Home / A-Level Maths 9709 / Integration / Integration by substitution
Cambridge International AS & A Level · Topic 3.5
A-Level Maths 9709 / Topic 3.5 / AS & A Level

Change the variable.

Paper 3 only. A well-chosen new variable \(u\) turns an awkward integral into a standard one. The examiner gives you the substitution; your job is to convert every part of the integral, \(dx\) and the limits included, and then integrate in \(u\).

Topic 3.5All integration lessons →
The key idea

With a substitution \(u = g(x)\), replace three things: the integrand (in terms of \(u\)), \(dx\) (using \(\dfrac{du}{dx}\)), and, for a definite integral, the limits (\(x\)-values become \(u\)-values). Then the whole integral is in \(u\) and no \(x\) may remain.

x u 0π/6π/201/21 u = sin x du = cos x dx
Fig. 1: with u = sin x, the x-limits 0, π/6 and π/2 map to the u-limits 0, 1/2 and 1 (drawn to scale on each line); a definite integral in u uses the new limits
Section 01

The method: convert everything.

Treat \(\dfrac{du}{dx}\) like a fraction to swap \(dx\) for \(du\). If the substitution is \(u = 2x + 1\), then \(\dfrac{du}{dx} = 2\), so \(dx = \dfrac{1}{2}\,du\), and any leftover \(x\) is written as \(x = \dfrac{u-1}{2}\).

Worked example A linear substitution
Use the substitution \(u = 2x + 1\) to find the exact value of \(\displaystyle\int_0^4 x\sqrt{2x+1}\,dx\).
  1. \(\dfrac{du}{dx} = 2\), so \(dx = \dfrac{1}{2}\,du\); and \(x = \dfrac{u-1}{2}\).
  2. Limits: \(x = 0 \Rightarrow u = 1\); \(x = 4 \Rightarrow u = 9\).
  3. The integral becomes \(\displaystyle\int_1^9 \dfrac{u-1}{2}\,u^{1/2}\times\dfrac{1}{2}\,du = \dfrac{1}{4}\int_1^9 \left(u^{3/2} - u^{1/2}\right) du\).
  4. \(= \dfrac{1}{4}\left[\dfrac{2}{5}u^{5/2} - \dfrac{2}{3}u^{3/2}\right]_1^9 = \left[\dfrac{1}{10}u^{5/2} - \dfrac{1}{6}u^{3/2}\right]_1^9\).
  5. \(9^{5/2} = 243\) and \(9^{3/2} = 27\): \(\dfrac{1}{10}(243 - 1) - \dfrac{1}{6}(27 - 1) = \dfrac{121}{5} - \dfrac{13}{3}\).
Answer: \(\dfrac{298}{15}\)
Your turn
Use the substitution \(u = x + 1\) to find the exact value of \(\displaystyle\int_0^3 \dfrac{x}{\sqrt{x+1}}\,dx\).
Examiner trap

Forgetting to replace \(dx\) is the most expensive slip: with \(u = 2x + 1\), \(dx = \dfrac{1}{2}du\), not \(du\). The second is mixing old and new limits: once the integral is in \(u\), the limits must be \(u\)-values.

Section 02

Trigonometric substitutions: spot the derivative.

With \(u = \sin x\) the group \(\cos x\,dx\) becomes \(du\); with \(u = \cos x\) the group \(\sin x\,dx\) becomes \(-du\). Use \(\sin^2 x + \cos^2 x = 1\) to convert any other powers.

Worked example The syllabus example
Use the substitution \(u = \sin x\) to find the exact value of \(\displaystyle\int_0^{\pi/2} \sin^2 2x\cos x\,dx\).
  1. \(\sin 2x = 2\sin x\cos x\), so \(\sin^2 2x = 4\sin^2 x\cos^2 x\).
  2. Integrand \(= 4\sin^2 x\cos^2 x \cdot \cos x\,dx = 4\sin^2 x\,(1 - \sin^2 x)\cos x\,dx\).
  3. \(du = \cos x\,dx\), so it becomes \(4u^2(1 - u^2)\,du\). Limits: \(0 \to 0\), \(\dfrac{\pi}{2} \to 1\).
  4. \(\displaystyle\int_0^1 \left(4u^2 - 4u^4\right) du = \left[\dfrac{4u^3}{3} - \dfrac{4u^5}{5}\right]_0^1 = \dfrac{4}{3} - \dfrac{4}{5}\).
Answer: \(\dfrac{8}{15}\)
Worked example A tangent substitution
Use the substitution \(x = \tan\theta\) to find the exact value of \(\displaystyle\int_0^1 \dfrac{1}{(1+x^2)^2}\,dx\).
  1. \(\dfrac{dx}{d\theta} = \sec^2\theta\), so \(dx = \sec^2\theta\,d\theta\). Limits: \(x = 0 \Rightarrow \theta = 0\); \(x = 1 \Rightarrow \theta = \dfrac{\pi}{4}\).
  2. \(1 + \tan^2\theta = \sec^2\theta\), so \(\dfrac{1}{(1+x^2)^2} = \dfrac{1}{\sec^4\theta}\).
  3. Integral \(= \displaystyle\int_0^{\pi/4} \dfrac{\sec^2\theta}{\sec^4\theta}\,d\theta = \int_0^{\pi/4} \cos^2\theta\,d\theta\).
  4. \(= \left[\dfrac{\theta}{2} + \dfrac{1}{4}\sin 2\theta\right]_0^{\pi/4} = \dfrac{\pi}{8} + \dfrac{1}{4}\).
Answer: \(\dfrac{\pi}{8} + \dfrac{1}{4}\)
Examiner trap

