AQA A-Level Mathematics Paper 1, June 2025: Question 17

10 marks · Medium difficulty · Multi-step Problem

Use integration by substitution to evaluate an integral involving exponentials, and solve a separable first-order differential equation with initial conditions.

Practise this question

Question

Question 17 consists of two parts. Part (a) asks to use the substitution u = e^x + 1 to show that the integral of (e^(2x) / (e^x + 1)) dx equals e^x - ln(e^x + 1) + k, worth 5 marks. Part (b) asks to solve the differential equation ((e^x + 1)/e^(2x)) dy/dx = cos^2(y) given that y = pi when x = 0, expressing the answer in the form tan(y) = e^x + ln(A / (e^x + 1)) + B, where A and B are constants to be found, worth 5 marks.
Question text

17 (a) Use the substitution u = ex + 1 to show that

e2x

∫ dx = ex – ln(ex + 1) + k

ex + 1

[5 marks]

17 (b) Solve the differential equation

(ex + 1 ) dy

= cos2 y

(26) e2x dx

given that y = π when x = 0

Write your answer in the form

tan y = ex + ln( A ) + B

ex + 1

where A and B are constants to be found.

[5 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 17. Part (a) awards B1 for du/dx = e^x, M1 for substituting into the integral, A1 for obtaining the integral of ((u-1)/u) du, M1 for integrating to get u - ln(u) + c, and R1 for completing the argument to obtain e^x - ln(e^x + 1) + k. Part (b) awards B1 for separating variables to (1/cos^2(y)) dy/dx = e^(2x)/(e^x + 1), M1 for integrating either side, A1 for tan(y) = e^x - ln(e^x + 1) + k, M1 for substituting the initial condition y = pi at x = 0 to find k = ln(2) - 1, and R1 for writing the final expression as tan(y) = e^x + ln(2/(e^x + 1)) - 1.

Q Marking instructions AO Marks Typical solution

17(a) du u = ex +1

Obtains = ex

dx 1.1b B1 du

= ex

OE

dx

Makes a partial substitution to 2x 2x

e2x 1 e e 1

obtain du or better x dx = x du

u ex e + 1 u e

ex

If they simplify the fraction first

x = du

x e 3.1a M1 u

e − x dx need to see

e +1 u −1

x = du

x e 1 u

e − x du or better for this

u e 1

= 1− du

mark u

u −1 = u −lnu + c

Obtains du OE 1.1b A1

u x ( x )

= e +1−ln e +1 + c

Integrates their two-term

integrand, which is in terms of u, = ex − ln(ex +1) + k

1.1a M1

with at least one of their terms

integrated correctly.

Completes reasoned argument

ex − ln(ex +1) + k

to obtain

Must see constant of integration 2.1 R1

introduced as they integrate and

dealt with consistently – A-LEVEL MATHEMATICS – –

AG

Subtotal 5

17(b) 1 dy e2x e2x

Obtains = sec2 y dy = dx

2 y dx e +x 1 3.1a B1 e +x 1

cos

or better tan y = ex − ln(ex +1) + k

Integrates to obtain one correct

tan π = e0 − ln(e0 +1) + k

side. Uses given answer from 1.1a M1

sec2 y dy = tany

(a) or recalls k = ln 2() −1

Obtains

x x x 2

tany = e −ln(e +1) + k tan y = e +ln x −1

1.1b A1 e +1

Condone missing or extra

constants of integration.

Substitutes y = π and x = 0 into

their integrated equation which

must be formed from

exponential functions of x and 1.1a M1

trigonometric function(s) of y to

obtain the constant of

integration

Completes reasoned argument

to obtain

22.1 R1

tan y = ex + ln −1

x

e +1

Subtotal 5

Question 17 Total 10

How to answer it

Integration by Substitution & First-Order Differential Equations

EXAM SPECIFICATION FOCUS

What this question tests

  • Integration by substitution: Changing variables completely, rewriting terms in terms of u, dividing algebraic expressions, and integrating standard forms including 1/u.
  • Rigorous proof & constants: Showing clear algebraic progression in a "show that" question, including proper handling of the arbitrary constant ( k = c + 1 ).
  • Separation of variables: Rearranging first-order differential equations into integrable functions of x and y.
  • Trigonometric integration: Recognising reciprocal identities such as 1 / cos²y = sec²y and knowing its standard integral tan y .
  • Boundary conditions & log laws: Finding the exact constant of integration using given initial conditions ( x = 0, y = π ) and combining logarithmic terms using ln a - ln b = ln(a/b) .

