1. C4 Chapter 6: Integration
Dr J Frost
Last modified: 9th February 2016

2. Overview
You haven’t probably seen integration since C1! You’ll be able to integrate a significantly greater variety of expressions, and be able to solve differential equations (which we encountered earlier in C4).
Integration by ‘reverse chain’ rule
(We imagine what would have differentiated to get the expression.)
Integration by standard result
(There’s certain expressions you’re expected to know straight off.)
sec 2 𝑥 𝑑𝑥= tan 𝑥 +𝐶
sin 3 𝑥 cos 𝑥 𝑑𝑥= sin 4 𝑥 +𝐶
Integration by parts
(Allows us to integrate a product, just as the product rule allowed us to differentiate one)
Integration by substitution
(We make a substitution to hopefully make the expression easier to integrate)
𝑥 cos 𝑥 𝑑𝑥
𝑢=𝑥 𝑑𝑣 𝑑𝑥 = cos 𝑥 𝑑𝑢 𝑑𝑥 = 𝑣= sin 𝑥 𝑥 cos 𝑥 𝑑𝑥=𝑥 sin 𝑥 − sin 𝑥 𝑑𝑥 =𝑥 sin 𝑥 + cos 𝑥
𝑥 2𝑥+5 𝑑𝑥
Let 𝑢=2𝑥 → 𝑥= 𝑢−5 2
𝑑𝑢 𝑑𝑥 =2 → 𝑑𝑥= 1 2 𝑢
𝑥 2𝑥+5 𝑑𝑥= 𝑢− 𝑑𝑢=…

3. Overview Volumes of revolution
(Find the volume of a solid formed by spinning a curve around the 𝑥 axis)
Finding an area using parametric equations
(Find the volume of a solid formed by spinning a curve around the 𝑥 axis)
𝑉=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
𝑎 𝑏 𝑦 𝑑𝑥 = 𝑡 2 𝑡 2 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡
Solving Differential Equations
(Solving here means to find one variable in terms of another without derivatives present)
𝑑𝑉 𝑑𝑡 =−𝑘𝑉 ∫ 1 𝑉 𝑑𝑉= −𝑘 𝑑𝑡 ln 𝑉 =−𝑘𝑡+𝐶 𝑉= 𝑒 −𝑘𝑡+𝐶 =𝐴 𝑒 −𝑘𝑡

4. SKILL #1: Integrating Standard Functions
There’s certain results you should be able to integrate straight off, by just thinking about the opposite of differentiation. 𝒚
∫𝒚 𝒅𝒙
𝑥 𝑛
1 𝑛+1 𝑥 𝑛+1 +𝐶
𝑒 𝑥
𝑒 𝑥 +𝐶
1 𝑥
ln |𝑥| +𝐶
cos 𝑥
sin 𝑥 +𝐶
sin 𝑥
− cos 𝑥 +𝐶
sec 2 𝑥
tan 𝑥 +𝐶
𝑐𝑜𝑠𝑒𝑐 𝑥 cot 𝑥
−𝑐𝑜𝑠𝑒𝑐 𝑥+𝐶
𝑐𝑜𝑠𝑒 𝑐 2 𝑥
− cot 𝑥 +𝐶
sec 𝑥 tan 𝑥
sec 𝑥 +𝐶 ? ?
The |𝑥| has to do with problems when 𝑥 is negative (as ln 𝑥 is not defined in the real domain) ?
Remember my memorisation trick of picturing sin above cos from C3, so that ‘going down’ is differentiating and ‘going up’ is integrating, and we change the sign if the wrong way round. ? ? ?
It’s vital you remember this one. ? ? ?
Have a good stare at this slide before turning your paper over – let’s see how many you remember…

5. Quickfire Questions (without cheating!)
sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 +𝐶 ?
sin 𝑥 𝑑𝑥 =− cos 𝑥 +𝐶 ?
𝑐𝑜𝑠𝑒 𝑐 2 𝑥 𝑑𝑥 =− cot 𝑥 +𝐶 ?
− cos 𝑥 𝑑𝑥 =− sin 𝑥 +𝐶 ?

6. Quickfire Questions (without cheating!)
sec 2 𝑥 𝑑𝑥 = tan 𝑥 +𝐶
𝑐𝑜𝑠𝑒𝑐 𝑥 cot 𝑥 𝑑𝑥 =𝑐𝑜𝑠𝑒𝑐 𝑥+𝐶
1 𝑥 𝑑𝑥 = ln |𝑥| +𝐶
− sin 𝑥 𝑑𝑥 = cos 𝑥 +𝐶

7. Quickfire Questions (without cheating!)
𝑐𝑜𝑠𝑒 𝑐 2 𝑥 𝑑𝑥 =− cot 𝑥 +𝐶 ?
sin 𝑥 𝑑𝑥 =− cos 𝑥 +𝐶 ?
sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 +𝐶 ?
cos 𝑥 𝑑𝑥 = sin 𝑥 +𝐶 ?

8. Test Your Understanding
2 cos 𝑥 + 3 𝑥 − 𝑥 𝑑𝑥 =2 sin 𝑥 +3 ln |𝑥| − 2 3 𝑥 𝐶 ?
cos 𝑥 sin 2 𝑥 𝑑𝑥 = cosec 𝑥 cot 𝑥 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐 𝑥+𝐶 ?
Hint: What ‘reciprocal’ trig functions does this simplify to?

9. Exercise 6A ? ? ? ? ? ? ? ? ? ? 1 Integrate the following.
3 sec 2 𝑥 + 5 𝑥 + 2 𝑥 𝟑 𝐭𝐚𝐧 𝒙+𝟓 𝐥𝐧 𝒙 − 𝟐 𝒙 +𝑪 2 sin 𝑥 − cos 𝑥 +𝑥 −𝟐 𝐜𝐨𝐬 𝒙 −𝟐 𝐬𝐢𝐧 𝒙 + 𝒙 𝟐 +𝑪 5 𝑒 𝑥 +4 cos 𝑥 − 2 𝑥 𝟓 𝒆 𝒙 +𝟒 𝐬𝐢𝐧 𝒙 + 𝟐 𝒙 +𝑪 1 𝑥 + 1 𝑥 𝑥 𝐥𝐧 𝒙 − 𝟏 𝒙 − 𝟏 𝟐 𝒙 𝟐 +𝑪
2𝑐𝑜𝑠𝑒𝑐 𝑥 cot 𝑥 − sec 2 𝑥 −𝟐𝒄𝒐𝒔𝒆𝒄 𝒙− 𝐭𝐚𝐧 𝒙 +𝑪
Calculate the following.
∫ 1 cos 2 𝑥 𝑥 2 𝑑𝑥=𝒕𝒂𝒏 𝒙− 𝟏 𝒙 +𝑪
∫ 1+ cos 𝑥 sin 2 𝑥 + 1+𝑥 𝑥 2 𝑑𝑥=− 𝐜𝐨𝐭 𝒙 −𝒄𝒐𝒔𝒆𝒄 𝒙− 𝟏 𝒙 + 𝐥𝐧 𝒙 +𝑪 𝑠𝑖𝑛𝑥 1+ sec 2 𝑥 𝑑𝑥 =− 𝐜𝐨𝐬 𝒙 + 𝐬𝐞𝐜 𝒙 +𝑪 𝑐𝑜𝑠𝑒 𝑐 2 𝑥 1+ tan 2 𝑥 𝑑𝑥=− 𝐜𝐨𝐭 𝒙 + 𝐭𝐚𝐧 𝒙 +𝑪 sec 2 𝑥 1+ 𝑒 𝑥 cos 2 𝑥 𝑑𝑥 = 𝐭𝐚𝐧 𝒙 + 𝒆 𝒙 +𝑪 ? a c ? ? e ? g i ? 2 ? a ? c ? e ? g ? i

10. SKILL #2: Integrating by Reverse Chain Rule
𝑑 𝑑𝑥 sin 3𝑥+1 =3 cos 3𝑥+1
Therefore:
cos 3𝑥+1 𝑑𝑥 = 𝟏 𝟑 𝐬𝐢 𝐧 𝟑𝒙+𝟏 +𝑪 ? ?
! For any expression where inner function is 𝑎𝑥+𝑏, integrate as before and ÷𝑎. 𝑓 ′ 𝑎𝑥+𝑏 𝑑𝑥 = 1 𝑎 𝑓 𝑎𝑥+𝑏 +𝐶
Quickfire:
𝑒 3𝑥 𝑑𝑥 = 1 3 𝑒 3𝑥 +𝐶
1 5𝑥+2 𝑑𝑥 = 1 5 ln 5𝑥+2 +𝐶
2sec 2 3𝑥−2 𝑑𝑥 = 2 3 tan 3𝑥−2 +𝐶 ?
3𝑥 𝑑𝑥 = 𝑥 𝐶
sin 1−5𝑥 𝑑𝑥 = 1 5 cos 1−5𝑥 +𝐶
1 3 4𝑥−2 2 𝑑𝑥=− 𝑥−2 −1 +𝐶
10𝑥 = 𝑥 𝐶 ? ? ? ? ? ?

11. Check Your Understanding
𝑒 3𝑥+1 𝑑𝑥 = 𝟏 𝟑 𝒆 𝟑𝒙+𝟏 +𝑪
1 1−2𝑥 𝑑𝑥 =− 𝟏 𝟐 𝐥𝐧 𝟏−𝟐𝒙 +𝑪
4−3𝑥 5 𝑑𝑥 =− 𝟏 𝟏𝟖 𝟒−𝟑𝒙 𝟔 +𝑪
sec 3𝑥 tan⁡(3𝑥) 𝑑𝑥 = 𝟏 𝟑 𝐬𝐞𝐜 𝟑𝒙 +𝑪 ? ? ? ?

