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8.5

Carry out simple cases of integration by substitution and integration by parts; understand these methods as the inverse processes of the chain and product rules respectively.

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Integration by substitution and parts

Worked answers and methods for 8.5 on Edexcel A-level Maths 9MA0.

Explanation

  • Substitution reverses the chain rule: choose u=g(x)u=g(x) so the remaining factor supplies du=g′(x) dx\mathrm du=g'(x)\,\mathrm dx. In particular, recognise ∫f′(x)f(x) dx=ln⁡∣f(x)∣+C\int\frac{f'(x)}{f(x)}\,\mathrm dx=\ln|f(x)|+C.
  • Integration by parts reverses the product rule: ∫u dv=uv−∫v du\int u\,\mathrm dv=uv-\int v\,\mathrm du; choose uu so that differentiating it simplifies the integral.
  • For a definite integral, either change the limits to the new variable or return fully to the original variable before applying the old limits.
  • A common error is to omit the minus sign in integration by parts or to mix xx and uu in the same transformed integral.
  • A required standard result is ∫ln⁡x dx=xln⁡x−x+c\int\ln x\,dx=x\ln x-x+c, obtained by integration by parts with u=ln⁡xu=\ln x and dv=dxdv=dx.

Worked example

Find ∫xe2x dx\int xe^{2x}\,\mathrm dx.

  1. 1.Use integration by parts with u=xu=x and dv=e2x dx\mathrm dv=e^{2x}\,\mathrm dx.
  2. 2.Then du=dx\mathrm du=\mathrm dx and v=12e2xv=\frac12e^{2x}.
  3. 3.Hence ∫xe2x dx=x2e2x−12∫e2x dx=x2e2x−14e2x+C\int xe^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac12\int e^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac14e^{2x}+C.

Answer: e2x(x2−14)+Ce^{2x}\left(\frac x2-\frac14\right)+C

Common mistakes

  • Don't change variable in a definite integral but keep the original xx-limits.
  • Don't choose integration by parts but reverse the selected u and dv, creating a harder integral.

Exam tip

For a product of an algebraic and exponential term, choose the algebraic factor as u and retain the boundary-free constant.

Worked practice

Q1
Tier 1 · Easy

1.

Find ∫6x3x2+5 dx\int\frac{6x}{3x^2+5}\,\mathrm dx.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • ln⁡(3x2+5)+C\ln(3x^2+5)+C
3
Notes
The numerator is the derivative of the denominator. Therefore this has the form ∫f′(x)/f(x) dx\int f'(x)/f(x)\,\mathrm dx, giving ln⁡∣3x2+5∣+C\ln|3x^2+5|+C. Since 3x2+5>03x^2+5>0, this is ln⁡(3x2+5)+C\ln(3x^2+5)+C.

(3 marks)

Q2
Tier 2 · Standard

2.

Use a substitution to evaluate ∫012xx2+4 dx\displaystyle\int_0^1\dfrac{2x}{x^2+4}\,\mathrm dx.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • ln⁡ ⁣(54)\ln\!\left(\dfrac54\right)
4
Notes
Let u=x2+4u=x^2+4, so du=2x dxdu=2x\,dx. The limits become u=4u=4 when x=0x=0 and u=5u=5 when x=1x=1. Therefore the integral is ∫45u−1 du=[ln⁡u]45=ln⁡5−ln⁡4=ln⁡(5/4)\int_4^5u^{-1}\,du=[\ln u]_4^5=\ln5-\ln4=\ln(5/4).

(4 marks)

Q3
Tier 3 · Hard

3.

Evaluate exactly ∫01xln⁡(1+x2) dx\int_0^1x\ln(1+x^2)\,\mathrm dx.

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • ln⁡2−12\ln2-\frac12
7
Notes
Let u=1+x2u=1+x^2, so du=2x dx\mathrm du=2x\,\mathrm dx and the limits become u=1u=1 to u=2u=2. The integral is 12∫12ln⁡u du\frac12\int_1^2\ln u\,\mathrm du. By parts, ∫ln⁡u du=uln⁡u−u\int\ln u\,\mathrm du=u\ln u-u. Therefore the value is 12[uln⁡u−u]12=12(2ln⁡2−2+1)=ln⁡2−12\frac12[u\ln u-u]_1^2=\frac12(2\ln2-2+1)=\ln2-\frac12.

(7 marks)

Q4
Tier 1 · Easy

4.

Find ∫8x(2x2+1)3 dx\displaystyle\int8x(2x^2+1)^3\,\mathrm dx.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • 12(2x2+1)4+C\dfrac12(2x^2+1)^4+C
3
Notes
Let u=2x2+1u=2x^2+1, so du=4x dx\mathrm du=4x\,\mathrm dx and 8x dx=2 du8x\,\mathrm dx=2\,\mathrm du. Then the integral is 2∫u3 du=12u4+C=12(2x2+1)4+C2\int u^3\,\mathrm du=\frac12u^4+C=\frac12(2x^2+1)^4+C.

(3 marks)

Q5
Tier 2 · Standard

5.

