=1−2sin²x
∫ cos²(x) dx
x/2 + sin(2x)/4 + C
Use cos(2x)=2cos²x−1
∫ u(dv/dx)dx
uv − ∫v(du/dx)dx
Integration by parts
∫ f(g(x))g'(x)dx
F(g(x)) + C
Substitution: u=g(x)
∫ A/(x+a) dx
A·ln|x+a| + C
After partial fractions
∫ ln(x) dx
x·ln(x) − x + C
Parts with dv=1
∫ x·eˣ dx
eˣ(x−1) + C
Parts
∫ x·sin(x) dx
−x·cos(x) + sin(x) + C
Parts
∫ x·cos(x) dx
x·sin(x) + cos(x) + C
Parts
∫ x²·eˣ dx
eˣ(x²−2x+2) + C
Parts twice
∫ eˣcos(x) dx
eˣ(sin x+cos x)/2 + C
Cyclic parts
∫ eˣsin(x) dx
eˣ(sin x−cos x)/2 + C
Cyclic parts
Proof Bank
Proof 1 — Derivation of Integration by Parts from the Product Rule
If u and v are differentiable functions of x, the product rule states:
d(uv)/dx = u·(dv/dx) + v·(du/dx)
Integrating both sides with respect to x:
∫ d(uv)/dx dx = ∫ u·(dv/dx) dx + ∫ v·(du/dx) dx
The left side integrates trivially:
uv = ∫ u·(dv/dx) dx + ∫ v·(du/dx) dx
We cannot integrate sin²(x) directly since it is not a standard form. We use the cosine double angle identity:
cos(2x) = 1 − 2sin²(x)
Rearranging: sin²(x) = (1 − cos(2x))/2
Now integrate:
∫ sin²(x) dx = ∫ (1 − cos(2x))/2 dx
= (1/2)∫ 1 dx − (1/2)∫ cos(2x) dx
= x/2 − (1/2)·sin(2x)/2 + C = x/2 − sin(2x)/4 + C QED
Proof 3 — ∫ (1/x) dx = ln|x| + C
We need to find a function whose derivative is 1/x.
For x > 0: We know d/dx[ln(x)] = 1/x (from the definition of the natural logarithm as the inverse of eˣ). Therefore ∫ (1/x) dx = ln(x) + C for x > 0.
For x < 0: Let x = −t where t > 0. Then:
d/dx[ln(−x)] = d/dx[ln(t)] · (dt/dx) = (1/t)·(−1) · (−1) = 1/t = 1/(−x) = 1/x (using chain rule and substitution).
Therefore d/dx[ln|x|] = 1/x for all x ≠ 0. Hence ∫ (1/x) dx = ln|x| + C QED
Note: the absolute value ensures the expression is defined for both positive and negative x; we always write ln|x|, never ln(x), when x could be negative.
Integration Visualiser — Area Under the Curve
Select a function, set limits a and b, and see the shaded area computed numerically.
Area will appear here after clicking Compute & Draw.
f(x) = x·eˣ
Area ≈ —
Method: Numerical (1000 rectangles)
Exercise 1 — Standard Integrals (Definite)
Evaluate each definite integral. Give answers to 3 decimal places where not exact.
Exercise 2 — Integration by Substitution
Use substitution. Give numerical answers to 3 dp, or write expressions for indefinite integrals.
Exercise 3 — Integration by Parts
Use integration by parts. Give numerical answers to 3 dp.
Exercise 4 — Partial Fraction Integrals
Decompose then integrate. Give numerical answers to 3 dp.
Exercise 5 — Trig Integrals using Double Angle
Use identities to integrate. Give answers to 3 dp or exact form.
Practice — Mixed Methods (30 Questions)
Mixed integration questions. Identify the method first, then compute.
Challenge — Higher Order Questions (15 Questions)
Harder problems combining multiple techniques.
Exam Style Questions
Cambridge A-Level 9709 style. Click "Show Mark Scheme" after attempting each question.
Q1 [4 marks]
Find ∫ x·e^(3x) dx.
M1: Identify parts: u = x, dv = e^(3x)dx
M1: Correct v = e^(3x)/3
A1: x·e^(3x)/3 − ∫ e^(3x)/3 dx = x·e^(3x)/3 − e^(3x)/9 + C
A1: Final answer: e^(3x)(3x−1)/9 + C
Q2 [5 marks]
Use the substitution u = 2x+1 to find ∫₀¹ x·√(2x+1) dx.
(c) B1: The answer π represents the area between the curve y=x·sin(x) and the x-axis from 0 to π.
B1: Since x·sin(x) ≥ 0 on [0,π], the integral gives the exact area = π square units.
Past Paper Questions
Authentic Cambridge 9709 style questions. Attempt each fully before revealing the mark scheme.
PP1 — 9709/22/O/N/18 Style [4 marks]
Use integration by parts to find ∫ x·cos(2x) dx.
M1: u = x, dv = cos(2x)dx → du=dx, v=sin(2x)/2
M1: = x·sin(2x)/2 − ∫ sin(2x)/2 dx
A1: = x·sin(2x)/2 − (−cos(2x)/4)
A1: = x·sin(2x)/2 + cos(2x)/4 + C
PP2 — 9709/23/M/J/19 Style [5 marks]
Given that ∫₀^a x·e^(2x) dx = 1, find the value of a correct to 3 significant figures.