12. Exercise 6B
Page 86 ?
1 2𝑥+1 𝑑𝑥= 𝟏 𝟐 𝐥𝐧 𝟐𝒙+𝟏 +𝑪 𝑥 𝑑𝑥=− 𝟏 𝟐 𝟐𝒙+𝟏 +𝑪 𝑥 𝑑𝑥= 𝟏 𝟔 𝟐𝒙+𝟏 𝟑 +𝑪 𝑥−1 𝑑𝑥 = 𝟑 𝟒 𝐥𝐧 𝟒𝒙−𝟏 +𝑪
3 1−4𝑥 2 𝑑𝑥= 𝟑 𝟒 𝟏−𝟒𝒙 +𝑪 −2𝑥 3 𝑑𝑥 = 𝟑 𝟒 𝟏−𝟐𝒙 𝟐 +𝑪 −2𝑥 𝑑𝑥 =− 𝟓 𝟐 𝐥𝐧 𝟑−𝟐𝒙 +𝑪
3 sin 2𝑥 𝑥+1 𝑑𝑥 =− 𝟑 𝟐 𝐜𝐨𝐬 𝟐𝒙+𝟏 +𝟐 𝐥𝐧 𝟐𝒙+𝟏 +𝑪
1 sin 2 2𝑥 𝑥 𝑥 2 𝑑𝑥 =− 𝟏 𝟐 𝐜𝐨𝐭 𝟐𝒙 + 𝟏 𝟐 𝐥𝐧 𝟏+𝟐𝒙 − 𝟏 𝟐 𝟏+𝟐𝒙 1 a
sin 2𝑥+1 𝑑𝑥=− 𝟏 𝟐 𝒄𝒐𝒔 𝟐𝒙+𝟏 +𝑪 4 𝑒 𝑥+5 𝑑𝑥=𝟒 𝒆 𝒙+𝟓 +𝑪 𝑐𝑜𝑠𝑒 𝑐 2 3𝑥 𝑑𝑥 =− 𝟏 𝟑 𝐜𝐨𝐭 𝟑𝒙 +𝑪 sec 4𝑥 tan 4𝑥 𝑑𝑥 = 𝟏 𝟒 𝐬𝐞𝐜 𝟒𝒙 +𝑪 3 sin 𝑥+1 𝑑𝑥 =−𝟔 𝐜𝐨𝐬 𝟏 𝟐 𝒙+𝟏 𝑐𝑜𝑠𝑒𝑐 2𝑥 cot 2𝑥 𝑑𝑥=− 𝟏 𝟐 𝒄𝒐𝒔𝒆𝒄 𝟐𝒙+𝑪
𝑒 2𝑥 − 1 2 sin 2𝑥−1 𝑑𝑥 = 𝟏 𝟐 𝒆 𝟐𝒙 + 𝟏 𝟒 𝐜𝐨𝐬 𝟐𝒙−𝟏 +𝑪 𝑒 𝑥 𝑑𝑥= 𝟏 𝟐 𝒆 𝟐𝒙 +𝟐 𝒆 𝒙 +𝒙+𝑪
sec 2 2𝑥 1+ sin 2𝑥 𝑑𝑥= 𝟏 𝟐 𝐭𝐚𝐧 𝟐𝒙 + 𝟏 𝟐 𝐬𝐞𝐜 𝟐𝒙 −2 cos 𝑥 sin 𝑥 𝑑𝑥=−𝟔 𝐜𝐨𝐭 𝟏 𝟐 𝒙 +𝟒𝒄𝒐𝒔𝒆𝒄 𝟏 𝟐 𝒙
𝑒 3−𝑥 + sin 3−𝑥 + cos 3−𝑥 𝑑𝑥 =− 𝒆 𝟑−𝒙 + 𝐜𝐨𝐬 𝟑−𝒙 − 𝐬𝐢𝐧 𝟑−𝒙 +𝑪 3 a ? c ? ? b e ? ? c f ? ? g ? d ? h ? f ? h 2 a ? ? j ? b ? 4 a c ? ? d c e ? ?

13. SKILL #3: Integrating using Trig Identities
Some expressions, such as sin 2 𝑥 and sin 𝑥 cos 𝑥 can’t be integrated directly, but we can use one of our trig identities to replace it with an expression we can easily integrate. Q
Find sin 2 𝑥 𝑑𝑥 Q
Find sin 3𝑥 cos 3𝑥 𝑑𝑥 ? ?
Do you know a trig identity involving sin 2 𝑥 ?
cos 2𝑥 =1−2 sin 2 𝑥 sin 2 𝑥 = 1 2 − 1 2 cos 2𝑥 sin 2 𝑥 𝑑𝑥 = 1 2 𝑥− 1 4 sin 2𝑥
sin 6𝑥≡2 sin 3𝑥 cos 3𝑥 sin 3𝑥 cos 3𝑥 = 1 2 sin 6𝑥
sin 3𝑥 cos 3𝑥 𝑑𝑥 =− cos 6𝑥 +𝐶 Q
Find sin 3𝑥 cos 2𝑥 𝑑𝑥 Q
Find cos 2 𝑥 𝑑𝑥
From formula booklet:
sin 𝐴 + sin 𝐵 =2 sin 𝐴+𝐵 2 cos 𝐴−𝐵 𝐴+𝐵=6𝑥, 𝐴−𝐵=4𝑥 𝐴=5𝑥, 𝐵=𝑥 sin 3𝑥 cos 2𝑥 = sin 5𝑥 + sin 𝑥 sin 3𝑥 cos 2𝑥 𝑑𝑥 =− cos 5𝑥 − 1 2 cos 𝑥 +𝐶 ? ?
cos 2𝑥 =2 cos 2 𝑥 −1 cos 2 𝑥 = cos 2𝑥 cos 2 𝑥 𝑑𝑥 = 1 2 𝑥 sin 2𝑥

14. Check Your UnderstandingQ
Find tan 2 𝑥 𝑑𝑥
Hint: What identity do you know involving tan 2 𝑥 ? ?
Recall that 1+ tan 2 𝑥 ≡ sec 2 𝑥
tan 2 𝑥 𝑑𝑥 = sec 2 𝑥 −1 𝑑𝑥 = tan 𝑥 −𝑥+𝐶 Q
Find sec 𝑥 + tan 𝑥 2 𝑑𝑥 Q
Find 4 cos 7𝑥 cos 3𝑥 𝑑𝑥 ? ?
sec 𝑥 + tan 𝑥 2 𝑑𝑥 = sec 2 𝑥 +2 sec 𝑥 tan 𝑥 + tan 2 𝑥 𝑑𝑥 = tan 𝑥 +2 sec 𝑥 + tan 𝑥 −𝑥+𝐶 =2 tan 𝑥 −2 sec 𝑥 −𝑥+𝐶
From formula booklet:
cos 𝐴 + cos 𝐵 =2 cos 𝐴+𝐵 2 cos 𝐴−𝐵 𝐴+𝐵=14𝑥 𝐴−𝐵=6𝑥 𝐴=10𝑥, 𝐵=4𝑥
4 cos 7𝑥 cos 3𝑥 𝑑𝑥 = 2( cos 10𝑥 + cos 4𝑥 ) 𝑑𝑥 = 1 5 sin 10𝑥 cos 4𝑥
Bro Tip: The quick way is this.
𝐴 is the sum of the two nums.
𝐵 is the difference.

15. Exercise 6C
Page 88 1
cot 2 𝑥 𝑑𝑥 =− cot 𝑥 −𝑥+𝐶
cos 2 𝑥 𝑑𝑥 = 1 2 𝑥 sin 2𝑥 +𝐶
sin 2𝑥 cos 2𝑥 𝑑𝑥 =− 1 8 cos 4𝑥 +𝐶
1+ sin 𝑥 2 𝑑𝑥 = 3 2 𝑥−2 cos 𝑥 − 1 4 sin 2𝑥 +𝐶
tan 2 3𝑥 𝑑𝑥 = 1 3 tan 3𝑥 −𝑥+𝐶
cot 𝑥 −𝑐𝑜𝑠𝑒𝑐 𝑥 2 𝑑𝑥 =−2 cot 𝑥 −𝑥+2 cosec 𝑥 +𝐶
sin 𝑥 + cos 𝑥 2 𝑑𝑥 =𝑥− 1 2 cos 2𝑥 +𝐶
sin 2 𝑥 cos 2 𝑥 𝑑𝑥 = 1 8 𝑥− sin 4𝑥 +𝐶
1 sin 2 𝑥 cos 2 𝑥 𝑑𝑥=−2 cot 2𝑥 +𝐶 ?
1− sin 𝑥 cos 2 𝑥 𝑑𝑥 = tan 𝑥 − sec 𝑥 +𝐶 cos 2𝑥 cos 2 2𝑥 𝑑𝑥 =2𝑥− tan 𝑥 +𝐶 cos 𝑥 sin 2 𝑥 𝑑𝑥=−2 cot 𝑥 −𝑥−2𝑐𝑜𝑠𝑒𝑐 𝑥+𝐶 cot 𝑥 − tan 𝑥 2 𝑑𝑥 =− cot 𝑥 −4𝑥+ tan 𝑥 +𝐶 cos 𝑥 − sec 𝑥 2 𝑑𝑥 =− 3 2 𝑥 sin 2𝑥 + tan 𝑥 +𝐶
cos 2𝑥 cos 𝑥 𝑑𝑥 = 𝟏 𝟔 𝐬𝐢𝐧 𝟑𝒙 + 𝟏 𝟐 𝐬𝐢𝐧 𝒙
2 sin 3𝑥 cos 5𝑥 𝑑𝑥 =− 1 8 cos 8𝑥 cos 2𝑥 cos 3𝑥 cos 7𝑥 𝑑𝑥 = 1 5 sin 10𝑥 sin 4𝑥
2 cos 4𝑥 sin 4𝑥 𝑑𝑥=− 1 8 cos 8𝑥 +𝐶 a ? 2 a ? ? b c ? ? c e ? ? g d i e ? ? ? f ? ? g 3 a ? c h ? ? i e ? ? g

