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8.2

Integrate xⁿ (excluding n = −1) and related sums, differences and constant multiples; integrate e^(kx), 1/x, sin kx, cos kx and related sums, differences and constant multiples.

Draft — not yet indexed

Standard integrals

Worked answers and methods for 8.2 on Edexcel A-level Maths 9MA0.

Explanation

  • For n≠−1n\ne-1, $\int xn\,\mathrm dx=\frac{x^{n+1}}{n+1}+C$; rewrite roots and reciprocals as powers before applying the rule. Standard forms include ∫ekx dx=1kekx+C\int e^{kx}\,\mathrm dx=\frac1k e^{kx}+C and ∫1x dx=ln⁡∣x∣+C\int\frac1x\,\mathrm dx=\ln|x|+C on intervals not crossing 00.
  • Divide by the inner coefficient in trigonometric integrals.
  • Also use ∫sec⁡2(kx) dx=1ktan⁡(kx)+C\int\sec^2(kx)\,\mathrm dx=\frac1k\tan(kx)+C, ∫tan⁡x dx=−ln⁡∣cos⁡x∣+C\int\tan x\,\mathrm dx=-\ln|\cos x|+C, and identities for powers such as sin⁡2x=12(1−cos⁡2x)\sin^2x=\frac12(1-\cos2x).
  • A common error is to forget the constant of integration or to multiply by kk rather than divide by it when integrating a function of kxkx.
  • Integrate term by term: the first three terms give 2e3x+4ln⁡∣x∣+52cos⁡(2x)2e^{3x}+4\ln|x|+\frac52\cos(2x).

Worked example

Find ∫(6e3x+4x−5sin⁡(2x)+3tan⁡x) dx\int\left(6e^{3x}+\frac4x-5\sin(2x)+3\tan x\right)\,\mathrm dx.

  1. 1.Integrate term by term: the first three terms give 2e3x+4ln⁡∣x∣+52cos⁡(2x)2e^{3x}+4\ln|x|+\frac52\cos(2x).
  2. 2.Since ∫tan⁡x dx=−ln⁡∣cos⁡x∣\int\tan x\,\mathrm dx=-\ln|\cos x|, the final term gives −3ln⁡∣cos⁡x∣-3\ln|\cos x|.
  3. 3.Add CC.

Answer: 2e3x+4ln⁡∣x∣+52cos⁡(2x)−3ln⁡∣cos⁡x∣+C2e^{3x}+4\ln|x|+\frac52\cos(2x)-3\ln|\cos x|+C

Common mistakes

  • Don't integrate ekxe^{kx} as kekxke^{kx} instead of ekx/ke^{kx}/k.
  • Don't treat the integral of 1/x as a power-rule case or miss reciprocal chain factors.

Exam tip

Match each term to its standard antiderivative and include the integration constant after combining the results.

Worked practice

Q1
Tier 1 · Easy

1.

Find ∫(6x2−4x1/2+3sec⁡2(2x)) dx\int(6x^2-4x^{1/2}+3\sec^2(2x))\,\mathrm dx.

(4)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • 2x3−83x3/2+32tan⁡(2x)+C2x^3-\frac83x^{3/2}+\frac32\tan(2x)+C
4
Notes
The power terms give 2x3−83x3/22x^3-\frac83x^{3/2}. Since ∫sec⁡2(2x) dx=12tan⁡(2x)\int\sec^2(2x)\,\mathrm dx=\frac12\tan(2x), the final term gives 32tan⁡(2x)\frac32\tan(2x). Add CC.

(4 marks)

Q2
Tier 2 · Standard

2.

For x>0x>0, find ∫(1+x)2x dx\displaystyle\int\dfrac{(1+\sqrt{x})^2}{x}\,\mathrm dx.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • ln⁡x+4x+x+C\ln x+4\sqrt{x}+x+C
4
Notes
First expand and divide by xx: (1+x)2/x=(1+2x+x)/x=x−1+2x−1/2+1(1+\sqrt{x})^2/x=(1+2\sqrt{x}+x)/x=x^{-1}+2x^{-1/2}+1. Integrating term by term gives ln⁡x+4x1/2+x+C\ln x+4x^{1/2}+x+C.

(4 marks)

Q3
Tier 3 · Hard

3.