Evaluate exactly ∫01xcos⁡(πx) dx\displaystyle\int_0^1x\cos(\pi x)\,\mathrm dx.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • −2π2-\dfrac{2}{\pi^2}
4
Notes
Use integration by parts with u=xu=x and dv=cos⁡(πx) dx\mathrm dv=\cos(\pi x)\,\mathrm dx, so v=sin⁡(πx)/πv=\sin(\pi x)/\pi. Thus the integral is [xsin⁡(πx)/π]01−1π∫01sin⁡(πx) dx=0−2/π2[x\sin(\pi x)/\pi]_0^1-\frac1\pi\int_0^1\sin(\pi x)\,\mathrm dx=0-2/\pi^2.

(4 marks)

Q6
Tier 3 · Hard

6.

Evaluate exactly ∫0π/2x2sin⁡x dx\displaystyle\int_0^{\pi/2}x^2\sin x\,\mathrm dx.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • π−2\pi-2
6
Notes
Integrating by parts with u=x2u=x^2 and dv=sin⁡x dx\mathrm dv=\sin x\,\mathrm dx gives [−x2cos⁡x]0π/2+2∫0π/2xcos⁡x dx[-x^2\cos x]_0^{\pi/2}+2\int_0^{\pi/2}x\cos x\,\mathrm dx. A second integration by parts gives ∫xcos⁡x dx=xsin⁡x+cos⁡x\int x\cos x\,\mathrm dx=x\sin x+\cos x. Therefore the value is 2[xsin⁡x+cos⁡x]0π/2=π−22[x\sin x+\cos x]_0^{\pi/2}=\pi-2.

(6 marks)

Q7
Tier 2 · Standard

7.

Use integration by parts to evaluate exactly ∫1e3ln⁡x dx\displaystyle\int_1^{e^3}\ln x\,\mathrm dx.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • 2e3+12e^3+1
4
Notes
Choose u=ln⁡xu=\ln x and dv=dx\mathrm dv=\mathrm dx. Then du=dx/x\mathrm du=\mathrm dx/x and v=xv=x, so ∫ln⁡x dx=xln⁡x−x+C\int\ln x\,\mathrm dx=x\ln x-x+C. Applying the limits gives [xln⁡x−x]1e3=2e3−(−1)=2e3+1[x\ln x-x]_1^{e^3}=2e^3-(-1)=2e^3+1.

(4 marks)

Q8
Tier 3 · Hard

8.

Use integration by parts twice to find ∫excos⁡x dx\displaystyle\int e^x\cos x\,\mathrm dx.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • 12ex(sin⁡x+cos⁡x)+C\dfrac12e^x(\sin x+\cos x)+C
6
Notes
Let I=∫excos⁡x dxI=\int e^x\cos x\,\mathrm dx. Integration by parts gives I=excos⁡x+∫exsin⁡x dxI=e^x\cos x+\int e^x\sin x\,\mathrm dx. Applying integration by parts to the remaining integral gives ∫exsin⁡x dx=exsin⁡x−I\int e^x\sin x\,\mathrm dx=e^x\sin x-I. Hence 2I=ex(sin⁡x+cos⁡x)2I=e^x(\sin x+\cos x), so I=12ex(sin⁡x+cos⁡x)+CI=\frac12e^x(\sin x+\cos x)+C.

(6 marks)

Q9
Tier 3 · Hard

9.

Make the change of variable u=x2+1u=x^2+1 to evaluate exactly ∫03x3x2+1 dx\displaystyle\int_0^{\sqrt3}\dfrac{x^3}{x^2+1}\,\mathrm dx.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • 32−ln⁡2\dfrac32-\ln2
5
Notes
Let u=x2+1u=x^2+1, so du=2x dx\mathrm du=2x\,\mathrm dx, x2=u−1x^2=u-1, and the limits become 11 and 44. The integral becomes 12∫14(u−1)/u du=12∫14(1−1/u) du\frac12\int_1^4(u-1)/u\,\mathrm du=\frac12\int_1^4(1-1/u)\,\mathrm du. Hence its value is 12[u−ln⁡u]14=3/2−ln⁡2\frac12[u-\ln u]_1^4=3/2-\ln2.

(5 marks)

Q10
Tier 3 · Hard

10.

Use integration by parts to evaluate exactly ∫01x2ln⁡(1+x) dx\displaystyle\int_0^1x^2\ln(1+x)\,\mathrm dx.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • 23ln⁡2−518\dfrac23\ln2-\dfrac5{18}
6
Notes
Take u=ln⁡(1+x)u=\ln(1+x) and dv=x2 dx\mathrm dv=x^2\,\mathrm dx. The integral is [x3ln⁡(1+x)/3]01−13∫01x3/(1+x) dx[x^3\ln(1+x)/3]_0^1-\frac13\int_0^1x^3/(1+x)\,\mathrm dx. Since x3/(1+x)=x2−x+1−1/(1+x)x^3/(1+x)=x^2-x+1-1/(1+x), this becomes 13ln⁡2−13[x3/3−x2/2+x−ln⁡(1+x)]01=23ln⁡2−518\frac13\ln2-\frac13[x^3/3-x^2/2+x-\ln(1+x)]_0^1=\frac23\ln2-\frac5{18}.

(6 marks)

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