16. SKILL #4: Using Partial Fractions
We saw earlier that we can split some expressions into partial fractions. This allows us to differentiate some expressions with more complicated denominators.
Find 𝑥 2 −1 𝑑𝑥 ?
2 𝑥 2 −1 = 2 𝑥+1 𝑥−1 = 𝐴 𝑥+1 + 𝐵 𝑥−1 2=𝐴 𝑥−1 +𝐵 𝑥+1 2=𝐴𝑥−𝐴+𝐵𝑥+𝐵 𝐴+𝐵= 𝐵−𝐴=2 𝐵=1 𝐴=−1
∫ 2 𝑥 2 −1 𝑑𝑥= − 1 𝑥 𝑥−1 𝑑𝑥 =− 𝐥𝐧 𝒙+𝟏 + 𝐥𝐧 𝒙−𝟏 +𝑪 =𝒍𝒏 𝒙−𝟏 𝒙+𝟏 +𝑪

17. Test Your Understanding
Find 𝑥−5 𝑥+1 𝑥−2 𝑑𝑥
Find 𝑥 2 −19𝑥+1 2𝑥+1 𝑥− 𝑑𝑥 ? ?
8 𝑥 2 −19𝑥+1 2𝑥+1 𝑥−2 2 = 𝐴 2𝑥+1 + 𝐵 𝑥− 𝐶 𝑥−2
𝐴=2, 𝐵=−1, 𝐶=3
8 𝑥 2 −19𝑥+1 2𝑥+1 𝑥− 𝑑𝑥 = 𝐥𝐧 𝟐𝒙+𝟏 𝒙−𝟐 𝟑 + 𝟏 𝒙−𝟐 +𝑪
𝑥−5 𝑥+1 𝑥−2 ≡ 𝐴 𝑥+1 + 𝐵 𝑥−2 𝐴=2 𝑎𝑛𝑑 𝐵=−1
𝑥−5 𝑥+1 𝑥−2 𝑑𝑥 =∫ 2 𝑥+1 − 1 𝑥−2 𝑑𝑥 =2 ln 𝑥+1 − ln 𝑥−2 +𝐶 = 𝐥𝐧 𝒙+𝟏 𝟐 𝒙−𝟐 +𝑪

18. Exercise 6D
Page 92
Use partial fractions to integrate the following.
3𝑥+5 𝑥+1 𝑥+2 𝑑𝑥 = 𝐥𝐧 𝒙+𝟏 𝟐 𝒙+𝟐 +𝑪 2𝑥−6 𝑥+3 𝑥−1 𝑑𝑥 = 𝐥𝐧 𝒙+𝟑 𝟑 𝒙−𝟏 +𝑪 𝑥+1 1−2𝑥 𝑑𝑥 = 𝐥𝐧 𝟐𝒙+𝟏 𝟏−𝟐𝒙 +𝑪 3−5𝑥 1−𝑥 2−3𝑥 𝑑𝑥 = 𝐥𝐧 𝟐−𝟑𝒙 𝟏 𝟑 𝟏−𝒙 𝟐 +𝑪 𝑥 𝑥+2 𝑥 𝑑𝑥 = 𝐥𝐧 𝒙+𝟏 𝒙+𝟐 − 𝟐 𝒙+𝟏 +𝑪
2 𝑥 2 +3𝑥−1 𝑥+1 2𝑥−1 𝑑𝑥 =𝒙+ 𝐥𝐧 𝒙+𝟏 𝟐 𝟐𝒙−𝟏 +𝑪
𝑥 2 𝑥 2 −4 𝑑𝑥 =𝒙+ 𝐥𝐧 𝒙−𝟐 𝒙+𝟐 +𝑪
6+3𝑥− 𝑥 2 𝑥 3 +2 𝑥 2 𝑑𝑥=− 𝟑 𝒙 − 𝐥𝐧 𝒙+𝟐 +𝑪 ? 1 a ? c ? e ? g ? i 2 a ? c ? e ?

19. SKILL #5: ‘Consider and Scale’
There’s certain more complicated expressions which look like the result of having applied the chain rule. The process is then simply:
Consider some expression that will differentiate to something similar to it.
Differentiate, and adjust for any scale difference.
𝑥 𝑥 𝑑𝑥
cos 𝑥 sin 2 𝑥 𝑑𝑥
2𝑥 𝑥 𝑑𝑥
The first 𝑥 looks like it arose from differentiating the 𝑥 2 inside the brackets.
Consider 𝒚= 𝒙 𝟐 +𝟓 𝟒
𝒅𝒚 𝒅𝒙 =𝟒 𝒙 𝟐 +𝟓 𝟑 ×𝟐𝒙 =𝟖𝒙 𝒙 𝟐 +𝟓 𝟑
Thus 𝒙 𝒙 𝟐 +𝟓 𝟑 𝒅𝒙 = 𝟏 𝟖 𝒙 𝟐 +𝟓 𝟒 +𝑪
The cos 𝑥 probably arose from differentiating the 𝑠𝑖𝑛.
Consider 𝒚= 𝐬𝐢𝐧 𝟑 𝒙
𝒅𝒚 𝒅𝒙 =𝟑 𝐬𝐢𝐧 𝟐 𝒙 𝐜𝐨𝐬 𝒙
Thus 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝟐 𝒙 𝒅𝒙 = 𝟏 𝟑 𝐬𝐢𝐧 𝟑 𝒙+𝑪
The 2𝑥 probably arose from differentiating the 𝑥 2 .
Consider 𝒚=𝒍𝒏| 𝒙 𝟐 +𝟏|
𝒅𝒚 𝒅𝒙 = 𝟐𝒙 𝒙 𝟐 +𝟏
Thus 𝟐𝒙 𝒙 𝟐 +𝟏 𝒅𝒙 =𝒍𝒏 𝒙 𝟐 +𝟏 +𝑪 ? ? ?

20. SKILL #5: ‘Consider and Scale’
! Integration by Inspection: Use common sense to consider some expression that would differentiate to the expression given. Then scale appropriately.
Common patterns:
𝑘 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥 → 𝑇𝑟𝑦 ln 𝑓 𝑥 ∫𝑘 𝑓 ′ 𝑥 𝑓 𝑥 𝑛 → 𝑇𝑟𝑦 𝑓 𝑥 𝑛+1
In words: “If the bottom of a fraction differentiates to give the top (forgetting scaling), try ln of the bottom”.
𝑥 2 𝑥 𝑑𝑥
𝑥 𝑒 𝑥 𝑑𝑥 ? ?
Consider 𝑦=ln⁡| 𝑥 3 +1|
𝑑𝑦 𝑑𝑥 = 3 𝑥 2 𝑥 3 +1
Therefore:
𝑥 2 𝑥 𝑑𝑥 = 1 3 𝑙𝑛 𝑥 𝐶
Consider 𝑦= 𝑒 𝑥 2 +1
𝑑𝑦 𝑑𝑥 =2𝑥 𝑒 𝑥 2 +1
Therefore:
𝑥 𝑒 𝑥 2 +1 𝑑𝑥 = 1 2 𝑒 𝑥 𝐶

21. Quickfire
In your head!
4 𝑥 3 𝑥 4 −1 𝑑𝑥 = 𝐥𝐧 𝒙 𝟒 −𝟏 +𝑪 cos 𝑥 sin 𝑥 +2 𝑑𝑥 = 𝒍𝒏 𝐬𝐢𝐧 𝒙 +𝟐 +𝑪 cos 𝑥 𝑒 sin 𝑥 𝑑𝑥= 𝒆 𝐬𝐢𝐧 𝒙 +𝑪 cos 𝑥 sin 𝑥 −5 7 𝑑𝑥 = 𝟏 𝟖 𝐬𝐢𝐧 𝒙 −𝟓 𝟖 𝑥 2 𝑥 = 𝟏 𝟐𝟒 𝒙 𝟑 +𝟓 𝟖 +𝑪 ? ? ? ? ?
Not in your head…
Bro Tip: If there’s as power around the whole denominator, DON’T use 𝑙𝑛: reexpress the expression as a product.
e.g. 𝑥 𝑥 −3
𝑥 𝑥 𝑑𝑥 =− 𝟏 𝟒 𝒙 𝟐 +𝟓 −𝟐 +𝑪 ?

