1. Integration

2. After completing this chapter you should be able to integrate:
A number of standard mathematical functions
Using the reverse of the chain rule
Using trigonometrical identities
Using partial fractions
By substitution
By parts
To find areas and volumes
To solve differential equations
In addition some functions are too difficult to integrate and hence you should be able to
Use the trapezium rule to approximate their value

3. 6.1 You need to be able to integrate standard functions
There are 13 of these so please look at the sheet and then practise with exercise 6A page 89

4. 6.2 You can integrate some functions using the reverse of the chain rule
Basically you look to see what might have differentiated to the function you want to integrate!
𝑓 ′ 𝑎𝑥+𝑏 𝑑𝑥= 1 𝑎 𝑓 𝑎𝑥+𝑏 +𝐶
Lets look at some examples to see have this is applied

5. Find 𝑠𝑒𝑐 2 4𝑥+1 𝑑𝑥
Let I = 𝑠𝑒𝑐 2 4𝑥+1 𝑑𝑥
Then from (x) on our sheet we can see
I = tan 4𝑥+1 +𝐶

6. This is 3 2 x the function we want to integrate So I = 3 2 ln 1+2𝑥 +𝐶
Find 𝑥 𝑑𝑥
Let I = 𝑥 𝑑𝑥
Now consider y = ln 1+2𝑥
Then 𝑑𝑦 𝑑𝑥 = 𝑥
This is x the function we want to integrate
So I = ln 1+2𝑥 +𝐶
Exercise 6B page 92

7. 6.3 You can use trigonometric identities in integration
It is sometimes easier to change an expression, using trigonometrical identities, into one you know how to integrate.
For example you could change sin² x into 1 2 (1 − cos 2𝑥)
Or sin3x cos 3x into 𝑠𝑖𝑛6𝑥 (sin2A = 2 sinA cosA)
Or (sec x + tan x)²
= sec²x + 2 sec x tan x + tan²x
= sec²x + 2 sec x tan x + sec²x – 1
= 2sec²x + 2 sec x tan x – 1
All these are in the list, so you can take it from here !

8. Example
Find 𝑐𝑜𝑡 2 3𝑥 𝑑𝑥
Use cosec²A = cot²A + 1
So I = 𝑐𝑜𝑡 2 3𝑥 𝑑𝑥= ( 𝑐𝑜𝑠𝑒𝑐 2 3𝑥 −1) 𝑑𝑥
This is on the list and gives us
I = − 1 3 cot 3𝑥 −1 +𝐶
Exercise 6C page 94

9. 6.4 You can use partial fractions to integrate expressions
It can be useful to split some expressions into partial fractions before integrating.
𝑥+8 (𝑥 −1)(𝑥+2) 𝑑𝑥= 3 𝑥 −1 𝑑𝑥 − 𝑥+2 𝑑𝑥
= 𝑙𝑛 𝑥 −1 − 𝑙𝑛 𝑥+2 +𝐶
Exercise 6D page 98

10. 6.5 You can use standard patterns to integrate some expressions
To integrate an expression is of the form 𝑘 𝑓 ′ (𝑥) 𝑓(𝑥) 𝑑𝑥
try ln 𝑓(𝑥) and differentiate to check and adjust any constants
To integrate an expression of the form 𝑘 𝑓 ′ 𝑥 [𝑓 𝑥 ] 𝑛 𝑑𝑥
Try [𝑓 𝑥 ] 𝑛+1 and differentiate to check and adjust any constants

11. Example
Find 𝑥+1 𝑥 2 +2𝑥+5 𝑑𝑥
Let I = 𝑥+1 𝑥 2 +2𝑥+5 𝑑𝑥
This looks like it could be of the form 𝑘 𝑓 ′ (𝑥) 𝑓(𝑥) 𝑑𝑥
So we try y = 𝑙𝑛 𝑥 2 +2𝑥+5
𝑑𝑦 𝑑𝑥 = 2𝑥+2 𝑥 2 +2𝑥+5 = 2(𝑥+1) 𝑥 2 +2𝑥+5
So I = 1 2 𝑙𝑛 𝑥 2 +2𝑥+5 + C