Use trigonometric identities to find ∫(4sin⁡2x+3tan⁡2x) dx\int(4\sin^2x+3\tan^2x)\,\mathrm dx.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • 3tan⁡x−x−sin⁡(2x)+C3\tan x-x-\sin(2x)+C
6
Notes
Use sin⁡2x=12(1−cos⁡2x)\sin^2x=\frac12(1-\cos2x) and tan⁡2x=sec⁡2x−1\tan^2x=\sec^2x-1. The integrand becomes 2−2cos⁡2x+3sec⁡2x−3=3sec⁡2x−1−2cos⁡2x2-2\cos2x+3\sec^2x-3=3\sec^2x-1-2\cos2x. Integrating gives 3tan⁡x−x−sin⁡(2x)+C3\tan x-x-\sin(2x)+C.

(6 marks)

Q4
Tier 1 · Easy

4.

For x>0x>0, find ∫2x3−5x+4x dx\displaystyle\int\dfrac{2x^3-5x+4}{x}\,\mathrm dx.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • 23x3−5x+4ln⁡x+C\dfrac23x^3-5x+4\ln x+C
3
Notes
Divide each term by xx to obtain 2x2−5+4/x2x^2-5+4/x. Integrating term by term gives 23x3−5x+4ln⁡x+C\frac23x^3-5x+4\ln x+C.

(3 marks)

Q5
Tier 2 · Standard

5.

Find ∫(7e−x/2+4cos⁡(3x)−5sec⁡2(4x)) dx\displaystyle\int\left(7e^{-x/2}+4\cos(3x)-5\sec^2(4x)\right)\,\mathrm dx.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • −14e−x/2+43sin⁡(3x)−54tan⁡(4x)+C-14e^{-x/2}+\dfrac43\sin(3x)-\dfrac54\tan(4x)+C
4
Notes
Integrating each term and dividing by its inner coefficient gives −14e−x/2-14e^{-x/2}, 43sin⁡(3x)\frac43\sin(3x) and −54tan⁡(4x)-\frac54\tan(4x) respectively. Adding the constant gives the stated result.

(4 marks)

Q6
Tier 3 · Hard

6.

Find ∫(2sin⁡x+cos⁡x)2 dx\displaystyle\int(2\sin x+\cos x)^2\,\mathrm dx.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • 52x−34sin⁡(2x)−cos⁡(2x)+C\dfrac52x-\dfrac34\sin(2x)-\cos(2x)+C
5
Notes
Expand and use double-angle identities: (2sin⁡x+cos⁡x)2=4sin⁡2x+4sin⁡xcos⁡x+cos⁡2x=52−32cos⁡(2x)+2sin⁡(2x)(2\sin x+\cos x)^2=4\sin^2x+4\sin x\cos x+\cos^2x=\frac52-\frac32\cos(2x)+2\sin(2x). Integration gives 52x−34sin⁡(2x)−cos⁡(2x)+C\frac52x-\frac34\sin(2x)-\cos(2x)+C.

(5 marks)

Q7
Tier 2 · Standard

7.

Evaluate exactly ∫0π/6tan⁡(2x) dx\displaystyle\int_0^{\pi/6}\tan(2x)\,\mathrm dx.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • 12ln⁡2\dfrac12\ln2
4
Notes
Since ∫tan⁡(2x) dx=−12ln⁡∣cos⁡(2x)∣+C\int\tan(2x)\,\mathrm dx=-\frac12\ln|\cos(2x)|+C, the definite integral is [−12ln⁡∣cos⁡(2x)∣]0π/6=−12ln⁡(1/2)+12ln⁡1=12ln⁡2[-\frac12\ln|\cos(2x)|]_0^{\pi/6}=-\frac12\ln(1/2)+\frac12\ln1=\frac12\ln2.

(4 marks)

Q8
Tier 3 · Hard

8.

By writing sin⁡(3x)cos⁡x\sin(3x)\cos x in terms of sin⁡(3x+x)\sin(3x+x) and sin⁡(3x−x)\sin(3x-x), using the sine addition and subtraction formulae, find ∫8sin⁡(3x)cos⁡x dx\displaystyle\int 8\sin(3x)\cos x\,\mathrm dx.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • −cos⁡(4x)−2cos⁡(2x)+C-\cos(4x)-2\cos(2x)+C
5
Notes
Adding the formulae for sin⁡(3x+x)\sin(3x+x) and sin⁡(3x−x)\sin(3x-x) gives 2sin⁡(3x)cos⁡x=sin⁡(4x)+sin⁡(2x)2\sin(3x)\cos x=\sin(4x)+\sin(2x). Hence 8sin⁡(3x)cos⁡x=4sin⁡(4x)+4sin⁡(2x)8\sin(3x)\cos x=4\sin(4x)+4\sin(2x). Integrating term by term gives −cos⁡(4x)−2cos⁡(2x)+C-\cos(4x)-2\cos(2x)+C.

(5 marks)

Q9
Tier 3 · Hard

9.