Part (a)

Integration by Substitution [5 Marks]

📐 Step-by-Step Solution

  1. B1 Differentiate the substitution:
    Given u = eˣ + 1 , differentiate with respect to x:
    du/dx = eˣ  ⇒  dx = du / eˣ
  2. M1 Perform partial substitution:
    Rewrite the integral by substituting dx and the denominator:
    ∫ (e²ˣ / (eˣ + 1)) dx = ∫ (e²ˣ / u) · (du / eˣ) = ∫ (eˣ / u) du
  3. A1 Convert completely into terms of u:
    Since u = eˣ + 1 , we have eˣ = u - 1 .
    Substitute this into the numerator:
    ∫ ((u - 1) / u) du
  4. M1 Split the fraction and integrate:
    ∫ (u/u - 1/u) du = ∫ (1 - 1/u) du = u - ln|u| + c
    (At least one term must be correctly integrated with respect to u)
  5. R1 Substitute back and reconcile the constant:
    Replace u with eˣ + 1 :
    = (eˣ + 1) - ln(eˣ + 1) + c
    Combine constants: let k = c + 1 (since 1 is a constant):
    = eˣ - ln(eˣ + 1) + k  (as required)

💡 Key Knowledge

  • Laws of Indices: Recognise that e²ˣ = (eˣ)² = eˣ · eˣ so that dividing by eˣ leaves eˣ .
  • Standard Integrals: ∫ 1 du = u and ∫ (1/u) du = ln|u| .
  • Since eˣ + 1 > 0 for all real x, modulus signs are not strictly required: ln|eˣ + 1| = ln(eˣ + 1) .

❌ Common Errors & Pitfalls

  • Losing the +1 constant: Writing u - ln u + c = eˣ - ln(eˣ + 1) + k without acknowledging where the +1 went will forfeit the final R1 mark. Top candidates explicitly state k = c + 1 .
  • Mixing variables: Leaving eˣ inside the integral alongside du during final integration. All variables must become u before integrating.

Part (b)

Differential Equations & Boundary Conditions [5 Marks]

📐 Step-by-Step Solution

  1. B1 Separate the variables:
    The differential equation is:
    ((eˣ + 1) / e²ˣ) (dy/dx) = cos²y
    Divide both sides by cos²y and multiply by e²ˣ / (eˣ + 1) :
    (1 / cos²y) dy = (e²ˣ / (eˣ + 1)) dx
  2. M1 A1 Integrate both sides (using Part a):
    Recall the reciprocal identity 1 / cos²y = sec²y :
    ∫ sec²y dy = ∫ (e²ˣ / (eˣ + 1)) dx
    Left-hand side: ∫ sec²y dy = tan y
    Right-hand side (from part a): eˣ - ln(eˣ + 1) + k
    General equation: tan y = eˣ - ln(eˣ + 1) + k
  3. M1 Apply initial boundary conditions:
    Substitute x = 0 and y = π :
    tan(π) = e⁰ - ln(e⁰ + 1) + k
    Since tan(π) = 0 and e⁰ = 1 :
    0 = 1 - ln(1 + 1) + k
    0 = 1 - ln(2) + k  ⇒  k = ln(2) - 1
  4. R1 Rearrange into requested form:
    Substitute k back into the equation:
    tan y = eˣ - ln(eˣ + 1) + ln(2) - 1
    Combine the log terms using ln(2) - ln(eˣ + 1) = ln( 2 / (eˣ + 1) ) :
    tan y = eˣ + ln( 2 / (eˣ + 1) ) - 1

✅ Final Answer

The equation in the required form:

tan y = eˣ + ln( 2 / (eˣ + 1) ) - 1

Constants: A = 2, B = -1

🧠 Exam Technique

  • Spot the link to Part (a): In multi-part exam questions, the right-hand side of part (b) almost always utilizes the integral proven in part (a). Do not re-integrate from scratch!
  • Reciprocal Trig Identities: Dividing by cos²y immediately gives sec²y , which is a standard derivative of tan y provided on the formula sheet.
  • Matching the target form: Notice the plus sign in front of the natural log in tan y = eˣ + ln(...) + B . This gives an immediate hint that the subtraction rule -ln(b) + ln(a) = ln(a/b) is required.

Topics

Pure Mathematics · H: Integration · F: Exponentials and logarithms

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.