1. FP3 Chapter 4 Integration
Dr J Frost
Last modified: 25th January 2017

2. FP3 Integration Overview
This chapter is long and perilous, but you will learn lots of interesting new techniques as well as reprising existing ones…
Section C: Arc lengths and surface area
Section B: Reduction Formulae
Length of a curve.
A technique for dealing with large powers in integration.
Surface area of volumes of revolution.
Using substitutions
Use 𝑥=𝑎 tan 𝜃 to find
1 9 𝑥 2 +6𝑥+5 𝑑𝑥
Completing the square
1 9 𝑥 2 +6𝑥+5 𝑑𝑥
Standard results
sinh 𝑥 𝑑𝑥
Section A: General Skills
By parts
0 1 𝑒 2𝑥 𝑠𝑖𝑛ℎ 𝑥 𝑑𝑥

3. SECTION A PART 1 :: Standard Integrals
Same as non-hyperbolic version?
sinh 𝑥 𝑑𝑥 = cosh 𝑥 +𝐶 cosh 𝑥 𝑑𝑥 = sinh 𝑥 +𝐶 sech 2 𝑥 𝑑𝑥 = tanh 𝑥 +𝐶 cosech 2 𝑥 𝑑𝑥 = −coth 𝑥 +𝐶 sech 𝑥 tanh 𝑥 𝑑𝑥 = − sech 𝑥 +𝐶 cosech 𝑥 coth 𝑥 𝑑𝑥 = −cosech 𝑥 +𝐶 − 𝑥 2 𝑑𝑥 = arcsin 𝑥 +𝐶, 𝑥 < 𝑥 2 𝑑𝑥 = arctan 𝑥 +𝐶 𝑥 2 𝑑𝑥 = arcsinℎ 𝑥 +𝐶 𝑥 2 −1 𝑑𝑥 = arccosh 𝑥 +𝐶, 𝑥>1 ?  ?  ?  ? 
Not in formula booklet. ?   ? ? ? ? ?

4. Click only if you’ve forgotten them.
Quickfire Examples – Do From Memory!
Recall from C4 that:
𝒇 ′ 𝒂𝒙+𝒃 𝒅𝒙 = 𝟏 𝒂 𝒇 𝒂𝒙+𝒃 +𝑪
e.g. 𝑒 3𝑥+2 𝑑𝑥 = 1 3 𝑒 3𝑥+2
cosh 4𝑥−1 𝑑𝑥 = 𝟏 𝟒 𝐬𝐢𝐧𝐡 𝟒𝒙−𝟏 +𝑪 𝑠𝑖𝑛ℎ 𝑥 𝑑𝑥= 𝟑 𝟐 𝐜𝐨𝐬𝐡 𝟐 𝟑 𝒙 +𝑪 𝑐𝑜𝑠𝑒𝑐ℎ 3𝑥 coth 3𝑥 𝑑𝑥=− 𝟏 𝟑 𝒄𝒐𝒔𝒆𝒄𝒉 𝟑𝒙+𝑪 𝑠𝑒𝑐 ℎ 2 𝑥+1 𝑑𝑥= 𝐭𝐚𝐧𝐡 𝒙+𝟏 +𝑪 sech 4𝑥 tanh 4𝑥 𝑑𝑥=− 𝟏 𝟒 𝐬𝐞𝐜𝐡 𝟒𝒙 +𝑪 𝑐𝑜𝑠𝑒𝑐 ℎ 2 2𝑥−1 𝑑𝑥=− 𝟏 𝟐 𝐜𝐨𝐭𝐡 𝟐𝒙−𝟏 +𝑪 ?
sinh 𝑥 𝑑𝑥 = cosh 𝑥 +𝐶 cosh 𝑥 𝑑𝑥 = sinh 𝑥 +𝐶 sech 2 𝑥 𝑑𝑥 = tanh 𝑥 +𝐶 cosech 2 𝑥 𝑑𝑥 = −coth 𝑥 +𝐶 sech 𝑥 tanh 𝑥 𝑑𝑥 = − sech 𝑥 +𝐶 cosech 𝑥 coth 𝑥 𝑑𝑥 = −cosech 𝑥 +𝐶 − 𝑥 2 𝑑𝑥 = arcsin 𝑥 +𝐶, 𝑥 < 𝑥 2 𝑑𝑥 = arctan 𝑥 +𝐶 𝑥 2 𝑑𝑥 = arcsinℎ 𝑥 +𝐶 𝑥 2 −1 𝑑𝑥 = arccosh 𝑥 +𝐶, 𝑥>1 ?
Click only if you’ve forgotten them. ? ? ? ? ?

5. Integration by Recognition
Recall from C4 that:
𝑓 ′ 𝑥 𝑓 𝑥 𝑛 𝑑𝑥= 𝑓 𝑥 𝑛+1 𝑛+1 +𝐶 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥 = ln 𝑓 𝑥 +𝐶
4𝑥 1+ 𝑥 2 𝑑𝑥 =𝟐 𝟐𝒙 𝟏+ 𝒙 𝟐 𝒅𝒙 =𝟐 𝐥𝐧 𝟏+ 𝒙 𝟐 +𝑪 5𝑥 1+ 𝑥 2 𝑑𝑥= 𝟓𝒙 𝟏+ 𝒙 𝟐 − 𝟏 𝟐 𝒅𝒙
Try 𝒚= 𝟏+ 𝒙 𝟐 𝟏 𝟐 𝒅𝒚 𝒅𝒙 = 𝟐𝒙 𝟏 𝟐 𝟏+ 𝒙 𝟐 − 𝟏 𝟐 =𝒙 𝟏+𝒙 − 𝟏 𝟐 ∴ 𝟓𝒙 𝟏+ 𝒙 𝟐 𝒅𝒙= 𝟓 𝟏+ 𝒙 𝟐 +𝑪
sin 2𝑥 cos 𝑛 𝑥 𝑑𝑥 = 𝟐 𝒏−𝟐 𝐜𝐨𝐬 𝟐−𝒏 𝒙 +𝒄 ? ? ?
Bro Tip: If there’s a power outside in the denominator, always reexpress as product first. ?