With \(u = \cos x\), \(du = -\sin x\,dx\). The minus sign is easy to drop; it is usually absorbed by swapping the limits, because \(x = 0\) gives \(u = 1\), the larger value.

Stage 1 · Learn

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With the substitution \(u = 3x + 1\), \(dx\) is replaced by

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With \(u = \sin x\), the limits \(x = 0\) and \(x = \dfrac{\pi}{6}\) become

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Using \(u = \sin x\), find \(\displaystyle\int \cos x\sin^4 x\,dx\).

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With \(u = x^2 + 1\), the group \(x\,dx\) is replaced by

Section 03

Exponential substitutions and back-substituting.

For \(u = e^{x} + 1\), \(du = e^{x}\,dx\). Split \(e^{2x}\) as \(e^{x}\cdot e^{x}\): one factor goes with \(dx\), the other becomes \(u - 1\). For an indefinite integral, write the final answer back in terms of \(x\).

Worked example Splitting the exponential
Use the substitution \(u = e^{x} + 1\) to find the exact value of \(\displaystyle\int_0^1 \dfrac{e^{2x}}{e^{x}+1}\,dx\).
  1. \(du = e^{x}\,dx\), and \(e^{2x}\,dx = e^{x}\cdot e^{x}\,dx = (u - 1)\,du\).
  2. Limits: \(x = 0 \Rightarrow u = 2\); \(x = 1 \Rightarrow u = e + 1\).
  3. \(\displaystyle\int_2^{e+1} \dfrac{u-1}{u}\,du = \int_2^{e+1} \left(1 - \dfrac{1}{u}\right) du = \left[u - \ln u\right]_2^{e+1}\).
  4. \(= (e + 1) - \ln(e+1) - 2 + \ln 2\).
Answer: \(e - 1 - \ln\dfrac{e+1}{2}\) (about 1.10)
Worked example Indefinite: return to x
Use the substitution \(u = 2x - 1\) to find \(\displaystyle\int x(2x-1)^5\,dx\).
  1. \(dx = \dfrac{1}{2}du\) and \(x = \dfrac{u+1}{2}\).
  2. \(\displaystyle\int \dfrac{u+1}{2}\,u^5\cdot\dfrac{1}{2}\,du = \dfrac{1}{4}\int \left(u^6 + u^5\right) du = \dfrac{u^7}{28} + \dfrac{u^6}{24}\).
  3. Substitute back \(u = 2x - 1\).
Answer: \(\dfrac{(2x-1)^7}{28} + \dfrac{(2x-1)^6}{24} + c\)
Examiner trap

An indefinite answer left in \(u\) loses the final mark. A definite answer evaluated with the old \(x\)-limits in a \(u\)-integral is simply wrong.

Stage 2 · Exam

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Use the substitution \(u = 1 - x\) to find the exact value of \(\displaystyle\int_0^1 x(1-x)^4\,dx\).

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Use the substitution \(u = \cos x\) to find the exact value of \(\displaystyle\int_0^{\pi/3} \dfrac{\sin x}{\cos^3 x}\,dx\).

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Use the substitution \(u = 1 + e^{x}\) to find the exact value of \(\displaystyle\int_0^{\ln 2} \dfrac{e^{x}}{(1+e^{x})^2}\,dx\).

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Use the substitution \(x = 2\sin\theta\) to find the exact value of \(\displaystyle\int_0^{\sqrt{3}} \dfrac{x^2}{\sqrt{4-x^2}}\,dx\).

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Six original Cambridge-style questions
Linear, root, trigonometric and exponential substitutions, a show-that, and the area under a curve with a maximum point.