Home / A-Level Maths 9709 / Integration / Integrating exponentials, 1/(ax + b) and trigonometric functions
Cambridge International AS & A Level · Topic 3.5
A-Level Maths 9709 / Topic 3.5 / AS & A Level

Reverse the chain, divide by a.

Papers 2 and 3. Every derivative you learned in Pure 2 and 3 runs backwards into an integral: \(e^{ax+b}\), \(\dfrac{1}{ax+b}\), \(\sin\), \(\cos\) and \(\sec^2\). The one extra move is dividing by the coefficient of \(x\). (The \(\dfrac{1}{x^2+a^2}\) result is Paper 3 only.)

Topic 3.5All integration lessons →
The key idea

If \(\dfrac{d}{dx}F(x) = f(x)\), then \(\displaystyle\int f(ax+b)\,dx = \dfrac{1}{a}F(ax+b) + c\). Integrate as if the bracket were just \(x\), then divide by \(a\), the coefficient of \(x\). Angles are always in radians.

x y 01234 10.5 y = 1 / (2x + 1) area = ln 3 ≈ 1.10
Fig. 1: the area under y = 1/(2x + 1) from x = 0 to x = 4 is the integral of 1/(2x + 1), which is (1/2) ln(2x + 1); between the limits this gives (1/2) ln 9 = ln 3, about 1.10 square units
Section 01

Exponentials and logarithms: divide by a.

Each result below is a derivative read backwards. Check any of them by differentiating the right-hand side: the chain rule multiplies by \(a\), and the \(\dfrac{1}{a}\) in front cancels it.

\(f(x)\)\(\displaystyle\int f(x)\,dx\)
\(e^{ax+b}\)\(\dfrac{1}{a}e^{ax+b} + c\)
\(\dfrac{1}{ax+b}\)\(\dfrac{1}{a}\ln|ax+b| + c\)
\(\dfrac{1}{x}\)\(\ln|x| + c\)
Worked example Indefinite integral
Find \(\displaystyle\int \left(e^{3x-1} + \dfrac{4}{2x+5}\right) dx\).
  1. For \(e^{3x-1}\), \(a = 3\): \(\displaystyle\int e^{3x-1}\,dx = \dfrac{1}{3}e^{3x-1}\).
  2. For \(\dfrac{4}{2x+5}\), take the 4 outside and use \(a = 2\): \(4 \times \dfrac{1}{2}\ln|2x+5| = 2\ln|2x+5|\).
  3. Add one constant of integration for the whole answer.
Answer: \(\dfrac{1}{3}e^{3x-1} + 2\ln|2x+5| + c\)
Worked example Definite integral, exact answer
Find the exact value of \(\displaystyle\int_0^2 \dfrac{4}{2x+1}\,dx\).
  1. \(\displaystyle\int \dfrac{4}{2x+1}\,dx = 4 \times \dfrac{1}{2}\ln(2x+1) = 2\ln(2x+1)\) (no modulus needed: \(2x+1 > 0\) on \([0, 2]\)).
  2. Substitute the limits: \(2\ln 5 - 2\ln 1 = 2\ln 5 - 0\).
  3. Use \(k\ln a = \ln a^{k}\) to tidy: \(2\ln 5 = \ln 25\).
Answer: \(\ln 25\) (about 3.22)
Your turn
Find \(\displaystyle\int \left(e^{x} + 1\right)^{2} dx\).
Examiner trap

\(\displaystyle\int \dfrac{1}{2x+1}\,dx\) is NOT \(\ln|2x+1|\). Differentiate \(\ln|2x+1|\) and you get \(\dfrac{2}{2x+1}\), twice too big, so the answer needs \(\dfrac{1}{2}\) in front. Equally, never multiply by \(a\): that is differentiating, not integrating.

Section 02

Sine, cosine and secant squared: watch the signs.

The trigonometric derivatives \(\dfrac{d}{dx}\sin x = \cos x\), \(\dfrac{d}{dx}\cos x = -\sin x\) and \(\dfrac{d}{dx}\tan x = \sec^2 x\) only hold in radians, so these integrals do too. Put your calculator in radian mode before evaluating any limits.