6. Exercise 4A
Integrate the following with respect to 𝑥.
sinh 𝑥 +3 cosh 𝑥 → 𝐜𝐨𝐬𝐡 𝒙 +𝟑 𝐬𝐢𝐧𝐡 𝒙 +𝑪 5 sech 2 𝑥 →𝟓 𝐭𝐚𝐧𝐡 𝒙 +𝑪
1 sinh 2 𝑥 →− 𝐜𝐨𝐭𝐡 𝒙 +𝑪
cosh 𝑥 − 1 cosh 2 𝑥 → 𝐬𝐢𝐧𝐡 𝒙 − 𝐭𝐚𝐧𝐡 𝒙 +𝑪
sinh 𝑥 cosh 2 𝑥 →− 𝐬𝐞𝐜𝐡 𝒙 +𝑪
3 sinh 𝑥 tanh 𝑥 →−𝟑𝒄𝒐𝒔𝒆𝒄𝒉 𝒙+𝑪
sech 𝑥 sech 𝑥 + tanh 𝑥 → 𝐭𝐚𝐧𝐡 𝒙 − 𝐬𝐞𝐜𝐡 𝒙 +𝑪
sech 𝑥 +𝑐𝑜𝑠𝑒𝑐ℎ 𝑥 sech 𝑥 −𝑐𝑜𝑠𝑒𝑐ℎ 𝑥
→ 𝐭𝐚𝐧𝐡 𝒙 + 𝐜𝐨𝐭𝐡 𝒙 +𝑪
Find
sinh 2𝑥 𝑑𝑥 = 𝟏 𝟐 𝐜𝐨𝐬𝐡 𝒙 +𝑪
cosh 𝑥 3 𝑑𝑥 =𝟑 𝐬𝐢𝐧𝐡 𝒙 𝟑 +𝑪
sech 2 2𝑥−1 𝑑𝑥 = 𝟏 𝟐 𝐭𝐚𝐧𝐡 𝟐𝒙−𝟏 +𝑪
cosech 2 5𝑥 𝑑𝑥 =− 𝟏 𝟓 𝐜𝐨𝐭𝐡 𝟓𝒙 +𝑪
cosech 2𝑥 coth 2𝑥 𝑑𝑥 =− 𝟏 𝟐 𝒄𝒐𝒔𝒆𝒄𝒉 𝟐𝒙+𝑪
sech 𝑥 tanh 𝑥 𝑑𝑥 =− 𝟐 𝐬𝐞𝐜𝐡 𝒙 𝟐
(5 sinh 5𝑥 −4 cosh 4𝑥 +3 sech 2 𝑥 2 𝑑𝑥 = 𝐜𝐨𝐬𝐡 𝟓𝒙 − 𝐬𝐢𝐧𝐡 𝟒𝒙 +𝟔 𝐭𝐚𝐧𝐡 𝒙 𝟐 +𝑪 1 3
Write down the results of the following. (This is a recognition exercise and involves the integrals from C4)
1 1+ 𝑥 2 𝑑𝑥 = 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 +𝑪
𝑥 2 𝑑𝑥 =𝒂𝒓𝒔𝒊𝒏𝒉 𝒙+𝑪
1 1+𝑥 𝑑𝑥 = 𝐥𝐧 |𝟏+𝒙| +𝑪
2𝑥 1+ 𝑥 2 𝑑𝑥 = 𝐥𝐧 (𝟏+ 𝒙 𝟐 ) +𝑪
1 1− 𝑥 2 𝑑𝑥 =𝒂𝒓𝒔𝒊𝒏 𝒙+𝑪
1 𝑥 2 −1 𝑑𝑥 =𝒂𝒓𝒄𝒐𝒔𝒉 𝒙+𝑪
3𝑥 𝑥 2 −1 𝑑𝑥 =𝟑 𝒙 𝟐 −𝟏 +𝑪
3 1+𝑥 2 𝑑𝑥 =− 𝟑 𝟏+𝒙 +𝑪
Find
2𝑥+1 1− 𝑥 2 𝑑𝑥 =−𝟐 𝟏− 𝒙 𝟐 + 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙 +𝑪
1+𝑥 𝑥 2 −1 𝑑𝑥 =𝒂𝒓𝒄𝒐𝒔𝒉 𝒙+ 𝒙 𝟐 −𝟏 +𝑪
𝑥− 𝑥 2 𝑑𝑥 = 𝟏+ 𝒙 𝟐 −𝟑 𝐚𝐫𝐜𝐬𝐢𝐧𝐡 𝒙 +𝑪 ? a ? b ? c ? ? d ? ? e ? ? f ? g ? h ? ? 2 ? ? a ? ? b ? ? c ? d 4 ? e ? ? ? f g ? ?

7. Exercise 4A ? 5 Show that 𝑥 2 1+ 𝑥 2 =1− 1 1+ 𝑥 2
Hence find 𝑥 𝑥 2 𝑑𝑥
𝒙− 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 +𝑪 ?