12. = 12 𝑠𝑖𝑛 2 4𝑥 −1 cos 4𝑥−1 So I = 1 12 𝑠𝑖𝑛 3 4𝑥 −1 +𝐶 Example
Find cos 4𝑥 −1 𝑠𝑖𝑛 2 4 𝑥 2 −1 𝑑𝑥
Let I = cos 4𝑥 −1 𝑠𝑖𝑛 2 4 𝑥 2 −1 𝑑𝑥
This looks like it could be of the form 𝑘 𝑓 ′ 𝑥 [𝑓 𝑥 ] 𝑛 𝑑𝑥
So we try y = sin³(4x – 1)
𝑑𝑦 𝑑𝑥 =3 𝑠𝑖𝑛 2 4𝑥 −1 cos 4𝑥−1 ×4
= 12 𝑠𝑖𝑛 2 4𝑥 −1 cos 4𝑥−1
So I = 𝑠𝑖𝑛 3 4𝑥 −1 +𝐶
Exercise 6E page 100

13. 6. 6 Sometimes you can simplify an integral by changing the variable
6.6 Sometimes you can simplify an integral by changing the variable. This process is similar to using the chain rule in differentiation and is called ‘integration by substitution’.
When you try to find 𝑥 (2𝑥+5) 𝑑𝑥
it is easier to do if we substitute for 2x + 5
Let u = 2x then x = 𝑢 − and 𝑑𝑢 𝑑𝑥 =2 this gives dx = 𝑑𝑢
Replace these elements into the original integral
I = 𝑢 −5 2 𝑢 𝑑𝑢
= 𝑢 −5 𝑢 𝑑𝑢 which we can hopefully do and get
I = (2𝑥+5) − 5 (2𝑥+5) C

14. Let I = 𝑠𝑒𝑐 2 𝑥 𝑒 𝑡𝑎𝑛𝑥 dx u = tanx
Example
Find 𝑠𝑒𝑐 2 𝑥 𝑒 𝑡𝑎𝑛𝑥 𝑑𝑥 by using the substitution u = tanx
Let I = 𝑠𝑒𝑐 2 𝑥 𝑒 𝑡𝑎𝑛𝑥 dx u = tanx
𝑑𝑢 𝑑𝑥 = 𝑠𝑒𝑐 2 𝑥 giving du = sec²x dx
Substituting this information in gives
I = 𝑒 𝑢 𝑑𝑢
I = 𝑒 𝑢 +𝐶 So
I = 𝑒 𝑡𝑎𝑛𝑥 +𝐶
The substitutions suggested by the question writers should make life easier, if they don’t you’re doing something wrong!

15. Example
Use the substitution u = 1 + sin x to integrate 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 (1+𝑠𝑖𝑛𝑥) 5 𝑑𝑥
I = 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 (1+𝑠𝑖𝑛𝑥) 5 𝑑𝑥
u = 1 + sin x sinx = u 𝑑𝑢 𝑑𝑥 =𝑐𝑜𝑠𝑥 so du = cosx dx
Substituting these in gives
I = 𝑢 −1 𝑢 5 𝑑𝑢
I = (𝑢 6 − 𝑢 5 )𝑑𝑢
I = 𝑢 𝑢 𝐶 we can rearrange this to give I = 𝑢 𝑢 −7 +𝐶
Now substitute for U I = (1+𝑠𝑖𝑛𝑥) 𝑠𝑖𝑛𝑥+6 −7 +𝐶
I = (1+𝑠𝑖𝑛𝑥) 𝑠𝑖𝑛𝑥 −1 +𝐶

16. Exercise 6F page 105
HINT if you see u² -1 in any shape or form factorise it to (u + 1)(u – 1)

17. 6.7 You can use integration by parts to integrate some expressions
The product rule gives
𝑑 𝑑𝑥 𝑢𝑣 =𝑣 𝑑𝑢 𝑑𝑥 +𝑢 𝑑𝑣 𝑑𝑥
Rearranging this gives us
Integrating each term with respect to x gives
𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥 = 𝑑 𝑑𝑥 𝑢𝑣 𝑑𝑥 − 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥
𝑢 𝑑𝑣 𝑑𝑥 = 𝑑 𝑑𝑥 𝑢𝑣 −𝑣 𝑑𝑢 𝑑𝑥
This simplifies to give us the integration by parts formula
𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥

22. 𝐼= −𝑥𝑐𝑜𝑡𝑥 𝜋 3 - 𝜋 6 𝜋 3 (−𝑐𝑜𝑡𝑥)𝑑𝑥
Use integration by parts to find the exact value of 𝜋 6 𝜋 3 𝑥 𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝑑𝑥
𝑑𝑢 𝑑𝑥 =1
u = x
𝑑𝑣 𝑑𝑥 = cosec²x v = -cotx
substituting all of this back into the formula
𝐼= −𝑥𝑐𝑜𝑡𝑥 𝜋 𝜋 6 𝜋 3 (−𝑐𝑜𝑡𝑥)𝑑𝑥
𝜋 6
𝐼= 𝜋 𝑙𝑛 −𝑙𝑛 1 2
𝐼= 𝜋 𝑙𝑛 3

23. 6.8 You can use numerical integration
You may be asked to use the trapezium rule for integrals involving some of the new functions we have met in C3 and C4
Remember the trapezium rule looks like this:
𝑎 𝑏 𝑦 𝑑𝑥= 1 2 ℎ 𝑦 𝑦 1 + 𝑦 𝑦 𝑛−1 + 𝑦 𝑛
Where ℎ= 𝑏 −𝑎 𝑛 and 𝑦 𝑖 =𝑓 𝑎+𝑖ℎ

24. 𝑎 𝑏 𝑦 𝑑𝑥= 1 2 ℎ 𝑦 0 +2 𝑦 1 + 𝑦 2 + . . . + 𝑦 𝑛−1 + 𝑦 𝑛
Where ℎ= 𝑏 −𝑎 𝑛 and 𝑦 𝑖 =𝑓 𝑎+𝑖ℎ
Example
Complete the table below, giving your answers to 3 significant figures
Use the trapezium rule, with 4 intervals, to estimate the value of 𝐼= ln 2𝑥 𝑑𝑥 ln 2𝑥 𝑑𝑥
Find the exact value of I
Calculate the percentage error in using your answer from part (b) to estimate the value of I x 1
1.5 2
2.5 3
1 + ln 2x
2.10
2.61
2.79

25. 𝑎 𝑏 𝑦 𝑑𝑥= 1 2 ℎ 𝑦 0 +2 𝑦 1 + 𝑦 2 + . . . + 𝑦 𝑛−1 + 𝑦 𝑛
Where ℎ= 𝑏 −𝑎 𝑛 and 𝑦 𝑖 =𝑓 𝑎+𝑖ℎ a)
Use the trapezium rule, with 4 intervals, to estimate the value of 𝐼= ln 2𝑥 𝑑𝑥 ln 2𝑥 𝑑𝑥
𝐼 ≈ ×0.5 ×
So I =
I = 4.67 x 1
1.5 2
2.5 3
1 + ln 2x
1.69
2.10
2.39
2.61
2.79

26. I = ln 108 Find the exact value of I 𝐼= 𝑥 3 + 1 3 ln 2𝑥 𝑑𝑥
Use integration by parts to deal with ln 2𝑥 𝑑𝑥
𝑢= ln 2𝑥 𝑑𝑥 ⇒ 𝑑𝑢 𝑑𝑥 = 1 𝑥
𝑑𝑣 𝑑𝑥 = ⟹ 𝑣=𝑥
1 3 ln 2𝑥 𝑑𝑥= 𝑥 ln 2𝑥 − 1 3 𝑥 1 𝑥 𝑑𝑥
= 3𝑙𝑛6 − 𝑙𝑛2 − 𝑥 1 3
𝐼=2+𝑙𝑛 −2
I = ln 108

27. Percentage error = 100 x ln 108 −4.67 ln 108 =0.26%
Calculate the percentage error in using your answer from part (b) to estimate the value of I
Percentage error = 100 x ln 108 − ln =0.26%
Exercise 6H page 110

28. 6.9 You can use integration to find areas and volumes
Area of a region between y = f(x), the x-axis and x = a and x = b is given by
Area = 𝑎 𝑏 𝑦 𝑑𝑥
Volume of revolution formed when y = f(x) is rotated about the x-axis between x + a and x + b is given by:
Volume = 𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Now we use these with all the nice new integrals we have