22. Test Your Understanding
sin 𝑥 cos 𝑥 𝑑𝑥
𝑐𝑜𝑠𝑒 𝑐 2 𝑥 2+ cot 𝑥 3 𝑑𝑥 ? ?
Consider 𝑦= cos 𝑥 +1 6
𝑑𝑦 𝑑𝑥 =−6 sin 𝑥 cos 𝑥 +1 5
Therefore:
sin 𝑥 cos 𝑥 𝑑𝑥 =− 𝟏 𝟔 𝐜𝐨𝐬 𝒙 +𝟏 𝟔 +𝑪
Consider 𝑦= 2+ cot 𝑥 −2
𝑑𝑦 𝑑𝑥 =−2 2+ cot 𝑥 −3 ×−𝑐𝑜𝑠𝑒 𝑐 2 𝑥 =2𝑐𝑜𝑠𝑒 𝑐 2 𝑥 2+ cot 𝑥 −3
Thus 𝑐𝑜𝑠𝑒 𝑐 2 𝑥 2+ cot 𝑥 3 𝑑𝑥 = 𝟏 𝟐 𝟐+ 𝐜𝐨𝐭 𝒙 −𝟐 +𝑪
sec 2 2𝑥 tan 2𝑥 +1 𝑑𝑥
5 tan 𝑥 sec 2 𝑥 𝑑𝑥
Note that 𝑑 𝑑𝑥 sec 𝑥 = sec 𝑥 tan 𝑥 , so the power of sec 𝑥 stays the same.
Try 𝑦= sec 2 𝑥
Then 𝑑𝑦 𝑑𝑥 =2 sec 𝑥 × sec 𝑥 tan 𝑥 =2 sec 2 𝑥 tan 𝑥
Thus 5 tan 𝑥 sec 2 𝑥 𝑑𝑥 = sec 2 𝑥 +𝐶 ? ?
= 1 2 ln | tan 2𝑥 +1| +𝐶

23. Exercise 6E
Page 94
𝑥+1 𝑥 2 +2𝑥+3 4 𝑑𝑥 = = 𝟏 𝟏𝟎 𝒙 𝟐 +𝟐𝒙+𝟑 𝟓 𝑐𝑜𝑠𝑒 𝑐 2 2𝑥 cot 2𝑥 𝑑𝑥 =− 𝟏 𝟒 𝐜𝐨𝐭 𝟐 𝟐𝒙 +𝑪 sin 5 3𝑥 cos 3𝑥 𝑑𝑥 = 𝟏 𝟏𝟖 𝐬𝐢𝐧 𝟔 𝟑𝒙 +𝑪
cos 𝑥 𝑒 sin 𝑥 𝑑𝑥 = 𝒆 𝐬𝐢𝐧 𝒙 +𝑪
𝑥 𝑥 𝑑𝑥= 𝟏 𝟓 𝒙 𝟐 +𝟏 𝟓 𝟐 +𝑪
sin 𝑥 cos 𝑥 cos 2𝑥 +3 𝑑𝑥 =− 𝟏 𝟒 𝐥𝐧 𝐜𝐨𝐬 𝟐𝒙 +𝟑 +𝑪
𝑥 𝑥 2 +4 𝑑𝑥 = 𝟏 𝟐 𝐥𝐧 𝒙 𝟐 +𝟒 +𝑪
𝑒 2𝑥 𝑒 2𝑥 +1 𝑑𝑥 = 𝟏 𝟐 𝐥𝐧 𝒆 𝟐𝒙 +𝟏 +𝑪
𝑥 𝑥 𝑑𝑥 = − 𝟏 𝟒 𝒙 𝟐 +𝟒 −𝟐 +𝑪
𝑒 2𝑥 𝑒 2𝑥 𝑑𝑥 =− 𝟏 𝟒 𝒆 𝟐𝒙 +𝟏 −𝟐 +𝑪
cos 2𝑥 3+ sin 2𝑥 𝑑𝑥 = 𝟏 𝟐 𝐥𝐧 𝟑+ 𝐬𝐢𝐧 𝟐𝒙 +𝑪
𝑥 𝑒 𝑥 2 𝑑𝑥 = 𝟏 𝟐 𝒆 𝒙 𝟐 +𝑪
cos 2𝑥 1+ sin 2𝑥 4 𝑑𝑥= 𝟏 𝟏𝟎 𝟏+ 𝐬𝐢𝐧 𝟐𝒙 𝟓 +𝑪
∫ sec 2 𝑥 tan 2 𝑥 𝑑𝑥= 𝟏 𝟑 𝐭𝐚𝐧 𝟑 𝒙 +𝑪 ? 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ?

24. SKILL #6: Integration by Substitution
For some integrations involving a complicated expression, we can make a substitution to turn it into an equivalent integration that is simpler. We wouldn’t be able to use ‘reverse chain rule’ on the following: Q
Use the substitution 𝑢=2𝑥+5 to find 𝑥 2𝑥+5 𝑑𝑥
The aim is to completely remove any reference to 𝑥, and replace it with 𝑢. We’ll have to work out 𝑥 and 𝑑𝑥 so that we can replace them.
STEP 1: Using substitution, work out 𝑥 and 𝑑𝑥 (or variant) ?
𝑑𝑢 𝑑𝑥 =2 → 𝑑𝑥= 1 2 𝑑𝑢
Bro Tip: Be careful about ensuring you reciprocate when rearranging.
𝑥= 𝑢−5 2
𝑥 2𝑥+5 𝑑𝑥= 𝑢−5 2 𝑢 1 2 𝑑𝑢 = 𝑢 𝑢−5 𝑑𝑢 = 𝑢 −5 𝑢 1 2
= 𝑢 − 𝑢 𝐶
= 2𝑥 − 5 2𝑥 𝐶 ?
STEP 2: Substitute these into expression.
Bro Tip: If you have a constant factor, factor it out of the integral.
STEP 3: Integrate simplified expression. ?
STEP 4: Write answer in terms of 𝑥. ?

25. How can we tell what substitution to use?
In Edexcel you will usually be given the substitution!
However in some other exam boards, and in STEP, you often aren’t.
There’s no hard and fast rule, but it’s often helpful to replace to replace expressions inside roots, powers or the denominator of a fraction.
Sensible substitution:
cos 𝑥 1+ sin 𝑥 𝑑𝑥 ?
𝒖=𝟏+ 𝐬𝐢𝐧 𝒙
6𝑥 𝑒 𝑥 2 𝑑𝑥 ?
But this can be integrated by inspection.
𝒖= 𝒙 𝟐
𝑥 𝑒 𝑥 1+𝑥 𝑑𝑥 ?
𝒖=𝟏+𝒙
𝑒 1−𝑥 1+𝑥 𝑑𝑥
𝒖= 𝟏−𝒙 𝟏+𝒙 ?

26. Another Example Q
Use the substitution 𝑢= sin 𝑥 +1 to find cos 𝑥 sin 𝑥 1+ sin 𝑥 𝑑𝑥
STEP 1: Using substitution, work out 𝑥 and 𝑑𝑥 (or variant) ?
𝑢= sin 𝑥 +1 𝑑𝑢 𝑑𝑥 = cos 𝑥 → 𝑑𝑢= cos 𝑥 𝑑𝑥 sin 𝑥 =𝑢−1
Notice this time we didn’t find 𝒙 or 𝒅𝒙. We could, but then 𝒙= 𝐚𝐫𝐜𝐬𝐢𝐧 (𝒖−𝟏) , and it would be slightly awkward simplifying the expression (although is still very much a valid method!)
STEP 2: Substitute these into expression. ?
= 𝑢−1 𝑢 3 𝑑𝑢 = 𝑢 4 − 𝑢 3 𝑑𝑢 = 1 5 𝑢 5 − 1 4 𝑢 4 +𝐶
= sin 𝑥 − sin 𝑥 𝐶
STEP 3: Integrate simplified expression. ?
STEP 4: Write answer in terms of 𝑥. ?

27. Using substitutions involving implicit differentiation
When a root is involved, it makes thing much tidier if we use 𝑢 2 =… Q
Use the substitution 𝑢 2 =2𝑥+5 to find 𝑥 2𝑥+5 𝑑𝑥
STEP 1: Using substitution, work out 𝑥 and 𝑑𝑥 (or variant) ?
2𝑢 𝑑𝑢 𝑑𝑥 =2 → 𝑑𝑥=𝑢 𝑑𝑢
𝑥= 𝑢 2 −5 2
𝑥 2𝑥+5 𝑑𝑥= 𝑢 2 −5 2 𝑢×𝑢𝑑𝑢 = 𝑢 4 − 5 2 𝑢 2 𝑑𝑢 = 𝑢 5 − 5 6 𝑢 3 +𝐶
= 2𝑥 − 5 2𝑥 𝐶 ?
STEP 2: Substitute these into expression.
STEP 3: Integrate simplified expression. ?
STEP 4: Write answer in terms of 𝑥. ?
This was marginally less tedious than when we used 𝑢=2𝑥+5, as we didn’t have fractional powers to deal with.

28. Test Your Understanding
Edexcel C4 Jan 2012 Q6c
Hint: You might want to use your double angle formula first. ?
𝑑𝑢 𝑑𝑥 =− sin 𝑥 → 𝑑𝑥=− 1 sin 𝑥 𝑑𝑢
cos 𝑥 =𝑢−1 (As before 𝑥= arccos 𝑢−1 is going to be messy)
2 sin 2𝑥 1+ cos 𝑥 𝑑𝑥 = 4 sin 𝑥 cos 𝑥 1+ cos 𝑥 𝑑𝑥 = − 4 sin 𝑥 𝑢−1 𝑢 1 sin 𝑥 𝑑𝑢 =−4 𝑢−1 𝑢 𝑑𝑢 =−4 1− 1 𝑢 𝑑𝑢 =−4 𝑢− ln 𝑢 +𝑐=−4 1+ cos 𝑥 − ln cos 𝑥 𝑐=…

29. Definite Integration ? ? ? ? Now consider:Q
Calculate 0 𝜋 2 cos 𝑥 1+ sin 𝑥 𝑑𝑥 ?
Use substitution: 𝒖=𝟏+ 𝐬𝐢𝐧 𝒙
𝑑𝑢 𝑑𝑥 = cos 𝑥 → 𝑑𝑢= cos 𝑥 𝑑𝑥
sin 𝑥 =𝑢−1
Now because we’ve changed from 𝒙 to 𝒖, we have to work out what values of 𝒖 would have given those limits for 𝒙:
When 𝑥= 𝜋 2 , 𝑢=2 When 𝑥=0, 𝑢=1
0 𝜋 2 cos 𝑥 1+ sin 𝑥 𝑑𝑥= 1 2 𝑢 𝑑𝑢 = 𝑢 = −1 ? ? ?