8. Integrating when not quite so standard
sech 6 𝑥 tanh 𝑥 𝑑𝑥
Method 1: “Consider and scale”
Method 2: “Put in form 𝑓 ′ 𝑥 𝑓 𝑥 𝑛 ” ?
Try 𝑦= sech 6 𝑥 = 𝑠𝑒𝑐ℎ 𝑥 𝑑𝑦 𝑑𝑥 =6 sech 5 𝑥 ×− sech 𝑥 tanh 𝑥 =−6 sech 5 𝑥 tanh 𝑥
∴ sech 6 𝑥 tanh 𝑥 𝑑𝑥 =− sech 6 𝑥 +𝐶 ?
sech 6 𝑥 tanh 𝑥 𝑑𝑥 =− − sech 𝑥 tanh 𝑥 sech 𝑥 5 =− sech 6 𝑥 +𝐶
Bro Note: For 𝑠𝑒𝑐ℎ, don’t increment power by 1, because we know differentiating 𝑠𝑒𝑐ℎ gives us another 𝑠𝑒𝑐ℎ
Reminder:
𝑓 ′ 𝑥 𝑓 𝑥 𝑛 𝑑𝑥= 𝑓 𝑥 𝑛+1 𝑛+1 +𝐶 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥 = ln 𝑓 𝑥 +𝐶

9. Integrating when not quite so standard
cosh 5 2𝑥 sinh 2𝑥 𝑑𝑥 ?
(Using Method 1)
Try 𝒚= 𝐜𝐨𝐬𝐡 𝟔 𝟐𝒙 𝒅𝒚 𝒅𝒙 =𝟔 𝐜𝐨𝐬𝐡 𝟓 𝟐𝒙 ×𝟐𝒔𝒊𝒏𝒉 𝟐𝒙 =𝟏𝟐 𝐜𝐨𝐬𝐡 𝟓 𝟐𝒙 𝒔𝒊𝒏𝒉 𝟐𝒙
∴ 𝒄𝒐𝒔𝒉 𝟓 𝟐𝒙 𝒔𝒊𝒏𝒉 𝟐𝒙 𝒅𝒙 = 𝟏 𝟏𝟐 𝐜𝐨𝐬𝐡 𝟔 𝟐𝒙 +𝑪
tanh 𝑥 𝑑𝑥
sech 2 𝑥 2+5 tanh 𝑥 𝑑𝑥 ? ?
= sinh 𝑥 cosh 𝑥 𝑑𝑥 = ln cosh 𝑥 +𝐶
= 1 5 ln 2+5 tanh 𝑥
Note that because 𝑐𝑜𝑠ℎ differentiates to positive 𝑠𝑖𝑛ℎ, unlike the non-hyperbolic version, we don’t have the minus.

10. Using Identities ? ? ? tanh 2 𝑥 𝑑𝑥 cosh 2 3𝑥 𝑑𝑥
Use double angle formulae for cos 2𝐴
= 1− sech 2 𝑥 𝑑𝑥 =𝑥− tanh 𝑥 +𝐶
Recap: If you forget a hyperbolic identity, use Osborn’s Rule.
= cosh 6𝑥 𝑑𝑥 = 1 2 𝑥 sinh 6𝑥 +𝐶
sinh 3 𝑥 𝑑𝑥 ?
= sinh 2 𝑥 sinh 𝑥 𝑑𝑥 = cosh 2 𝑥 −1 sinh 𝑥 𝑑𝑥 = cosh 2 𝑥 sinh 𝑥 − sinh 𝑥 𝑑𝑥 = cosh 3 𝑥 − cosh 𝑥 +𝐶
Use this approach in general for small odd powers of sinh and cosh.

11. When that doesn’t work…
Sometimes there are techniques which work on non-hyperbolic trig functions but doesn’t work on hyperbolic ones.
Find 𝑒 2𝑥 sinh 𝑥 𝑑𝑥
Find sech 𝑥 𝑑𝑥 ? ?
𝑒 2𝑥 sinh 𝑥 𝑑𝑥 = 𝑒 2𝑥 𝑒 𝑥 − 𝑒 −𝑥 2 = 𝑒 3𝑥 − 𝑒 𝑥 𝑑𝑥 = 𝑒 3𝑥 −3 𝑒 𝑥 +𝐶
= 𝑒 𝑥 + 𝑒 −𝑥 𝑑𝑥 = 2 𝑒 𝑥 𝑒 2𝑥 +1 𝑑𝑥
Use the substitution 𝑢= 𝑒 𝑥
𝑑𝑢 𝑑𝑥 = 𝑒 𝑥 ∴𝑑𝑢= 𝑒 𝑥 𝑑𝑥
2 𝑒 𝑥 𝑒 2𝑥 +1 𝑑𝑥= 𝑢 2 +1 𝑑𝑢 =2 arctan 𝑢 +𝐶 =2 arctan 𝑒 𝑥 +𝐶
(Integration by parts DOES also work, but requires a significantly greater amount of working!)
(Bro Exam Note: This very question appeared June 2014, except involving definite integration)