29. Example
the region R is bounded by the curve y = 4𝑥 − 6 𝑥 , 𝑥>0, the x-axis and the lines with equations x = 2 and x = 4.
This region is rotated through 2π radians about the x-axis.
Find the exact value of the volume of the solid generated.
𝑉= 𝜋 2 4 𝑦 2 𝑑𝑥
𝑦 2 =16 𝑥 2 −48+36 𝑥 −2
So 𝑉= 𝜋 2 4 (16 𝑥 2 −48+36 𝑥 −2 )𝑑𝑥
𝑉= 𝜋 16 𝑥 −48𝑥 −36 𝑥 −
𝑉= 𝜋 −192 −9 − −96 −18
𝑉= 𝜋

30. Exercise 6I page 113

31. When 𝒅𝒚 𝒅𝒙 =𝒇 𝒙 𝒈(𝒚) you can write
6.10 You can use integration to solve differential equations
We can solve first order differential equations by the process called separation of variables
When 𝒅𝒚 𝒅𝒙 =𝒇 𝒙 𝒈(𝒚) you can write
𝟏 𝒈(𝒚) 𝒅𝒚= 𝒇 𝒙 𝒅𝒙
Sometimes boundary conditions may be given in a question which means you can find an particular solution. In this case you take the general solution and use the boundary conditions to find the constant of integration.

32. 1 𝑦(𝑦+1) 𝑑𝑦= 1 𝑐𝑜𝑡𝑥 𝑑𝑥 1 𝑦 − 1 𝑦+1 𝑑𝑦= 𝑡𝑎𝑛𝑥 𝑑𝑥 Example
Obtain the solution of 𝑐𝑜𝑡𝑥 𝑑𝑦 𝑑𝑥 =𝑦 𝑦+1 , −1.3<𝑥<1.3
for which y = 1 when x = 𝜋 3 .
Give your answer in the form y = f(x).
Separate the variables and integrate
1 𝑦(𝑦+1) 𝑑𝑦= 𝑐𝑜𝑡𝑥 𝑑𝑥
Use partial fractions to simplify the integral in y and remember 1 𝑐𝑜𝑡𝑥 =𝑡𝑎𝑛𝑥
1 𝑦 − 1 𝑦+1 𝑑𝑦= 𝑡𝑎𝑛𝑥 𝑑𝑥

33. This gives us
ln y – ln (y + 1) = ln sec x + C
use log laws to combine the y terms and remember + C can be written as ln k
𝑙𝑛 𝑦 𝑦+1 = ln 𝑘𝑠𝑒𝑐 𝑥
getting out of log notation gives us
𝑦 𝑦+1 =𝑘 𝑠𝑒𝑐𝑥
use the given conditions, y = 1 when x = 𝜋 3 , to find k
⇒ 1 2 =2𝑘 so k = 1 4

34. This gives us
4y cos x = y + 1
rearrange to get in the required form
y(4 cos x – 1) = 1
so 𝑦= 1 4 cos 𝑥 −1
Exercise 6J page 116

35. 6.11 Sometimes the differential equation will arise out of a context and the solution may need interpreting in terms of that context
Example
The rate of increase of a population of P organisms at time t is given by 𝑑𝑃 𝑑𝑡 =𝑘𝑃, where k is a positive constant. Given that at t = 0 the population was of size 8, and at t + 1 the population is 56, find the size of the population at time t = 2.

36. start by separating the variables
𝑑𝑃 𝑑𝑡 =𝑘 𝑃
1 𝑃 𝑑𝑃= 𝑘 𝑑𝑡
ln 𝑃 =𝑘𝑡+𝐶 find C using t = 0 P = 8
ln 8 = C
so ln 𝑃 =𝑘𝑡+𝑙𝑛 simplify using log laws
𝑙𝑛 𝑃 8 =𝑘𝑡
use t = 1 and P = 56 to find out that k = ln7
this gives us
𝑙𝑛 𝑃 8 =𝑡 𝑙𝑛7

37. 𝑙𝑛 𝑃 8 =𝑡 𝑙𝑛7
now use t = 2
ln 𝑃 8 =2 ln 7
ln 𝑃 8 = ln 49
⟹ 𝑝 8 =49
so P = 392
Exercise 6K page 117