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Cambridge International AS & A Level · Topic 3.5
A-Level Maths 9709 / Topic 3.5 / AS & A Level

The product rule, run backwards.

Paper 3 only. When the integrand is a product, such as \(x\sin 2x\), \(x^2e^{-x}\), \(\ln x\) or \(x\tan^{-1}x\), integration by parts trades it for a simpler integral. The skill is choosing which factor to differentiate.

Topic 3.5All integration lessons →
The key idea

\(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\). Choose \(u\) to be the factor that gets simpler when differentiated (a power of \(x\), or \(\ln x\) and \(\tan^{-1}x\), which you cannot integrate directly), and \(\dfrac{dv}{dx}\) to be the factor you can integrate.

x y 12e3 1 y = ln x area = 1
Fig. 1: the area under y = ln x from x = 1 to x = e is found by parts with u = ln x and dv/dx = 1: [x ln x - x] from 1 to e = (e - e) - (0 - 1) = 1 square unit
Section 01

The formula and choosing u.

Integrate the product rule \(\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\) and rearrange to get the formula. It only helps if the new integral \(\displaystyle\int v\dfrac{du}{dx}\,dx\) is simpler than the one you started with, so choose \(u\) to be the factor that differentiates to something simpler.

Integrand\(u\)\(\dfrac{dv}{dx}\)
\(x e^{2x}\), \(x\sin 2x\), \(x^2\cos x\)the power of \(x\)the exponential or trig factor
\(\ln x\), \(x\ln x\), \(\dfrac{\ln x}{x^2}\)\(\ln x\)the rest (even if it is just 1)
\(x\tan^{-1}x\), \(\tan^{-1}x\)\(\tan^{-1}x\)the rest
Worked example Power times exponential
Find \(\displaystyle\int x e^{2x}\,dx\).
  1. \(u = x\), so \(\dfrac{du}{dx} = 1\). \(\dfrac{dv}{dx} = e^{2x}\), so \(v = \dfrac{1}{2}e^{2x}\).
  2. \(uv - \displaystyle\int v\dfrac{du}{dx}\,dx = \dfrac{x}{2}e^{2x} - \int \dfrac{1}{2}e^{2x}\,dx\).
  3. \(\displaystyle\int \dfrac{1}{2}e^{2x}\,dx = \dfrac{1}{4}e^{2x}\).
Answer: \(\dfrac{x}{2}e^{2x} - \dfrac{1}{4}e^{2x} + c = \dfrac{1}{4}(2x - 1)e^{2x} + c\)
Worked example The syllabus example, definite
Find the exact value of \(\displaystyle\int_0^{\pi/4} x\sin 2x\,dx\).
  1. \(u = x\), \(\dfrac{du}{dx} = 1\); \(\dfrac{dv}{dx} = \sin 2x\), \(v = -\dfrac{1}{2}\cos 2x\).
  2. \(\displaystyle\int x\sin 2x\,dx = -\dfrac{x}{2}\cos 2x + \int \dfrac{1}{2}\cos 2x\,dx = -\dfrac{x}{2}\cos 2x + \dfrac{1}{4}\sin 2x\).
  3. At \(\dfrac{\pi}{4}\): \(-\dfrac{\pi}{8}\cos\dfrac{\pi}{2} + \dfrac{1}{4}\sin\dfrac{\pi}{2} = 0 + \dfrac{1}{4}\). At 0: \(0 + 0\).
Answer: \(\dfrac{1}{4}\)
Examiner trap

Choosing \(u = e^{2x}\) and \(\dfrac{dv}{dx} = x\) makes things worse: the new integral contains \(\dfrac{x^2}{2}e^{2x}\). If the integral you produce is harder than the original, swap \(u\) and \(\dfrac{dv}{dx}\).

Section 02

Integrating ln x: the hidden 1.

You cannot integrate \(\ln x\) directly, but you can differentiate it to \(\dfrac{1}{x}\). So whenever \(\ln x\) appears, make it \(u\). On its own, write \(\ln x = 1\times\ln x\) and take \(\dfrac{dv}{dx} = 1\).