12. Exercise 4B 1
Find
sinh 3 𝑥 cosh 𝑥 𝑑𝑥 = 𝟏 𝟒 𝐬𝐢𝐧𝐡 𝟒 𝒙 +𝑪
tanh 4𝑥 𝑑𝑥 = 𝟏 𝟒 𝐥𝐧 𝐜𝐨𝐬𝐡 𝟒𝒙 +𝑪
tanh 5 𝑥 sech 2 𝑥 𝑑𝑥 = 𝟏 𝟔 𝐭𝐚𝐧𝐡 𝟔 𝒙 +𝑪
𝑐𝑜𝑠𝑒𝑐 ℎ 7 𝑥 coth 𝑥 𝑑𝑥 =− 𝟏 𝟕 𝒄𝒐𝒔𝒆𝒄 𝒉 𝟕 𝒙+𝑪
cosh 2𝑥 sinh 2𝑥 𝑑𝑥 = 𝟏 𝟑 𝐜𝐨𝐬𝐡 𝟐𝒙 𝟑 𝟐 +𝑪
sech 10 3𝑥 tanh 3𝑥 𝑑𝑥 =− 𝟏 𝟑𝟎 𝐬𝐞𝐜𝐡 𝟏𝟎 𝟑𝒙 +𝑪
sinh 𝑥 2+3 cosh 𝑥 𝑑𝑥 = 𝟏 𝟑 𝐥𝐧 𝟐+𝟑 𝐜𝐨𝐬𝐡 𝒙 +𝑪
1+ tanh 𝑥 cosh 2 𝑥 𝑑𝑥 = 𝐭𝐚𝐧𝐡 𝒙 + 𝟏 𝟐 𝐭𝐚𝐧𝐡 𝟐 𝒙 +𝑪
5 cosh 𝑥 +2 sinh 𝑥 cosh 𝑥 𝑑𝑥 =𝟓𝒙+𝟐 𝐥𝐧 𝐜𝐨𝐬𝐡 𝒙 +𝑪
Show that coth 𝑥 𝑑𝑥 = 𝐥𝐧 𝐬𝐢𝐧𝐡 𝒙 +𝑪
Show that coth 2𝑥 𝑑𝑥 = 𝐥𝐧 𝒆 𝟐 + 𝟏 𝒆 𝟐
Use integration by parts to find:
𝑥 sinh 3𝑥 𝑑𝑥 = 𝟏 𝟑 𝒙 𝐜𝐨𝐬𝐡 𝟑𝒙 − 𝟏 𝟗 𝐬𝐢𝐧𝐡 𝟑𝒙 +𝑪
𝑥𝑠𝑒𝑐 ℎ 2 𝑥 𝑑𝑥 =𝒙 𝐭𝐚𝐧𝐡 𝒙 − 𝐥𝐧 𝐜𝐨𝐬𝐡 𝒙 +𝑪 5
Find
𝑒 𝑥 cosh 𝑥 𝑑𝑥 = 𝟏 𝟒 𝒆 𝟐𝒙 + 𝟏 𝟐 𝒙+𝑪
𝑒 −2𝑥 sinh 3𝑥 𝑑𝑥 = 𝟏 𝟐 𝒆 𝒙 + 𝟏 𝟏𝟎 𝒆 −𝟓𝒙 +𝑪
cosh 𝑥 cosh 3𝑥 𝑑𝑥 = 𝟏 𝟏𝟔 𝒆 𝟒𝒙 − 𝟏 𝟏𝟔 𝒆 −𝟒𝒙 + 𝟏 𝟖 𝒆 𝟐𝒙 − 𝟏 𝟖 𝒆 −𝟐𝒙 +𝑪
By writing cosh 3𝑥 in exponential form, find cosh 2 3𝑥 𝑑𝑥 and show that it is equivalent to the result found in 5b.
Evaluate sinh 𝑥 + cosh 𝑥 𝑑𝑥 , giving your answer in terms of 𝑒.
𝟏− 𝟏 𝒆 ? ? ? ? ? ? ? ? ? 6 2 ? ? ? 7 3 ? ? ? 4 ? ?

13. Exercise 4B
Use appropriate identities to find:
sinh 2 𝑥 𝑑𝑥 = 𝟏 𝟒 𝐬𝐢𝐧𝐡 𝟐𝒙 − 𝟏 𝟐 𝒙+𝑪
sech 𝑥 − tanh 𝑥 2 𝑑𝑥=𝒙+𝟐 𝐬𝐞𝐜𝐡 𝒙 +𝑪
cosh 2 3𝑥 sinh 2 3𝑥 𝑑𝑥 =𝒙− 𝟏 𝟑 𝐜𝐨𝐭𝐡 𝟑𝒙 +𝑪
sinh 2 𝑥 cosh 2 𝑥 𝑑𝑥 =− 𝟏 𝟖 𝒙+ 𝟏 𝟑𝟐 𝐬𝐢𝐧𝐡 𝟒𝒙 +𝑪
cosh 5 𝑥 𝑑𝑥 = 𝐬𝐢𝐧𝐡 𝒙 + 𝟐 𝟑 𝐬𝐢𝐧𝐡 𝟑 𝒙 + 𝟏 𝟓 𝐬𝐢𝐧𝐡 𝟓 𝒙 +𝑪
tanh 3 2𝑥 𝑑𝑥 = 𝟏 𝟐 𝐥𝐧 𝐜𝐨𝐬𝐡 𝟐𝒙 − 𝟏 𝟒 𝐭𝐚𝐧𝐡 𝟐 𝟐𝒙 +𝑪
Show that 0 ln 2 cosh 2 𝑥 2 𝑑𝑥 = ln 16
The region bounded by the curve 𝑦= sinh 𝑥 , the line 𝑥=1 and the positive 𝑥-axis is rotated through 360° about the 𝑥-axis. Show that the volume of the solid of revolution formed is 𝜋 8 𝑒 2 𝑒 4 −4 𝑒 2 −1 11
Using the result for sech 𝑥 𝑑𝑥 given in the examples, find
2 cosh 𝑥 𝑑𝑥 =𝟒 𝐚𝐫𝐜𝐭𝐚𝐧 𝒆 𝒙 +𝑪
sech 2𝑥 𝑑𝑥 = 𝐚𝐫𝐜𝐭𝐚𝐧 𝒆 𝟐𝒙 +𝑪
1− tanh 2 𝑥 2 𝑑𝑥 =𝟒 𝐚𝐫𝐜𝐭𝐚𝐧 𝒆 𝒙 𝟐 +𝑪
Using the substitution 𝑢= 𝑥 2 or otherwise, find
𝑥 cosh 2 𝑥 2 𝑑𝑥= 𝟏 𝟖 𝐬𝐢𝐧𝐡 𝟐 𝒙 𝟐 + 𝒙 𝟐 𝟒 +𝑪
𝑥 cosh 2 𝑥 2 𝑑𝑥 = 𝟏 𝟐 𝐭𝐚𝐧𝐡 𝒙 𝟐 +𝑪 8 ? ? ? ? ? ? ? ? ? 12 ? 9 ? 10