\(f(x)\)\(\displaystyle\int f(x)\,dx\)
\(\sin(ax+b)\)\(-\dfrac{1}{a}\cos(ax+b) + c\)
\(\cos(ax+b)\)\(\dfrac{1}{a}\sin(ax+b) + c\)
\(\sec^2(ax+b)\)\(\dfrac{1}{a}\tan(ax+b) + c\)
Worked example Sine with a definite integral
Find the exact value of \(\displaystyle\int_0^{\pi/3} \sin 2x\,dx\).
  1. \(\displaystyle\int \sin 2x\,dx = -\dfrac{1}{2}\cos 2x\).
  2. Upper limit: \(-\dfrac{1}{2}\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}\left(-\dfrac{1}{2}\right) = \dfrac{1}{4}\).
  3. Lower limit: \(-\dfrac{1}{2}\cos 0 = -\dfrac{1}{2}\).
  4. Upper minus lower: \(\dfrac{1}{4} - \left(-\dfrac{1}{2}\right) = \dfrac{3}{4}\).
Answer: \(\dfrac{3}{4}\)
Worked example Secant squared
Find the exact value of \(\displaystyle\int_0^{\pi/12} \sec^2 3x\,dx\).
  1. \(\displaystyle\int \sec^2 3x\,dx = \dfrac{1}{3}\tan 3x\).
  2. At \(x = \dfrac{\pi}{12}\): \(3x = \dfrac{\pi}{4}\), and \(\tan\dfrac{\pi}{4} = 1\). At \(x = 0\): \(\tan 0 = 0\).
  3. So the value is \(\dfrac{1}{3}(1 - 0)\).
Answer: \(\dfrac{1}{3}\)
Worked example Mixed terms
Find \(\displaystyle\int \left(\cos 3x - \sec^2\dfrac{x}{2}\right) dx\).
  1. \(\cos 3x\): \(a = 3\), so \(\dfrac{1}{3}\sin 3x\).
  2. \(\sec^2\dfrac{x}{2}\): \(a = \dfrac{1}{2}\), and dividing by \(\dfrac{1}{2}\) means multiplying by 2, so \(2\tan\dfrac{x}{2}\).
Answer: \(\dfrac{1}{3}\sin 3x - 2\tan\dfrac{x}{2} + c\)
Examiner trap

\(\displaystyle\int \sin x\,dx = -\cos x\), with a MINUS. The most common lost mark is writing \(+\cos x\). A quick check: differentiate your answer and see whether you get back exactly what you started with.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Find \(\displaystyle\int e^{4x}\,dx\).

Quick check+10 XP

Find \(\displaystyle\int \dfrac{1}{3x+2}\,dx\).

Quick check+10 XP

Find \(\displaystyle\int \sin 2x\,dx\).

Quick check+10 XP

Find \(\displaystyle\int \sec^2 5x\,dx\).

Section 03

Paper 3 only: the inverse tan integral.

In Pure 3 you meet \(\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}\). Reversing it, with a scale factor \(a\), gives the last standard result. It produces an angle, so it often leads to an answer with \(\pi\) in it.

Worked example Straight from the formula
Find the exact value of \(\displaystyle\int_0^3 \dfrac{1}{x^2+9}\,dx\).
  1. Here \(a^2 = 9\), so \(a = 3\): \(\displaystyle\int \dfrac{1}{x^2+9}\,dx = \dfrac{1}{3}\tan^{-1}\dfrac{x}{3}\).
  2. Limits: \(\dfrac{1}{3}\tan^{-1}1 - \dfrac{1}{3}\tan^{-1}0 = \dfrac{1}{3}\times\dfrac{\pi}{4} - 0\).
Answer: \(\dfrac{\pi}{12}\)
Worked example Coefficient of x squared not 1
Find \(\displaystyle\int \dfrac{1}{4+9x^2}\,dx\).
  1. Take out the 9: \(\dfrac{1}{4+9x^2} = \dfrac{1}{9}\times\dfrac{1}{x^2+\frac{4}{9}}\), so \(a = \dfrac{2}{3}\).
  2. \(\dfrac{1}{9}\times\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} = \dfrac{1}{9}\times\dfrac{3}{2}\tan^{-1}\dfrac{3x}{2}\).
  3. Simplify \(\dfrac{1}{9}\times\dfrac{3}{2} = \dfrac{1}{6}\). Check: \(\dfrac{d}{dx}\dfrac{1}{6}\tan^{-1}\dfrac{3x}{2} = \dfrac{1}{6}\times\dfrac{3/2}{1+\frac{9x^2}{4}} = \dfrac{1}{4+9x^2}\).
Answer: \(\dfrac{1}{6}\tan^{-1}\left(\dfrac{3x}{2}\right) + c\)
Examiner trap

Do not forget the \(\dfrac{1}{a}\) in front of \(\tan^{-1}\), and do not leave \(\tan^{-1}1\) in degrees: \(\tan^{-1}1 = \dfrac{\pi}{4}\), never 45, inside a calculus answer.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Find the exact value of \(\displaystyle\int_0^{\ln 2} e^{2x}\,dx\).

Exam style+20 XP

Find the exact value of \(\displaystyle\int_1^3 \dfrac{2}{4x-3}\,dx\).

Exam style+20 XP

Find the exact value of \(\displaystyle\int_0^{\pi/6} \left(2\sin 2x + \sec^2 x\right) dx\).

Exam style+20 XP

(Paper 3) Find the exact value of \(\displaystyle\int_0^2 \dfrac{3}{4+3x^2}\,dx\).

Skill unlocked

Standard integrals, sorted.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
Exponentials, logarithms, sine, cosine and secant squared, exact definite integrals, an unknown limit and an area.