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AS & A Level · Practice questions · Integration by substitution

New variable, easier integral.

Six original Cambridge-style questions using a given substitution to evaluate definite and indefinite integrals, with exact answers.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Integrand, dx, limits: convert all three.

01
Recall
[4 marks]

Use the substitution \(u = x + 2\) to find \(\displaystyle\int \dfrac{x}{(x+2)^3}\,dx\), giving your answer in terms of \(x\).

  • \(dx = du\) and \(x = u - 2\). ✓
  • \(\displaystyle\int \dfrac{u-2}{u^3}\,du = \int \left(u^{-2} - 2u^{-3}\right) du = -u^{-1} + u^{-2}\). ✓
  • \(-\dfrac{1}{x+2} + \dfrac{1}{(x+2)^2} + c\). ✓
02
Calculation
[4 marks]

Use the substitution \(u = 1 + \sin x\) to find the exact value of \(\displaystyle\int_0^{\pi/2} \cos x\sqrt{1+\sin x}\,dx\).

  • \(du = \cos x\,dx\). Limits: \(x = 0 \Rightarrow u = 1\); \(x = \dfrac{\pi}{2} \Rightarrow u = 2\). ✓
  • \(\displaystyle\int_1^2 u^{1/2}\,du = \left[\dfrac{2}{3}u^{3/2}\right]_1^2\). ✓
  • \(= \dfrac{2}{3}\left(2\sqrt{2} - 1\right) = \dfrac{4\sqrt{2} - 2}{3}\). ✓
03
Calculation
[5 marks]

Use the substitution \(u = \sqrt{x}\) to show that \(\displaystyle\int_1^4 \dfrac{1}{\sqrt{x}\left(1+\sqrt{x}\right)}\,dx = 2\ln\dfrac{3}{2}\).

  • \(x = u^2\), so \(dx = 2u\,du\). Limits: \(x = 1 \Rightarrow u = 1\); \(x = 4 \Rightarrow u = 2\). ✓
  • Integrand: \(\dfrac{1}{u(1+u)}\times 2u\,du = \dfrac{2}{1+u}\,du\). ✓
  • \(\left[2\ln(1+u)\right]_1^2 = 2\ln 3 - 2\ln 2 = 2\ln\dfrac{3}{2}\). ✓
04
Proof
[5 marks]

Use the substitution \(u = \cos x\) to show that \(\displaystyle\int_0^{\pi/3} \sin^3 x\,dx = \dfrac{5}{24}\).

  • \(\sin^3 x = \sin x(1 - \cos^2 x)\), and \(du = -\sin x\,dx\). ✓
  • Limits: \(x = 0 \Rightarrow u = 1\); \(x = \dfrac{\pi}{3} \Rightarrow u = \dfrac{1}{2}\). ✓
  • \(\displaystyle -\int_1^{1/2} (1 - u^2)\,du = \int_{1/2}^1 (1 - u^2)\,du = \left[u - \dfrac{u^3}{3}\right]_{1/2}^1\). ✓
  • \(= \dfrac{2}{3} - \left(\dfrac{1}{2} - \dfrac{1}{24}\right) = \dfrac{16}{24} - \dfrac{11}{24} = \dfrac{5}{24}\). ✓
05
Calculation
[5 marks]

Use the substitution \(u = 1 + x^2\) to find the exact value of \(\displaystyle\int_0^{\sqrt{3}} \dfrac{x^3}{\sqrt{1+x^2}}\,dx\).

  • \(du = 2x\,dx\), and \(x^3\,dx = x^2\cdot x\,dx = (u - 1)\cdot\dfrac{1}{2}du\). Limits: 1 to 4. ✓
  • \(\dfrac{1}{2}\displaystyle\int_1^4 \left(u^{1/2} - u^{-1/2}\right) du = \dfrac{1}{2}\left[\dfrac{2}{3}u^{3/2} - 2u^{1/2}\right]_1^4\). ✓
  • \(= \dfrac{1}{2}\left[\left(\dfrac{16}{3} - 4\right) - \left(\dfrac{2}{3} - 2\right)\right] = \dfrac{1}{2}\times\dfrac{8}{3} = \dfrac{4}{3}\). ✓
06
Analysis
[8 marks]

The curve \(y = x\sqrt{1-x}\), for \(0 \le x \le 1\), has a maximum point \(M\).

(a) Find the exact coordinates of \(M\).

(b) Use the substitution \(u = 1 - x\) to find the exact area of the region between the curve and the \(x\)-axis.

  • (a) Product rule: \(\dfrac{dy}{dx} = \sqrt{1-x} - \dfrac{x}{2\sqrt{1-x}} = \dfrac{2(1-x) - x}{2\sqrt{1-x}} = \dfrac{2 - 3x}{2\sqrt{1-x}}\). ✓
  • \(\dfrac{dy}{dx} = 0 \Rightarrow x = \dfrac{2}{3}\); then \(y = \dfrac{2}{3}\sqrt{\dfrac{1}{3}} = \dfrac{2\sqrt{3}}{9}\). So \(M = \left(\dfrac{2}{3}, \dfrac{2\sqrt{3}}{9}\right)\). ✓
  • (b) \(dx = -du\), \(x = 1 - u\), limits \(x = 0 \to u = 1\), \(x = 1 \to u = 0\). ✓
  • Area \(= \displaystyle\int_0^1 (1-u)u^{1/2}\,du = \left[\dfrac{2}{3}u^{3/2} - \dfrac{2}{5}u^{5/2}\right]_0^1\). ✓
  • \(= \dfrac{2}{3} - \dfrac{2}{5} = \dfrac{4}{15}\) square units. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.