Worked example ln x on its own
Find the exact value of \(\displaystyle\int_1^{e} \ln x\,dx\).
  1. \(u = \ln x\), \(\dfrac{du}{dx} = \dfrac{1}{x}\); \(\dfrac{dv}{dx} = 1\), \(v = x\).
  2. \(\displaystyle\int \ln x\,dx = x\ln x - \int x\cdot\dfrac{1}{x}\,dx = x\ln x - x\).
  3. \(\left[x\ln x - x\right]_1^{e} = (e\ln e - e) - (1\ln 1 - 1) = (e - e) - (0 - 1)\).
Answer: \(1\) (the shaded area in Fig. 1)
Worked example ln x over a power
Find the exact value of \(\displaystyle\int_1^{e} \dfrac{\ln x}{x^2}\,dx\).
  1. \(u = \ln x\), \(\dfrac{du}{dx} = \dfrac{1}{x}\); \(\dfrac{dv}{dx} = x^{-2}\), \(v = -\dfrac{1}{x}\).
  2. \(\displaystyle\int \dfrac{\ln x}{x^2}\,dx = -\dfrac{\ln x}{x} + \int \dfrac{1}{x^2}\,dx = -\dfrac{\ln x}{x} - \dfrac{1}{x}\).
  3. \(\left[-\dfrac{\ln x}{x} - \dfrac{1}{x}\right]_1^{e} = \left(-\dfrac{1}{e} - \dfrac{1}{e}\right) - (0 - 1)\).
Answer: \(1 - \dfrac{2}{e}\)
Your turn
Find \(\displaystyle\int x\ln x\,dx\).
Examiner trap

With \(\ln x\) never choose \(\dfrac{dv}{dx} = \ln x\): you would need to integrate \(\ln x\) to find \(v\), which is the very thing you cannot yet do.

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

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Find \(\displaystyle\int x\cos x\,dx\).

Section 03

Parts twice, and inverse tan.

With \(x^2\) as \(u\), one application leaves \(x\) times something, so apply parts again. With \(\tan^{-1}x\) as \(u\), the second integral is a rational function that needs a little algebra.

Worked example Parts twice
Find the exact value of \(\displaystyle\int_0^1 x^2e^{-x}\,dx\).
  1. \(u = x^2\), \(v = -e^{-x}\): \(\displaystyle\int x^2e^{-x}\,dx = -x^2e^{-x} + \int 2xe^{-x}\,dx\).
  2. Again with \(u = 2x\), \(v = -e^{-x}\): \(\displaystyle\int 2xe^{-x}\,dx = -2xe^{-x} + \int 2e^{-x}\,dx = -2xe^{-x} - 2e^{-x}\).
  3. So \(\displaystyle\int x^2e^{-x}\,dx = -e^{-x}\left(x^2 + 2x + 2\right)\).
  4. \(\left[-e^{-x}(x^2 + 2x + 2)\right]_0^1 = -5e^{-1} - (-2) = 2 - \dfrac{5}{e}\).
Answer: \(2 - \dfrac{5}{e}\) (about 0.161)
Worked example The syllabus example x tan^-1 x
Find the exact value of \(\displaystyle\int_0^1 x\tan^{-1}x\,dx\).
  1. \(u = \tan^{-1}x\), \(\dfrac{du}{dx} = \dfrac{1}{1+x^2}\); \(\dfrac{dv}{dx} = x\), \(v = \dfrac{x^2}{2}\).
  2. \(\displaystyle\int x\tan^{-1}x\,dx = \dfrac{x^2}{2}\tan^{-1}x - \dfrac{1}{2}\int \dfrac{x^2}{1+x^2}\,dx\).
  3. \(\dfrac{x^2}{1+x^2} = 1 - \dfrac{1}{1+x^2}\), which integrates to \(x - \tan^{-1}x\).
  4. \(\left[\dfrac{x^2}{2}\tan^{-1}x - \dfrac{1}{2}x + \dfrac{1}{2}\tan^{-1}x\right]_0^1 = \dfrac{\pi}{8} - \dfrac{1}{2} + \dfrac{\pi}{8}\).
Answer: \(\dfrac{\pi}{4} - \dfrac{1}{2}\)
Examiner trap

The minus sign in front of the second integral applies to EVERYTHING it produces. When parts is used twice, bracket the second result: \(-x^2e^{-x} + \left(-2xe^{-x} - 2e^{-x}\right)\). Sign slips here cost the accuracy marks.

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Find the exact value of \(\displaystyle\int_0^1 xe^{x}\,dx\).

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Find the exact value of \(\displaystyle\int_1^{e} x\ln x\,dx\).

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Find the exact value of \(\displaystyle\int_0^{\pi/2} x\cos x\,dx\).

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Find the exact value of \(\displaystyle\int_0^{\pi/2} x^2\cos x\,dx\).

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Six original Cambridge-style questions
Power times exponential or trig, ln x times a power, parts twice, inverse tan, and the area under a curve with a maximum.