Let f(x)=asin⁡(2x)+bcos⁡(2x)f(x)=a\sin(2x)+b\cos(2x), where aa and bb are constants. Given that ∫0π/4f(x) dx=3\displaystyle\int_0^{\pi/4}f(x)\,\mathrm dx=3 and ∫π/4π/2f(x) dx=1\displaystyle\int_{\pi/4}^{\pi/2}f(x)\,\mathrm dx=1, find aa and bb. Hence find ∫f(x) dx\displaystyle\int f(x)\,\mathrm dx.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • a=4a=4, b=2b=2
  • ∫f(x) dx=−2cos⁡(2x)+sin⁡(2x)+C\displaystyle\int f(x)\,\mathrm dx=-2\cos(2x)+\sin(2x)+C
6
Notes
An antiderivative is −acos⁡(2x)/2+bsin⁡(2x)/2-a\cos(2x)/2+b\sin(2x)/2. The first given integral is (a+b)/2=3(a+b)/2=3, while the second is (a−b)/2=1(a-b)/2=1. Hence a+b=6a+b=6 and a−b=2a-b=2, so a=4a=4 and b=2b=2. Substitution into the antiderivative gives −2cos⁡(2x)+sin⁡(2x)+C-2\cos(2x)+\sin(2x)+C.

(6 marks)

Q10
Tier 3 · Hard

10.

Evaluate exactly ∫0π∣1+2cos⁡x∣ dx\displaystyle\int_0^{\pi}|1+2\cos x|\,\mathrm dx.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • π3+23\dfrac{\pi}{3}+2\sqrt3
6
Notes
On 0≤x≤π0\leq x\leq\pi, 1+2cos⁡x=01+2\cos x=0 only when x=2π/3x=2\pi/3. The expression is non-negative before this point and negative after it. Therefore the integral is ∫02π/3(1+2cos⁡x) dx−∫2π/3π(1+2cos⁡x) dx\int_0^{2\pi/3}(1+2\cos x)\,\mathrm dx-\int_{2\pi/3}^{\pi}(1+2\cos x)\,\mathrm dx. Using the antiderivative x+2sin⁡xx+2\sin x gives (2π/3+3)−(π/3−3)=π/3+23(2\pi/3+\sqrt3)-(\pi/3-\sqrt3)=\pi/3+2\sqrt3.

(6 marks)

Q11
Tier 1 · Easy

11.

Find ∫(6x2+4x13−3x3)dx\displaystyle\int\left(6x^2+4x^{\frac13}-\dfrac{3}{x^3}\right)\mathrm dx.

(3)

(Total for Question 11 is 3 marks)

Mark scheme

Mark scheme for question 11
QuestionSchemeMarks
11
  • 2x3+3x43+32x−2+C2x^3+3x^{\frac43}+\tfrac32x^{-2}+C
3
Notes
Write 3x3=3x−3\dfrac{3}{x^3}=3x^{-3} and integrate term by term with ∫xn dx=xn+1n+1+C\displaystyle\int x^n\,\mathrm dx=\dfrac{x^{n+1}}{n+1}+C for n≠−1n\neq-1: 6x33+4x4/34/3−3x−2−2+C=2x3+3x43+32x−2+C\dfrac{6x^3}{3}+\dfrac{4x^{4/3}}{4/3}-\dfrac{3x^{-2}}{-2}+C=2x^3+3x^{\frac43}+\tfrac32x^{-2}+C.

(3 marks)

Q12
Tier 2 · Standard

12.

Find ∫(2x−1)(x+3)x dx\displaystyle\int\dfrac{(2x-1)(x+3)}{\sqrt x}\,\mathrm dx, for x>0x>0, giving each term in its simplest form.

(5)

(Total for Question 12 is 5 marks)

Mark scheme

Mark scheme for question 12
QuestionSchemeMarks
12
  • 45x52+103x32−6x12+C\tfrac45x^{\frac52}+\tfrac{10}{3}x^{\frac32}-6x^{\frac12}+C
5
Notes
Expand the numerator: (2x−1)(x+3)=2x2+5x−3(2x-1)(x+3)=2x^2+5x-3. Divide each term by x12x^{\frac12}: the integrand is 2x32+5x12−3x−122x^{\frac32}+5x^{\frac12}-3x^{-\frac12}. Integrate term by term: 2x5/25/2+5x3/23/2−3x1/21/2+C=45x52+103x32−6x12+C\dfrac{2x^{5/2}}{5/2}+\dfrac{5x^{3/2}}{3/2}-\dfrac{3x^{1/2}}{1/2}+C=\tfrac45x^{\frac52}+\tfrac{10}{3}x^{\frac32}-6x^{\frac12}+C.

(5 marks)

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