30. Test Your Understanding
Edexcel C4 June 2011 Q4 ?
𝒅𝒖 𝒅𝒙 =𝟐𝒙 → 𝒅𝒖=𝟐𝒙 𝒅𝒙 𝒙 𝟐 =𝒖−𝟐
When 𝒙= 𝟐 , 𝒖= 𝟐 𝟐 +𝟐=𝟒
When 𝒙=𝟎, 𝒖= 𝟎 𝟐 +𝟐=𝟐
𝟎 𝟐 𝒙 𝟐 𝐥𝐧 𝒙 𝟐 +𝟐 𝒙 𝒅𝒙= 𝟐 𝟒 𝟏 𝟐 (𝒖−𝟐) 𝐥𝐧 𝒖 𝒅𝒖

31. Exercise 6F ? ? ? ? ? ? ? ? ? Use the given substitution to integrate.
Use an appropriate substitution.
𝑥 1+𝑥 𝑑𝑥 ; 𝑢=1+𝑥 = 𝟐 𝟓 𝟏+𝒙 𝟓 𝟐 − 𝟐 𝟑 𝟏+𝒙 𝟑 𝟐 +𝑪 sin 𝑥 cos 𝑥 𝑑𝑥;𝑢= sin 𝑥 =− 𝐥𝐧 𝟏− 𝐬𝐢𝐧 𝒙 +𝑪 sin 3 𝑥 𝑑𝑥 ; 𝑢= cos 𝑥 = 𝟏 𝟑 𝐜𝐨𝐬 𝟑 𝒙 − 𝐜𝐨𝐬 𝒙 +𝑪
𝑥 2+𝑥 𝑑𝑥 ; 𝑢 2 =2+𝑥
= 𝟐 𝟓 𝟐+𝒙 𝟓 𝟐 − 𝟒 𝟑 𝟐+𝒙 𝟑 𝟐 +𝑪
sec 2 𝑥 tan 𝑥 1+ tan 𝑥 𝑑𝑥 ; 𝑢 2 =1+ tan 𝑥
= 𝟐 𝟓 𝟏+ 𝐭𝐚𝐧 𝒙 𝟓 𝟐 − 𝟐 𝟑 𝟏+ 𝐭𝐚𝐧 𝒙 𝟑 𝟐 +𝑪
sec 4 𝑥 𝑑𝑥 ; 𝑢= tan 𝑥
= 𝐭𝐚𝐧 𝒙 + 𝟏 𝟑 𝐭𝐚𝐧 𝟑 𝒙 +𝑪 1 a
0 5 𝑥 𝑥+4 𝑑𝑥 = 𝟓𝟎𝟔 𝟏𝟑
𝑥−1 𝑑𝑥 =𝟐+𝟐 𝐥𝐧 𝟐 𝟑
0 1 𝑥 2+𝑥 3 𝑑𝑥 =𝟗.𝟕 ? 3 a ? c c ? ? e ? e ? 2 a ? c ? e ?

32. SKILL #7: Integration by Parts
𝑥 cos 𝑥 𝑑𝑥 =?
Just as the Product Rule was used to differentiate the product of two expressions, we can often use ‘Integration by Parts’ to integrate a product.
! To integrate by parts:
𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥=𝑢𝑣− 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥
Proof ?
(not needed for exam)
The Product Rule:
𝑑 𝑑𝑥 𝑢𝑣 =𝑣 𝑑𝑢 𝑑𝑥 +𝑢 𝑑𝑣 𝑑𝑥
On the right-hand-side, both 𝑣 𝑑𝑢 𝑑𝑥 and 𝑢 𝑑𝑣 𝑑𝑥 are the product of two expressions. So if we made either the subject, we could use 𝑢 𝑑𝑣 𝑑𝑥 say to represent 𝑥 cos 𝑥 in the example.
Rearranging:
𝑢 𝑑𝑣 𝑑𝑥 = 𝑑 𝑑𝑥 𝑢𝑣 −𝑣 𝑑𝑢 𝑑𝑥
Integrating both sides with respect to 𝑥, we get the desired formula.

33. SKILL #7: Integration by Parts
𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥=𝑢𝑣− 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥
𝑥 cos 𝑥 𝑑𝑥 =? ?
𝑢=𝑥 𝑑𝑣 𝑑𝑥 = cos 𝑥
𝑑𝑢 𝑑𝑥 = 𝑣= sin 𝑥
𝑥 cos 𝑥 𝑑𝑥 =𝑥 sin 𝑥 − 1 sin 𝑥 𝑑𝑥 =𝑥 sin 𝑥 + cos 𝑥 +𝐶
STEP 1: Decide which thing will be 𝑢 (and which 𝑑𝑣 𝑑𝑥 ).
You’re about to differentiate 𝑢 and integrate 𝑑𝑣 𝑑𝑥 , so the idea is to pick them so differentiating 𝑢 makes it ‘simpler’, and 𝑑𝑣 𝑑𝑥 can be integrated easily.
𝑢 will always be the 𝑥 𝑛 term UNLESS one term is ln 𝑥 . ?
STEP 2: Find 𝑑𝑢 𝑑𝑥 and 𝑣.
STEP 3: Use the formula. ?
I just remember it as “𝑢𝑣 minus the integral of the two new things timesed together”

34. Another Example ? Find 𝑥 2 ln 𝑥 𝑑𝑥 Q 𝑢= ln 𝑥 𝑑𝑣 𝑑𝑥 = 𝑥 2
𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑣= 1 3 𝑥 3
𝑥 2 ln 𝑥 𝑑𝑥 = 1 3 𝑥 3 ln 𝑥 − 𝑥 2 𝑑𝑥 = 𝟏 𝟑 𝒙 𝟑 𝐥𝐧 𝒙 − 𝟏 𝟗 𝒙 𝟑 +𝑪
This time, we choose ln 𝑥 to be the 𝑢 because it differentiates nicely.
STEP 1: Decide which thing will be 𝑢 (and which 𝑑𝑣 𝑑𝑥 ).
STEP 2: Find 𝑑𝑢 𝑑𝑥 and 𝑣.
STEP 3: Use the formula.

35. IBP twice!  Q
Find 𝑥 2 𝑒 𝑥 𝑑𝑥 ?
𝑢= 𝑥 𝑑𝑣 𝑑𝑥 = 𝑒 𝑥 𝑑𝑢 𝑑𝑥 =2𝑥 𝑣= 𝑒 𝑥 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2𝑥 𝑒 𝑥 𝑑𝑥
We have to apply IBP again!
𝑢=2𝑥 𝑑𝑣 𝑑𝑥 = 𝑒 𝑥 𝑑𝑢 𝑑𝑥 = 𝑣= 𝑒 𝑥
2𝑥 𝑒 𝑥 𝑑𝑥 =2𝑥 𝑒 𝑥 − 2 𝑒 𝑥 𝑑𝑥 =2𝑥 𝑒 𝑥 −2 𝑒 𝑥
Therefore
𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2𝑥 𝑒 𝑥 −2𝑥 𝑒 𝑥 +𝐶 = 𝑥 2 𝑒 𝑥 −2𝑥 𝑒 𝑥 +2 𝑒 𝑥 +𝐶
Bro Tip: I tend to write out my working for any second integral completely separately, and then put the result back into the original integral later.

36. Test Your UnderstandingQ
Find 𝑥 2 sin 𝑥 𝑑𝑥 ?
=2𝑥 sin 𝑥 − 𝑥 2 cos 𝑥 +2 cos 𝑥 +𝐶

37. Integrating ln 𝑥 and definite integrationQ
Find ln 𝑥 𝑑𝑥 , leaving your answer in terms of natural logarithms.
𝑢= ln 𝑥 𝑑𝑣 𝑑𝑥 =1 𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑣=𝑥 ln 𝑥 𝑑𝑥 =𝑥 ln 𝑥 − 1 𝑑𝑥 =𝑥 ln 𝑥 −𝑥+𝐶 ? Q
Find ln 𝑥 𝑑𝑥 , leaving your answer in terms of natural logarithms.
1 2 ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 −𝑥 1 2 =𝟐 𝐥𝐧 𝟐 −𝟏
If we were doing it from scratch:
𝑢= ln 𝑥 𝑑𝑣 𝑑𝑥 =1 𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑣=𝑥 ln 𝑥 𝑑𝑥 = 𝒙 𝒍𝒏 𝒙 𝟏 𝟐 − 𝑑𝑥 =2 ln 2 −1 ln 1 − 𝑥 =2 ln 2 −(2−1) =2 ln 2 −1 ?
In general:
𝑎 𝑏 𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥 = 𝑢𝑣 𝑎 𝑏 − 𝑎 𝑏 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥

38. Test Your UnderstandingQ
Find 0 𝜋 2 𝑥 sin 𝑥 𝑑𝑥 ?
𝑢=𝑥 𝑑𝑣 𝑑𝑥 = sin 𝑥 𝑑𝑢 𝑑𝑥 = 𝑣=− cos 𝑥 𝑥 sin 𝑥 𝑑𝑥 =−𝑥 cos 𝑥 − − cos 𝑥 𝑑𝑥 =−𝑥 cos 𝑥 + cos 𝑥 𝑑𝑥 =−𝑥 cos 𝑥 + sin 𝑥 𝑑𝑥 ∴ 0 𝜋 2 𝑥 sin 𝑥 𝑑𝑥 = −𝑥 cos 𝑥 + sin 𝑥 0 𝜋 2 = − 𝜋 2 cos 𝜋 2 + sin 𝜋 2 − 0+ sin 0 =1

39. Exercise 6G ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 3 a a c c e 4 a e c 2 e a g
𝑥 2 𝑒 −𝑥 𝑑𝑥=− 𝒆 −𝒙 𝒙 𝟐 −𝟐𝒙 𝒆 −𝒙 −𝟐 𝒆 −𝒙 +𝑪
12 𝑥 𝑥 5 𝑑𝑥 = 𝒙 𝟐 𝟑+𝟐𝒙 𝟔 − 𝟏 𝟕 𝒙 𝟑+𝟐𝒙 𝟕 + 𝟏 𝟏𝟏𝟐 𝟑+𝟐𝒙 𝟖 +𝑪
𝑥 2 2 sec 2 𝑥 tan 𝑥 𝑑𝑥 = 𝒙 𝟐 𝐬𝐞𝐜 𝟐 𝒙 −𝟐𝒙 𝐭𝐚𝐧 𝒙 +𝟐 𝐥𝐧 𝐬𝐞𝐜 𝒙 +𝑪
0 ln 2 𝑥 𝑒 2𝑥 𝑑𝑥=𝟐 𝐥𝐧 𝟐 − 𝟑 𝟒
0 𝜋 2 𝑥 cos 𝑥 𝑑𝑥 = 𝝅 𝟐 −𝟏
0 1 4𝑥 1+𝑥 3 𝑑𝑥=𝟗.𝟖
0 𝜋 3 sin 𝑥 ln⁡( sec 𝑥) 𝑑𝑥 = 𝟏 𝟐 𝟏− 𝐥𝐧 𝟐
𝑥 sin 𝑥 𝑑𝑥 =−𝒙 𝐜𝐨𝐬 𝒙 + 𝐬𝐢𝐧 𝒙 +𝑪 𝑥 sec 2 𝑥 𝑑𝑥 =𝒙 𝐭𝐚𝐧 𝒙 − 𝐥𝐧 𝐬𝐞𝐜 𝒙 +𝑪
𝑥 sin 2 𝑥 𝑑𝑥 =−𝒙 𝐜𝐨𝐭 𝒙 + 𝐥𝐧 𝐬𝐢𝐧 𝒙 +𝑪
𝑥 2 ln 𝑥 𝑑𝑥
= 𝟏 𝟑 𝒙 𝟑 𝐥𝐧 𝒙 − 𝟏 𝟗 𝒙 𝟑 +𝑪
ln 𝑥 𝑥 3 𝑑𝑥 =− 𝐥𝐧 𝒙 𝟐 𝒙 𝟐 − 𝟏 𝟒 𝒙 𝟐 +𝑪
𝑥 ln 𝑥 𝑑𝑥
= 𝟏 𝟑 𝒙 𝟑 𝐥𝐧 𝒙 − 𝟏 𝟗 𝒙 𝟑 +𝒙 𝐥𝐧 𝒙 −𝒙 3 a ? a ? c ? ? c e ? 4 ? a
Bonus Question:
arcsin 𝑥 𝑑𝑥 = 𝟏− 𝒙 𝟐 +𝒙 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙 +𝑪
(hint: 𝑑𝑣 𝑑𝑥 =1) e ? ? c ? ? ? 2 e a ? ? g c ?
You will need the following standard results (given in your formula booklet). We’ll prove them later. !
tan 𝑥 𝑑𝑥 = ln sec 𝑥 +𝐶 sec 𝑥 𝑑𝑥 = ln sec 𝑥+ tan 𝑥 +𝐶 cot 𝑥 𝑑𝑥 = ln s𝑖𝑛 𝑥 +𝐶 cosec 𝑥 𝑑𝑥 = ln cosec 𝑥+ cot 𝑥 +𝐶 e

40. Some manipulation to simplify
SKILL #8: Integrating top-heavy algebraic fractions
𝑥 2 𝑥+1 𝑑𝑥= ?
How would we deal with this? (the clue’s in the title)
Some manipulation to simplify ?
Now integrate ?
𝑥 −1
𝑥 2 𝑥+1 𝑑𝑥 = 𝑥−1+ 1 𝑥+1 𝑑𝑥 = 1 2 𝑥 2 −𝑥+ ln 𝑥 𝐶
𝑥+1
𝑥 2 +0𝑥+0
𝑥 2 +𝑥
−𝑥+0 −𝑥−1 1

41. Test Your Understanding
𝑥 𝑥−1 𝑑𝑥 =𝒙+ 𝐥𝐧 𝒙−𝟏 +𝑪
𝑥 3 +2 𝑥+1 𝑑𝑥 = 𝟏 𝟑 𝒙 𝟑 − 𝟏 𝟐 𝒙 𝟐 +𝒙 𝐥𝐧 𝒙+𝟏 +𝑪 ? ?

42. Trapezium Rule: 𝑎 𝑏 𝑦 𝑑𝑥 ≈ 1 2 ℎ 𝑦 0 +2 𝑦 1 +…+ 𝑦 𝑛−1 + 𝑦 𝑛
Trapezium Rule Revisited
Trapezium Rule: 𝑎 𝑏 𝑦 𝑑𝑥 ≈ 1 2 ℎ 𝑦 𝑦 1 +…+ 𝑦 𝑛−1 + 𝑦 𝑛
You already know the Trapezium Rule from C2. You may be asked similar questions in a C4 exam, except now the expressions involve integrations rules you’ve just learnt. Q
Given 𝐼= 0 𝜋 3 sec 𝑥 𝑑𝑥
Find the exact value of 𝐼.
Use the trapezium rule with two strips to estimate 𝐼.
Use the trapezium rule with four strips to find a second estimate of 𝐼.
Find the percentage error in using each estimate.
𝐼= ln sec 𝑥 + tan 𝑥 𝜋 3 = ln a ? c ? 𝒙
𝜋 12
𝜋 6
𝜋 4
𝜋 3 𝒚 1
1.035
1.155
1.414 2 b ? 𝒙
𝜋 6
𝜋 3 𝒚 1
1.155 2
𝐼≈ 1 2 𝜋 =1.34
𝐼≈ 1 2 𝜋 =1.39 d ?
Error is 5.6% and 1.5%.

43. Test Your UnderstandingQ
Find the exact value of 𝐼= 0 2 𝑥 2−𝑥 𝑑𝑥 .
Use the trapezium rule with 4 strips to estimate 𝐼.
Use the trapezium rule with 6 strips to find a second estimate of 𝐼.
Compare the percentage error in using each estimate. a
1.34
1.42
11.4% and 61.% ? ? b ? c d ?

44. SKILL #9: Volumes of Revolution
𝑎 𝑏 𝑦 𝑑𝑥 gives the area bounded between 𝑦=𝑓(𝑥), 𝑥=𝑎, 𝑥=𝑏 and the 𝑥-axis. Why?
If we split up the area into thin rectangular strips, each with width 𝑑𝑥 and each with height the 𝑦=𝑓 𝑥 for that particular value of 𝑥. Each has area 𝑓 𝑥 × 𝑑𝑥. 𝑦
If we had ‘discrete’ strips, the total area would be:
Σ 𝑥=𝑎 𝑏 (𝑓 𝑥 𝑑𝑥)
But because the strips are infinitely small and we have to think continuously, we use ∫ instead of Σ.
Integration therefore can be thought of as a continuous version of summation.
𝑦 1
𝑦 2 𝑥 𝑏 𝑎
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥

45. Click to Start Bromanimation
SKILL #9: Volumes of Revolution
Now suppose we spun the line 𝑦=𝑓(𝑥) about the 𝑥 axis to form a solid (known as a volume of revolution): 𝑦
Click to Start Bromanimation
How could we use the same principle as before to express the volume? What should we use instead of strips? 𝑦 𝑎 𝑏 𝑥
Reveal >>
We’re summing a bunch of infinitely thin cylinders, each of width 𝑑𝑥 and radius 𝑦.
Each has a volume of:
𝜋 𝑦 2 𝑑𝑥
Thus the volume is: !
𝑽=𝝅 𝒂 𝒃 𝒚 𝟐 𝒅𝒙
i.e. Square the function and slap a 𝜋 on front. 𝑑𝑥 ? ?