14. Using substitutions
From earlier:
1 1− 𝑥 2 𝑑𝑥 = arcsin 𝑥 +𝐶, 𝑥 < 𝑥 2 𝑑𝑥 = arctan 𝑥 +𝐶 𝑥 2 𝑑𝑥 = arcsinℎ 𝑥 +𝐶 𝑥 2 −1 𝑑𝑥 = arccosh 𝑥 +𝐶, 𝑥>1
Use an appropriate substitution to show that 𝑥 2 𝑑𝑥 = arctan 𝑥 +𝐶 ?
Bro Hint: Think what value of 𝑥 would make 1+ 𝑥 2 nicely simplify.
𝑥= tan 𝜃
𝑥= sinh 𝑢 would also work as 1+ sinh 2 𝑢 = cosh 2 𝑢 ?
𝑑𝑥 𝑑𝜃 = sec 2 𝜃 → 𝑑𝑥= sec 2 𝜃 𝑑𝜃 𝑥 2 𝑑𝑥 = tan 2 𝜃 sec 2 𝜃 𝑑𝜃 = 1 𝑑𝜃 =𝜃+𝐶 = arctan 𝑥 +𝐶

15. Dealing with 1/( 𝑎 2 − 𝑥 2 ) , 1/ 𝑎 2 − 𝑥 2 , ….
Sensible substitution and why? ?
𝑥=𝑎𝑠𝑖𝑛𝜃 (but 𝑥= acos 𝜃 would work fine)
The denominator changes from sin to cos but cancels with the cos obtained from differentiating the substitution.
1 𝑎 2 − 𝑥 2 𝑑𝑥 𝑥 <𝑎 ?
1 𝑎 2 + 𝑥 2 𝑑𝑥
𝑥= 𝑎tan 𝜃
As per example on previous slide.
𝑥= sinh 𝑢
tan wouldn’t work as well this time because the denominator would simplify to 𝑎 sec 𝜃 , but we’d be multiplying by 𝑎 sec 2 𝜃 , meaning not all the secs would cancel. With sinh 𝑢 the two cosh 𝑢 ’s obtained would fully cancel. ?
1 𝑎 2 + 𝑥 2 𝑑𝑥
1 𝑥 2 − 𝑎 2 𝑑𝑥 ?
𝑥= cosh 𝑢
sin 2 𝜃 + cos 2 𝜃 =1 1+ tan 2 𝜃 = sec 2 𝜃 1+ sinh 2 𝑢 = cosh 2 𝑢

16. Dealing with 1/( 𝑎 2 − 𝑥 2 ) , 1/ 𝑎 2 − 𝑥 2 , ….
Use an appropriate substitution to show that 𝑥 2 𝑑𝑥= 1 2 arctan 𝑥 2 +𝐶
Let 𝑥=2 tan 𝜃
𝑑𝑥 𝑑𝜃 =2 sec 2 𝜃 → 𝑑𝑥=2 sec 2 𝜃 𝑑𝜃
1 4+ 𝑥 2 𝑑𝑥 = tan 2 𝜃 2 sec 2 𝜃 𝑑𝜃 = sec 2 𝜃 4 sec 2 𝜃 𝑑𝜃 = 𝑑𝜃 = 1 2 𝜃+𝐶 = 1 2 arctan 𝑥 2 +𝐶 ?
! Standard results: (in formula booklet)
1 𝑎 2 − 𝑥 2 𝑑𝑥 = arcsin 𝑥 𝑎 +𝐶, 𝑥 <𝑎 𝑎 2 + 𝑥 2 𝑑𝑥 = 1 𝑎 arctan 𝑥 𝑎 +𝐶 𝑎 2 + 𝑥 2 𝑑𝑥 = arcsinℎ 𝑥 𝑎 +𝐶 𝑥 2 − 𝑎 2 𝑑𝑥 = arccosh 𝑥 𝑎 +𝐶, 𝑥>𝑎
Bro Note: Notice these are the same as results when 𝑎=1, except with 𝑥 replaced with 𝑥 𝑎
The one exception is 𝑎𝑟𝑐𝑡𝑎𝑛 one which is scaled by 1 𝑎

17. Dealing with 1/( 𝑎 2 − 𝑏 2 𝑥 2 ) , …. ? ? Find 1 25+9 𝑥 2 𝑑𝑥
= 𝑥 2 𝑑𝑥 = 1 9 × 𝑎𝑟𝑡𝑎𝑛 𝑥 𝐶 = arctan 3𝑥 5 +𝐶
= − − 𝑥 2 𝑑𝑥 = arcsin 2𝑥 − = 𝜋 12 − − 𝜋 12 = 𝜋 6 ?
Standard results: (in formula booklet)
1 𝑎 2 − 𝑥 2 𝑑𝑥 = arcsin 𝑥 𝑎 +𝐶, 𝑥 <𝑎 𝑎 2 + 𝑥 2 𝑑𝑥 = 1 𝑎 arctan 𝑥 𝑎 +𝐶 𝑎 2 + 𝑥 2 𝑑𝑥 = arcsinℎ 𝑥 𝑎 +𝐶 𝑥 2 − 𝑎 2 𝑑𝑥 = arccosh 𝑥 𝑎 +𝐶, 𝑥>𝑎

