What is 8.2: Standard integrals?

For n≠-1, x^n\, dx=x^n+1/n+1+C; rewrite roots and reciprocals as powers before applying the rule. Standard forms include e^kx\, dx=1/k e^kx+C and 1/x\, dx=|x|+C on intervals not crossing 0.

How many marks are 8.2 questions worth?

8.2 practice drills on this page are worth 3–6 marks each; marks in a real exam vary by question.

Which past exam papers has 8.2 appeared in?

We have not yet indexed a verified past-paper appearance for 8.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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 $n\ne-1$ , $\int xⁿ\,\mathrm dx=\frac{x^{n+1}}{n+1}+C$; rewrite roots and reciprocals as powers before applying the rule. Standard forms include $\int e^{kx}\,\mathrm dx=\frac1k e^{kx}+C$ and $\int\frac1x\,\mathrm dx=\ln|x|+C$ on intervals not crossing $0$ .
Divide by the inner coefficient in trigonometric integrals.
Also use $\int\sec^2(kx)\,\mathrm dx=\frac1k\tan(kx)+C$ , $\int\tan x\,\mathrm dx=-\ln|\cos x|+C$ , and identities for powers such as $\sin^2x=\frac12(1-\cos2x)$ .
A common error is to forget the constant of integration or to multiply by $k$ rather than divide by it when integrating a function of $kx$ .
Integrate term by term: the first three terms give $2e^{3x}+4\ln|x|+\frac52\cos(2x)$ .

Worked example

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

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

Answer: $2e^{3x}+4\ln|x|+\frac52\cos(2x)-3\ln|\cos x|+C$

Common mistakes

Don't integrate $e^{kx}$ as $ke^{kx}$ instead of $e^{kx}/k$ .
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

Tier 1 · Easy

Find

\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
Question	Scheme	Marks
1	$2x^3-\frac83x^{3/2}+\frac32\tan(2x)+C$	4
Notes
The power terms give $2x^3-\frac83x^{3/2}$ . Since $\int\sec^2(2x)\,\mathrm dx=\frac12\tan(2x)$ , the final term gives $\frac32\tan(2x)$ . Add $C$ .

(4 marks)

Tier 2 · Standard

For

x>0

, find

\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
Question	Scheme	Marks
2	$\ln x+4\sqrt{x}+x+C$	4
Notes
First expand and divide by $x$ : $(1+\sqrt{x})^2/x=(1+2\sqrt{x}+x)/x=x^{-1}+2x^{-1/2}+1$ . Integrating term by term gives $\ln x+4x^{1/2}+x+C$ .

(4 marks)

Tier 3 · Hard

Use trigonometric identities to find

\int(4\sin^2x+3\tan^2x)\,\mathrm dx

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$3\tan x-x-\sin(2x)+C$	6
Notes
Use $\sin^2x=\frac12(1-\cos2x)$ and $\tan^2x=\sec^2x-1$ . The integrand becomes $2-2\cos2x+3\sec^2x-3=3\sec^2x-1-2\cos2x$ . Integrating gives $3\tan x-x-\sin(2x)+C$ .

(6 marks)

Tier 1 · Easy

For

x>0

, find

\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
Question	Scheme	Marks
4	$\dfrac23x^3-5x+4\ln x+C$	3
Notes
Divide each term by $x$ to obtain $2x^2-5+4/x$ . Integrating term by term gives $\frac23x^3-5x+4\ln x+C$ .

(3 marks)

Tier 2 · Standard

Find

\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
Question	Scheme	Marks
5	$-14e^{-x/2}+\dfrac43\sin(3x)-\dfrac54\tan(4x)+C$	4
Notes
Integrating each term and dividing by its inner coefficient gives $-14e^{-x/2}$ , $\frac43\sin(3x)$ and $-\frac54\tan(4x)$ respectively. Adding the constant gives the stated result.

(4 marks)

Tier 3 · Hard

Find

\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
Question	Scheme	Marks
6	$\dfrac52x-\dfrac34\sin(2x)-\cos(2x)+C$	5
Notes
Expand and use double-angle identities: $(2\sin x+\cos x)^2=4\sin^2x+4\sin x\cos x+\cos^2x=\frac52-\frac32\cos(2x)+2\sin(2x)$ . Integration gives $\frac52x-\frac34\sin(2x)-\cos(2x)+C$ .

(5 marks)

Tier 2 · Standard

Evaluate exactly

\displaystyle\int_0^{\pi/6}\tan(2x)\,\mathrm dx

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\dfrac12\ln2$	4
Notes
Since $\int\tan(2x)\,\mathrm dx=-\frac12\ln\|\cos(2x)\|+C$ , the definite integral is $[-\frac12\ln\|\cos(2x)\|]_0^{\pi/6}=-\frac12\ln(1/2)+\frac12\ln1=\frac12\ln2$ .

(4 marks)

Tier 3 · Hard

By writing

\sin(3x)\cos x

in terms of

\sin(3x+x)

and

\sin(3x-x)

, using the sine addition and subtraction formulae, find

\displaystyle\int 8\sin(3x)\cos x\,\mathrm dx

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$-\cos(4x)-2\cos(2x)+C$	5
Notes
Adding the formulae for $\sin(3x+x)$ and $\sin(3x-x)$ gives $2\sin(3x)\cos x=\sin(4x)+\sin(2x)$ . Hence $8\sin(3x)\cos x=4\sin(4x)+4\sin(2x)$ . Integrating term by term gives $-\cos(4x)-2\cos(2x)+C$ .

(5 marks)

Tier 3 · Hard

Let

f(x)=a\sin(2x)+b\cos(2x)

, where

a

and

b

are constants. Given that

\displaystyle\int_0^{\pi/4}f(x)\,\mathrm dx=3

and

\displaystyle\int_{\pi/4}^{\pi/2}f(x)\,\mathrm dx=1

, find

a

and

b

. Hence find

\displaystyle\int f(x)\,\mathrm dx

(6)

(Total for Question 9 is 6 marks)

Mark scheme

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

(6 marks)

Q10

Tier 3 · Hard

10.

Evaluate exactly

\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
Question	Scheme	Marks
10	$\dfrac{\pi}{3}+2\sqrt3$	6
Notes
On $0\leq x\leq\pi$ , $1+2\cos x=0$ only when $x=2\pi/3$ . The expression is non-negative before this point and negative after it. Therefore the integral is $\int_0^{2\pi/3}(1+2\cos x)\,\mathrm dx-\int_{2\pi/3}^{\pi}(1+2\cos x)\,\mathrm dx$ . Using the antiderivative $x+2\sin x$ gives $(2\pi/3+\sqrt3)-(\pi/3-\sqrt3)=\pi/3+2\sqrt3$ .

(6 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 8.2. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 8 Integration

Want help turning this into marks?

Bring 8.2 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call