46. Examples Q
The region 𝑅 is bounded by the curve with equation 𝑦=𝑥, the 𝑥-axis and the lines 𝑥=0 and 𝑥=2.
Find the area of 𝑅.
Find the volume of the solid formed when the region 𝑅 is rotated through 2𝜋 radians about the 𝑥-axis.
0 2 𝑥 𝑑𝑥 = 𝑥 =2
𝜋 0 2 𝑥 2 𝑑𝑥=𝜋 𝑥 = 8𝜋 3 ? a ? b Q
Test Your Understanding: Do the same when 𝑦= sin 2𝑥 , bound between 𝑥=0 and 𝑥= 𝜋 2 .
0 𝜋 2 sin 2𝑥 𝑑𝑥 = − 1 2 cos 2𝑥 0 𝜋 2 =1
𝜋 0 𝜋 2 sin 2 2𝑥 𝑑𝑥=𝜋 0 𝜋 − 1 2 cos 4𝑥 𝑑𝑥 = 𝜋 2 4 ? a ? b

47. Area/Volume with Parametric Equations
I once again have parametric equations:
𝑥= 𝑡 2 𝑦=𝑡+1
I want the integration to be in terms of 𝑑𝑡 rather than 𝑑𝑥.
Bro Memory Tip: No need to remember the whole new formulae. All you need to remember is that 𝑑𝑥 𝑑𝑡 𝑑𝑡=𝑑𝑥 , which is obvious since the 𝑑𝑡’s cancel.
Area: 𝑦 𝑑𝑥 = 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡
Volume: 𝜋 𝑦 2 𝑑𝑥 =𝜋 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡 ? ?
Bro Tip: Remember that you have to change the bounds!
Example: Find the volume of revolution and the area bound by the curve and 𝑥=0 and 𝑥=2.
𝑑𝑥 𝑑𝑡 =2𝑡 𝑊ℎ𝑒𝑛 𝑥=2, 𝑡= 𝑊ℎ𝑒𝑛 𝑥=0, 𝑡=0 𝐴= 𝑡+1 2𝑡 𝑑𝑡 = 𝑡 2 +2𝑡 𝑑𝑡 = 𝑡 3 + 𝑡 = 4√ 𝑉=𝜋 𝑡 𝑡 𝑑𝑡 =𝜋 𝑡 3 +4 𝑡 2 +2𝑡 𝑑𝑡 =𝜋 ?
STEP 1: Find 𝒅𝒙 𝒅𝒕
STEP 2: Change limits
STEP 3: Integrate ? ? ?

48. Test Your Understanding
𝒅𝒙 𝒅𝜽 = 𝐬𝐞𝐜 𝟐 𝜽
When 𝒙= 𝟑 , 𝜽= 𝝅 𝟑
When 𝒙=𝟎, 𝜽=𝟎 𝝅 𝟎 𝝅 𝟑 𝒚 𝟐 𝒅𝒙 𝒅𝜽 𝒅𝜽 =𝝅 𝟎 𝝅 𝟑 𝒔𝒊𝒏 𝟐 𝜽 𝒔𝒆𝒄 𝟐 𝜽 𝒅𝜽 =𝝅 𝟎 𝝅 𝟑 𝒔𝒊𝒏 𝟐 𝜽 𝒄𝒐𝒔 𝟐 𝜽 𝒅𝜽 =𝝅 𝟎 𝝅 𝟑 𝒕𝒂𝒏 𝟐 𝜽 𝒅𝜽 =𝝅 𝟎 𝝅 𝟑 𝒔𝒆𝒄 𝟐 𝜽 −𝟏 𝒅𝜽 𝝅 𝐭𝐚𝐧 𝜽 −𝜽 𝟎 𝝅 𝟑 =𝝅 𝐭𝐚𝐧 𝝅 𝟑 − 𝝅 𝟑 =𝝅 𝟑 − 𝝅 𝟑 =𝝅 𝟑 − 𝟏 𝟑 𝝅 𝟐 ?
June 2011 Q7

49. Exercise 6I
Find the (i) area of region and (ii) solid of revolution bound by lines 𝑥=𝑎 and 𝑥=𝑏 for:
𝑓 𝑥 = 2 1+𝑥 ;𝑎=0, 𝑏= 𝑨=𝟐 𝐥𝐧 𝟐 𝑽=𝟐𝝅
𝑓 𝑥 = ln 𝑥 ;𝑎=1,𝑏= 𝑨=𝟐 𝒍𝒏 𝟐 −𝟏 𝑽= 𝟐 𝐥𝐧 𝟐 𝟐 −𝟒 𝐥𝐧 𝟐 𝜋
(Hint: use integration by parts with 𝑑𝑣 𝑑𝑥 =1 for ln 2 𝑥 )
𝑓 𝑥 =𝑥 4− 𝑥 2 ; 𝑎=0,𝑏= 𝑨= 𝟖 𝟑 𝑽= 𝟔𝟒 𝟏𝟓 𝝅
The region 𝑅 is bound by the curve 𝐶, the 𝑥-axis and the lines 𝑥=−8 and 𝑥=+8. The parametric equations for 𝐶 are 𝑥= 𝑡 3 and 𝑦= 𝑡 2 . Find:
The area of 𝑅. 𝟏𝟗𝟐 𝟓
The volume of the solid of revolution formed about the 𝑥-axis. 𝟕𝟔𝟖 𝟕 𝝅
The curve 𝐶 has parametric equations 𝑥= sin 𝑡 and 𝑦= sin 2𝑡, 0≤𝑡< 𝜋 2
Find the area of the region bounded by 𝐶 and the 𝑥-axis. 𝟐 𝟑
Find the volume of the solid of revolution. 𝟖 𝟏𝟓 𝝅 1 ? ? a ? ? c ? ? e 3 ? ? 4 ? ?

50. SKILL #10: Differential Equations
(We’re on the home straight!)
Earlier in C4 we encountered differential equations. These are equations involving a mix of variables and derivatives, e.g. 𝒚, 𝒙 and 𝒅𝒚 𝒅𝒙 .
‘Solving’ these equations means to get 𝑦 in terms of 𝑥 (with no 𝑑𝑦 𝑑𝑥 ). Q
Find the general solution to 𝑑𝑦 𝑑𝑥 =𝑥𝑦+𝑦 ?
STEP 1: Get 𝑦 to the side of 𝑑𝑦 𝑑𝑥 by dividing and 𝑥 to the other side.
(you may need to factorise)
𝑑𝑦 𝑑𝑥 =𝑦 𝑥 𝑦 𝑑𝑦 𝑑𝑥 =𝑥 𝑦 𝑑𝑦= 𝑥+1 𝑑𝑥 ln 𝑦 = 1 2 𝑥 2 +𝑥+𝐶 𝑦= 𝑒 𝑥 2 +𝑥+𝐶 = 𝑒 𝑥 2 +𝑥 𝑒 𝐶 =𝐴 𝑒 𝑥 2 +𝑥
STEP 2: Multiply both sides by 𝑑𝑥, then integrate both sides. ? ?
STEP 3: Make 𝑦 the subject, if the question asks.

51. Another Example Q
Find the general solution to 1+ 𝑥 2 𝑑𝑦 𝑑𝑥 =𝑥 tan 𝑦 ?
STEP 1: Get 𝑦 to the side of 𝑑𝑦 𝑑𝑥 by dividing and 𝑥 to the other side.
(you may need to factorise)
1 tan 𝑦 𝑑𝑦 𝑑𝑥 = 𝑥 1+ 𝑥 2
cot 𝑦 𝑑𝑦 = 𝑥 1+ 𝑥 2 𝑑𝑥 ln sin 𝑦 = 1 2 ln 1+ 𝑥 2 +𝐶 ln sin 𝑦 = 1 2 ln 1+ 𝑥 ln 𝑘 ln sin 𝑦 = ln 𝑘 1+ 𝑥 sin 𝑦 =𝑘 1+ 𝑥 2 𝑦= arcsin 𝑘 1+ 𝑥 2 ?
STEP 2: Multiply both sides by 𝑑𝑥, then integrate both sides. ?
STEP 2b: If possible, try to combine your constant of integration with other terms
(e.g. by letting 𝐶= ln 𝑘 where 𝑘 is another constant) ?
STEP 3: Make 𝑦 the subject, if the question asks.

52. Test Your UnderstandingQ
The rate of increase of a rabbit population (with population 𝑦, where time is 𝑥) is proportional to the current population.
Form a differential equation, and find its general solution. ?
𝑑𝑦 𝑑𝑥 =𝑘𝑦 1 𝑦 𝑑𝑦 𝑑𝑥 =𝑘 𝑦 𝑑𝑦 = 𝑘 𝑑𝑥 ln 𝑦 =𝑘𝑥+𝐶 𝑦= 𝑒 𝑘𝑥+𝐶 𝒚=𝑨 𝒆 𝒌𝒙 (𝑤ℎ𝑒𝑟𝑒 𝐴= 𝑒 𝐶 )
(Notice by the way that 𝑒 𝑘𝑥 = 𝑒 𝑘 𝑥 , and since 𝑒 𝑘 is a constant, we could always write 𝑦=𝐴⋅ 𝐵 𝑥 . i.e. The general solution is ‘any generic exponential function’, not just restricted to those with 𝑒 as the base. However it is customary to write 𝐴 𝑒 𝑘𝑥 )

53. Differential Equations with Boundary Conditions
Find the general solution to 𝑑𝑦 𝑑𝑥 =− 3 𝑦−2 2𝑥+1 𝑥+2
Given that 𝑥=1 when 𝑦=4. Leave your answer in the form 𝑦=𝑓 𝑥 ?
1 𝑦−2 𝑑𝑦 = −3 2𝑥+1 𝑥+2 𝑑𝑥
Use partial fractions to split up RHS.
−3 2𝑥+1 𝑥+2 ≡ 𝐴 2𝑥+1 + 𝐵 𝑥+2 𝐴=−2, 𝐵=1
1 𝑦−2 𝑑𝑦 = 1 𝑥+2 − 2 2𝑥+1 𝑑𝑥 ln 𝑦−2 = ln 𝑥+2 − ln 2𝑥+1 + ln 𝑘 ln 𝑦−2 = ln 𝑘 𝑥+2 2𝑥+1 𝑦−2= 𝑘 𝑥+2 2𝑥+1
When 𝑥=1,𝑦=4: 4−2= 𝑘 →𝑘=2
𝒚=𝟑+ 𝟑 𝟐𝒙+𝟏
Bro Tip: While it doesn’t matter when you determine your constant using the boundary conditions, usually it’s easiest to determine it as soon as possible.