18. ? Using a seemingly-sensible-but-turns-out-rather-nasty substitution
Harder Example
Show that 𝑥 2 𝑑𝑥 = 1 2 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+ 1 2 𝑥 1+ 𝑥 2 +𝐶.
(Hint: Use a sensible substitution)
? Using a seemingly-sensible-but-turns-out-rather-nasty substitution
Trying 𝑥=tan⁡𝜃:
𝑑𝑥 𝑑𝜃 = sec 2 𝜃 → 𝑑𝑥= sec 2 𝜃 𝑑𝜃
1+ 𝑥 2 𝑑𝑥 = 𝑠𝑒𝑐𝜃 sec 2 𝜃 𝑑𝜃
Using integration by parts:
𝑢=𝑠𝑒𝑐𝜃 𝑑𝑣 𝑑𝑥 = sec 2 𝜃 𝑑𝑢 𝑑𝑥 =𝑠𝑒𝑐𝜃 tan 𝜃 𝑣=𝑡𝑎𝑛𝜃 sec 3 𝜃 𝑑𝜃 = sec 𝜃 tan 𝜃 − sec 𝜃 tan 2 𝜃 𝑑𝜃+𝐶 sec 3 𝜃 𝑑𝜃 = sec 𝜃 tan 𝜃 − sec 𝜃 ( sec 2 𝜃−1) 𝑑𝜃 +𝐶 sec 3 𝜃 𝑑𝜃 = sec 𝜃 tan 𝜃 − sec 3 𝜃 + 𝑠𝑒𝑐𝜃 +𝐶 sec 3 𝜃 𝑑𝜃 =𝑠𝑒𝑐𝜃 tan 𝜃 + ln sec 𝜃 + tan 𝜃 +𝐶 sec 3 𝜃 𝑑𝜃 = 1 2 sec𝜃 tan 𝜃 ln sec 𝜃 + tan 𝜃 +𝐶′ = 1 2 𝑥 1+ 𝑥 ln 𝑥 2 +𝑥 +𝐶′ = 1 2 𝑥 1+ 𝑥 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+𝐶′
? Using the other-possible-substitution-that-turns-out-much-more-pretty-yay
Trying 𝑥= sinh 𝑢 :
𝑑𝑥 𝑑𝑢 = cosh 𝑢 → 𝑑𝑥= cosh 𝑢 𝑑𝑢
1+ 𝑥 2 𝑑𝑥 = cosh 𝑢 × cosh 𝑢 𝑑𝑢 = cosh 2 𝑢 𝑑𝑢 = cosh 2𝑢 𝑑𝑢 = 1 2 𝑢 sinh 2𝑢 +𝐶 = 1 2 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+ 1 2 𝑥 1+ 𝑥 2 +𝐶

19. Test Your Understanding?
Using a hyperbolic substitution, evaluate 𝑥 𝑥 𝑑𝑥 ? ?
Using 𝑥=3 sinh 𝑢 yields

20. Exercise 4C
Unless a substitution is given or asked for, use the standard results. Give numerical answers to 3 sf. 1
Use the substitution 𝑥= atan 𝜃 to show that 𝑎 2 + 𝑥 2 𝑑𝑥 = 1 𝑎 arctan 𝑥 𝑎 +𝐶
Use the substitution 𝑥= cos 𝜃 to show that − 𝑥 2 𝑑𝑥 =− arccos 𝑥 +𝐶
Use suitable substitutions to find
3 4− 𝑥 2 𝑑𝑥 = 𝑥 2 −9 𝑑𝑥 =
4 5+ 𝑥 2 𝑑𝑥 = 𝑥 𝑑𝑥 = … 2 3

21. Integrating by Completing the Square
Determine 𝑥 2 −8𝑥+8 𝑑𝑥
By completing the square, we can then use one of the standard results.
= 𝑥 2 −8𝑥+8 𝑑𝑥 = 𝑥−4 2 −8 𝑑𝑥
Let 𝑢=𝑥−4 → 𝑑𝑢=𝑑𝑥
= 𝑢 2 −8 𝑑𝑢 = ln 𝑢− 8 𝑢 𝐶 = ln 𝑥−4−2 2 𝑥− 𝐶 ?
This is not in the standard form yet, but a simple substitution would make it so.
Standard results: (in formula booklet)
1 𝑎 2 − 𝑥 2 𝑑𝑥 = arcsin 𝑥 𝑎 +𝐶, 𝑥 <𝑎 𝑎 2 + 𝑥 2 𝑑𝑥 = 1 𝑎 arctan 𝑥 𝑎 +𝐶 𝑎 2 + 𝑥 2 𝑑𝑥 = arcsinℎ 𝑥 𝑎 +𝐶 𝑥 2 − 𝑎 2 𝑑𝑥 = arccosh 𝑥 𝑎 +𝐶, 𝑥>𝑎 𝑎 2 − 𝑥 2 𝑑𝑥 = 1 2𝑎 ln 𝑎+𝑥 𝑎−𝑥 +𝐶 𝑥 2 − 𝑎 2 𝑑𝑥 = 1 2𝑎 ln 𝑥−𝑎 𝑥+𝑎 +𝐶
These can be obtained using C4 techniques: splitting into partial fractions first.

22. Further Example ? Determine 1 12𝑥+2 𝑥 2 𝑑𝑥
2 𝑥 2 +12𝑥=2 𝑥 2 +6𝑥 =2 𝑥+3 2 −9
1 12𝑥+2 𝑥 2 𝑑𝑥 = 𝑥+3 2 −9 𝑑𝑥 = 𝑥+3 2 −9 𝑑𝑥
Let 𝑢=𝑥+3 → 𝑑𝑢=𝑑𝑥
= 𝑢 2 −9 𝑑𝑢 = 𝑎𝑟𝑐𝑜𝑠ℎ 𝑢 3 +𝐶 = 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 𝐶
Standard results: (in formula booklet)
1 𝑎 2 − 𝑥 2 𝑑𝑥 = arcsin 𝑥 𝑎 +𝐶, 𝑥 <𝑎 𝑎 2 + 𝑥 2 𝑑𝑥 = 1 𝑎 arctan 𝑥 𝑎 +𝐶 𝑎 2 + 𝑥 2 𝑑𝑥 = arcsinℎ 𝑥 𝑎 +𝐶 𝑥 2 − 𝑎 2 𝑑𝑥 = arccosh 𝑥 𝑎 +𝐶, 𝑥>𝑎 𝑎 2 − 𝑥 2 𝑑𝑥 = 1 2𝑎 ln 𝑎+𝑥 𝑎−𝑥 +𝐶 𝑥 2 − 𝑎 2 𝑑𝑥 = 1 2𝑎 ln 𝑥−𝑎 𝑥+𝑎 +𝐶

23. Test Your Understanding
? b
? c

24. Exercise 4D

25. Integration by Parts ? Determine 𝑎𝑟𝑡𝑎𝑛ℎ 𝑥 𝑑𝑥
Remember how you found ln 𝑥 𝑑𝑥 by expressing as 1× ln 𝑥 𝑑𝑥 and using integration by parts?
Yeah, do that.
Determine 𝑎𝑟𝑡𝑎𝑛ℎ 𝑥 𝑑𝑥 ?
𝑢=𝑎𝑟𝑡𝑎𝑛ℎ 𝑥 𝑑𝑣 𝑑𝑥 =1 𝑑𝑢 𝑑𝑥 = 1 1− 𝑥 𝑣=𝑥
𝑎𝑟𝑡𝑎𝑛ℎ 𝑥 𝑑𝑥 =𝑥 arctanh 𝑥 − 𝑥 1− 𝑥 2 𝑑𝑥 =𝑥 arctanh 𝑥 ln 1− 𝑥 2 +𝐶
Bro Exam Note: This has never specifically come up in an exam, but could be tested.

26. Test Your Understanding
Determine − arcsin 𝑥 𝑑𝑥 ?
arcsin 𝑥 𝑑𝑥 =𝑥 arcsin 𝑥 + 1− 𝑥 2 +𝐶
− arcsin 𝑥 𝑑𝑥 =0.279 (3𝑠𝑓)

27. Exercise 4E
Show that ∫𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 𝑑𝑥=𝑥 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥− 1+ 𝑥 2 +𝐶
Evaluate 0 1 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 𝑑𝑥, giving your answer to 3sf.
Using the substitution 𝑢=2𝑥+1 and the result in a, or otherwise, find arcsinh 2𝑥+1 𝑑𝑥
Show that 𝑎𝑟𝑐𝑡𝑎𝑛 3𝑥 𝑑𝑥=𝑥 arctan 𝑥 − 1 6 ln 1+9 𝑥 2 +𝐶
Show that 𝑎𝑟𝑐𝑜𝑠ℎ 𝑑𝑥 =𝑥 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥− 𝑥 2 −1 +𝐶
Hence show that 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 𝑑𝑥= ln − 3
Show that 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 𝑑𝑥=𝑥 arctan 𝑥 − 1 2 ln 1+ 𝑥 2 +𝐶
Hence show that − arctan 𝑥 𝑑𝑥 = −3 𝜋 12 − 1 2 ln 2 .
The curve 𝐶 has equation 𝑦=2 arctan 𝑥 . The region 𝑅 is enclosed by 𝐶, the 𝑦-axis, the line 𝑦=𝜋 and the line 𝑥=3.
Find the area of 𝑅 giving your answer to 3sf.
Evaluate a) arcsin 𝑥 𝑑𝑥 b) 𝑥 arctan 𝑥 𝑑𝑥
Using the result that 𝑦=𝑥 arcsec 𝑥 , then 𝑑𝑦 𝑑𝑥 = 1 𝑥 𝑥 2 −1 , show that
arcsec 𝑥 𝑑𝑥 =𝑥 arcsec 𝑥 − ln 𝑥+ 𝑥 2 −1 +𝐶 1 a b c 2 3 a b 4 a b c 5 6

28. Section C: Arc lengths and surface area Section B: Reduction Formulae
Lengths of a curve.
A technique for dealing with large powers in integration.
Surface area of volumes of revolution.
Using substitutions
Use 𝑥=𝑎 tan 𝜃 to find
1 9 𝑥 2 +6𝑥+5 𝑑𝑥
Completing the square
1 9 𝑥 2 +6𝑥+5 𝑑𝑥
Standard results
sinh 𝑥 𝑑𝑥
Section A: General Skills
By parts
0 1 𝑒 2𝑥 𝑠𝑖𝑛ℎ 𝑥 𝑑𝑥

29. Reduction Formulae
Integrate 𝑥 3 𝑒 𝑥 𝑑𝑥 using a tedious and stupid method. ?
𝒖= 𝒙 𝟑 𝒅𝒗 𝒅𝒙 = 𝒆 𝒙 𝒅𝒖 𝒅𝒙 =𝟑 𝒙 𝟐 𝒗= 𝒆 𝒙
𝒙 𝟑 𝒆 𝒙 𝒅𝒙= 𝒙 𝟑 𝒆 𝒙 − 𝟑 𝒙 𝟐 𝒆 𝒙 𝒅𝒙
We then have to determine 𝟑 𝒙 𝟐 𝒆 𝒙 𝒅𝒙. We can see that each time we apply integration by parts we reduce the power of 𝒙 by 1.
This obviously becomes hugely tedious if the power was even larger.
Ideally we want a way to express an integral with some power 𝑛 in terms of the same expression but with power 𝑛−1. We can then iteratively apply this recurrence more rapidly.

30. Reduction Formulae
Given that 𝐼 𝑛 = 𝑥 𝑛 𝑒 𝑥 𝑑𝑥 where 𝑛 is a positive integer:
a) Show that 𝐼 𝑛 = 𝑥 𝑛 𝑒 𝑥 −𝑛 𝐼 𝑛−1 , 𝑛≥1.
b) Find 𝑥 4 𝑒 𝑥 𝑑𝑥.
𝑢= 𝑥 𝑛 , 𝑑𝑣 𝑑𝑥 = 𝑒 𝑥
𝑑𝑢 𝑑𝑥 =𝑛 𝑥 𝑛−1 , 𝑣= 𝑒 𝑥 𝐼 𝑛 = 𝑥 𝑛 𝑒 𝑥 − 𝑛 𝑥 𝑛−1 𝑒 𝑥 𝑑𝑥 +𝐶 = 𝑥 𝑛 𝑒 𝑥 −𝑛 𝑥 𝑛−1 𝑒 𝑥 𝑑𝑥+𝐶 = 𝑥 𝑛 𝑒 𝑥 −𝑛 𝐼 𝑛−1 ? a b ?
𝐼 4 = 𝑥 4 𝑒 𝑥 −4 𝐼 = 𝑥 4 𝑒 𝑥 −4 𝑥 3 𝑒 𝑥 −3 𝐼 = 𝑥 4 𝑒 𝑥 −4 𝑥 3 𝑒 𝑥 +12 𝑥 2 𝑒 𝑥 −2 𝐼 = 𝑥 4 𝑒 𝑥 −4 𝑥 3 𝑒 𝑥 +12 𝑥 2 𝑒 𝑥 −24 𝑥 𝑒 𝑥 − 𝐼 = 𝑥 4 𝑒 𝑥 −4 𝑥 3 𝑒 𝑥 +12 𝑥 2 𝑒 𝑥 −24𝑥 𝑒 𝑥 𝑒 𝑥 𝑑𝑥 = 𝒙 𝟒 𝒆 𝒙 −𝟒 𝒙 𝟑 𝒆 𝒙 +𝟏𝟐 𝒙 𝟐 𝒆 𝒙 −𝟐𝟒𝒙 𝒆 𝒙 +𝟐𝟒 𝒆 𝒙 +𝑪