54. Key Points ? Get 𝑦 on to LHS by dividing (possibly factorising first).
If after integrating you have 𝑙𝑛 on the RHS, make your constant of integration ln 𝑘 .
Be sure to combine all your 𝑙𝑛’s together just as you did in C2. E.g.: 2 ln 𝑥+1 − ln 𝑥 → ln 𝑥 𝑥
Sub in boundary conditions to work out your constant – better to do sooner rather than later.
Exam questions  partial fractions. ?

55. Exercise 6J 1
Find the general solution in the form 𝑦=𝑓(𝑥).
𝑑𝑦 𝑑𝑥 = 1+𝑦 1−2𝑥 𝒚=𝑨 𝒆 𝒙− 𝒙 𝟐 −𝟏
cos 2 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦 2 sin 2 𝑥 𝒚=−𝟏/( 𝐭𝐚𝐧 𝒙 −𝒙+𝑪)
𝑥 2 𝑑𝑦 𝑑𝑥 =𝑦+𝑥𝑦 𝒚=𝑨𝒙 𝒆 − 𝟏 𝒙
Find the general solution (no need to put in form 𝑦=𝑓 𝑥 )
𝑑𝑦 𝑑𝑥 = tan 𝑦 tan 𝑥 𝐬𝐢𝐧 𝒚 =𝒌 𝐬𝐞𝐜 𝒙
1+ 𝑥 2 𝑑𝑦 𝑑𝑥 =𝑥 1− 𝑦 𝟏+𝒚 𝟏−𝒚 =𝒌 𝟏+ 𝒙 𝟐
𝑒 𝑥+𝑦 𝑑𝑦 𝑑𝑥 =𝑥 2+ 𝑒 𝑦 𝐥𝐧 𝟐+ 𝒆 𝒚 =−𝒙 𝒆 −𝒙 − 𝒆 −𝒙 +𝑪
Find particular solutions using boundary conditions.
𝑑𝑦 𝑑𝑥 =𝑠𝑖𝑛𝑥 cos 2 𝑥 ;𝑦=0, 𝑥= 𝜋 𝟏 𝟐𝟒 − 𝟏 𝟑 𝐜𝐨𝐬 𝟑 𝒙
𝑑𝑦 𝑑𝑥 =2 cos 2 𝑦 cos 2 𝑥 ;𝑦= 𝜋 4 , 𝑥= 𝐭𝐚𝐧 𝒚 = 𝟏 𝟐 𝐬𝐢𝐧 𝟐𝒙 +𝒙+𝟏
2 1+𝑥 𝑑𝑦 𝑑𝑥 =1− 𝑦 2 ;𝑥=5,𝑦= 𝟏+𝒚 𝟏−𝒚 = 𝟏+𝒙 𝟐 ? a ? c ? e 2 ? a ? c e ? 4 ? a ? c ? e

56. Exercise 6K
The size of a certain population at time 𝑡 is given by 𝑃. The rate of increase of 𝑃 I given by 𝑑𝑃 𝑑𝑡 =2𝑃. Given that a time 𝑡=0, the population was 3, find the population at time 𝑡=2. 𝟑 𝒆 𝟒
The mass 𝑀 at time 𝑡 of the leaves of a certain plant varies according to:
𝑑𝑀 𝑑𝑡 =𝑀− 𝑀 2
a) Given that at time 𝑡=0, 𝑀=0.5, find an expression for 𝑀 in terms of 𝑡. 𝑴= 𝒆 𝒕 /(𝟏+ 𝒆 𝒕 )
b) Find a value for 𝑀 when 𝑡= ln 𝟐/𝟑
c) Explain what happens to the value of 𝑀 as a 𝑡 increases 𝑴 approaches 1.
The thickness of ice 𝑥 mm on a pond is increasing and 𝑑𝑥 𝑑𝑡 = 1 20 𝑥 2 , where 𝑡 is measured in hours. Find how long it takes the thickness of ice to increase from 1mm to 2mm.
𝟒𝟔 𝟐 𝟑
The rate of increase of the radius 𝑟 km of an oil slick is given by 𝑑𝑟 𝑑𝑡 =𝑘/ 𝑟 2 , where 𝑘 is a positive constant. When the slick was first observed the radius was 3km. Two days later it was 5km. Find, to the nearest day when the radius will be 6. 𝟒 1 ? 3 ? ? ? 5 ? 7 ?

57. Summary of Functions ? ? ? ? ? ? ? ? ? 𝒇(𝒙) How to deal with it
𝒇 𝒙 𝒅𝒙 (+constant)
Formula booklet?
𝒔𝒊𝒏 𝒙
Standard result
− cos 𝑥 No
𝒄𝒐𝒔 𝒙
sin 𝑥
𝒕𝒂𝒏 𝒙
In formula booklet, but use sin 𝑥 cos 𝑥 𝑑𝑥 which is of the form 𝑘 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥
ln sec 𝑥
Yes
𝒔𝒊 𝒏 𝟐 𝒙
For both sin 2 𝑥 and cos 2 𝑥 use identities for cos 2𝑥
cos 2𝑥 =1−2 sin 2 𝑥 sin 2 𝑥 = 1 2 − 1 2 cos 2𝑥
1 2 𝑥− 1 4 sin 2𝑥
𝒄𝒐 𝒔 𝟐 𝒙
cos 2𝑥 =2 cos 2 𝑥 −1 cos 2 𝑥 = cos 2𝑥
1 2 𝑥 sin 2𝑥
𝒕𝒂 𝒏 𝟐 𝒙
1+ tan 2 𝑥 ≡ sec 2 𝑥 tan 2 𝑥 ≡ sec 2 𝑥 −1
tan 𝑥 −𝑥
𝒄𝒐𝒔𝒆𝒄 𝒙
Would use substitution 𝑢=𝑐𝑜𝑠𝑒𝑐 𝑥+ cot 𝑥 , but too hard for exam.
−ln 𝑐𝑜𝑠𝑒𝑐 𝑥+ cot 𝑥
𝒔𝒆𝒄 𝒙
Would use substitution 𝑢= sec 𝑥 + tan 𝑥 , but too hard for exam.
ln sec 𝑥 + tan 𝑥
𝒄𝒐𝒕 𝒙
cos 𝑥 sin 𝑥 𝑑𝑥 which is of the form 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥
ln 𝑠𝑖𝑛 𝑥 ? ? ? ? ? ? ? ? ?

58. Summary of Functions ? ? ? ? ? ? ? ? 𝒇(𝒙) How to deal with it
𝒇 𝒙 𝒅𝒙 (+constant)
Formula booklet?
𝒄𝒐𝒔𝒆 𝒄 𝟐 𝒙
By observation.
− 𝐜𝐨𝐭 𝒙
No!
𝒔𝒆 𝒄 𝟐 𝒙
tan 𝑥
Yes
(but memorise)
𝒄𝒐 𝒕 𝟐 𝒙
1+ cot 2 𝑥 ≡𝑐𝑜𝑠𝑒 𝑐 2 𝑥
− cot 𝑥 −𝑥 No
𝒔𝒊𝒏 𝟐𝒙 𝒄𝒐𝒔 𝟐𝒙
For any product of sin and cos with same coefficient of 𝑥, use double angle.
sin 2𝑥 cos 2𝑥 ≡ 1 2 sin 4𝑥
− 1 8 cos 4𝑥
𝟏 𝒙
ln 𝑥
𝐥𝐧 𝒙
Use IBP, where 𝑢= ln 𝑥 , 𝑑𝑣 𝑑𝑥 = ln 𝑥
𝑥 ln 𝑥 −𝑥
𝒙 𝒙+𝟏
Use algebraic division. 𝑥 𝑥+1 ≡1− 1 𝑥+1
𝑥− ln 𝑥+1
𝟏 𝒙 𝒙+𝟏
Use partial fractions.
ln 𝑥 − ln 𝑥+1 ? ? ? ? ? ? ? ?

59. Summary of Functions ? ? ? ? ? 𝟒𝒙 𝒙 𝟐 +𝟏
𝒇(𝒙)
How to deal with it
𝒇 𝒙 𝒅𝒙 (+constant)
𝟒𝒙 𝒙 𝟐 +𝟏
Reverse chain rule. Of form 𝑘 𝑓 ′ 𝑥 𝑓 𝑥
2 ln 𝑥 2 +1
𝒙 𝒙 𝟐 +𝟏 𝟐
Power around denominator so NOT of form 𝑘 𝑓 ′ 𝑥 𝑓 𝑥 . Rewrite as product.
𝑥 𝑥 −2
Reverse chain rule (i.e. “Consider 𝑦= 𝑥 −1 " and differentiate)
− 𝑥 −1
𝒆 𝟐𝒙+𝟏 𝟏 𝟏−𝟑𝒙
For any function where ‘inner function’ is linear expression, divide by coefficient of 𝑥
1 2 𝑒 2𝑥+1 − 1 3 ln 1−3𝑥
𝒙 𝟐𝒙+𝟏
Use sensible substitution. 𝑢=2𝑥+1 or even better, 𝑢 2 =2𝑥+1.
𝑥 𝑥−1
𝐬𝐢𝐧 𝟓 𝒙 𝒄𝒐𝒔 𝒙
Reverse chain rule.
1 6 sin 6 𝑥 ? ? ? ? ?