31. Further Example
Show that if 𝐼 𝑛 = 0 1 𝑥 𝑛 1−𝑥 𝑑𝑥 then 𝐼 𝑛 = 2𝑛 2𝑛+3 𝐼 𝑛−1 , 𝑛≥1
𝑢= 𝑥 𝑛 , 𝑑𝑣 𝑑𝑥 = 1−𝑥 𝑑𝑢 𝑑𝑥 =𝑛 𝑥 𝑛−1 , 𝑣=− −𝑥 3 2
𝐼 𝑛 = − 2 3 𝑥 𝑛 1−𝑥 𝑛 𝑥 𝑛−1 1−𝑥 𝑑𝑥 = 0− 𝑛 0 1 𝑥 𝑛−1 1−𝑥 = 2 3 𝑛 0 1 𝑥 𝑛−1 1−𝑥 1−𝑥 𝑑𝑥 = 2 3 𝑛 0 1 𝑥 𝑛−1 1−𝑥 𝑑𝑥− 2 3 𝑛 0 1 𝑥 𝑛 1−𝑥 𝑑𝑥 𝐼 𝑛 = 2𝑛 3 𝐼 𝑛−1 − 2𝑛 3 𝐼 𝑛 𝑛 3 𝐼 𝑛 = 2𝑛 3 𝐼 𝑛−1 𝐼 𝑛 = 2𝑛 2𝑛+3 𝐼 𝑛−1 ?
Since 𝐼 𝑛 contains 1−𝑥 , it would seem sensible to write 1−𝑥 in terms of this.

32. A Trig Example ? Given that 𝐼 𝑛 = 0 𝜋 2 sin 𝑛 𝑥 𝑑𝑥 , 𝑛≥0
Derive the reduction formula: 𝑛 𝐼 𝑛 = 𝑛−1 𝐼 𝑛−2 , 𝑛≥2
Deduce the values of i) 0 𝜋 2 sin 5 𝑥 𝑑𝑥 and 0 𝜋 2 sin 6 𝑥 𝑑𝑥 a ?
Using integration by parts on sin 𝑥 × sin 𝑛−1 𝑥 :
𝑢= sin 𝑛−1 𝑥 𝑑𝑣 𝑑𝑥 = sin 𝑥 𝑑𝑢 𝑑𝑥 = 𝑛−1 sin 𝑛−2 𝑥 cos 𝑥 𝑣=− cos 𝑥
𝐼 𝑛 = − sin 𝑛−1 cos 𝑥 𝜋 𝜋 2 𝑛−1 sin 𝑛−2 𝑥 cos 2 𝑥 𝑑𝑥 = 0−0 + 𝑛−1 0 𝜋 2 sin 𝑛−2 𝑥 1− sin 2 𝑥 𝑑𝑥 = 𝑛−1 0 𝜋 2 sin 𝑛−2 𝑥 𝑑𝑥 − 𝑛−1 0 𝜋 2 sin 𝑛 𝑥 𝑑𝑥 𝐼 𝑛 = 𝑛−1 𝐼 𝑛−2 − 𝑛−1 𝐼 𝑛 𝑛 𝐼 𝑛 = 𝑛−1 𝐼 𝑛−2
We only want sin so that we can eventually replace with 𝐼 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔

33. A Trig Example ? Given that 𝐼 𝑛 = 0 𝜋 2 sin 𝑛 𝑥 𝑑𝑥 , 𝑛≥0
Derive the reduction formula: 𝑛 𝐼 𝑛 = 𝑛−1 𝐼 𝑛−2 , 𝑛≥2
Deduce the values of i) 0 𝜋 2 sin 5 𝑥 𝑑𝑥 and 0 𝜋 2 sin 6 𝑥 𝑑𝑥 b ?
𝐼 𝑛 = 𝑛−1 𝑛 𝐼 𝑛− 𝐼 5 = 4 5 𝐼 3 = 𝐼 1 = 𝜋 2 sin 𝑥 𝑑𝑥 = 8 15
𝐼 6 = 𝐼 4 = 𝐼 2 = 𝐼 0 = 𝜋 = 5𝜋 32
Note that sin 0 𝑥 =1

34. Test Your Understanding? ?

35. Exercise 4F Given that 𝐼 𝑛 = 𝑥 𝑛 𝑒 𝑥 2 𝑑𝑥, 1
Show that 𝐼 𝑛 =2 𝑥 𝑛 𝑒 𝑥 2 −2𝑛 𝐼 𝑛−1 , 𝑛≥1
Hence find 𝑥 3 𝑒 𝑥 2 𝑑𝑥
Given that 𝐼 𝑛 = 1 𝑒 𝑥 ln 𝑥 𝑛 𝑑𝑥 , 𝑛∈ℕ,
Show that 𝐼 𝑛 = 𝑒 2 2 − 𝑛 2 𝐼 𝑛−2 , 𝑛∈ℕ
Hence show that 1 𝑒 𝑥 ln 𝑥 4 𝑑𝑥= 𝑒 2 −3 4
If 𝐼 𝑛 = 0 1 𝑥 𝑛 1−𝑥 𝑑𝑥, then 𝐼 𝑛 = 2𝑛 2𝑛+3 𝐼 𝑛−1 , 𝑛≥1.
Use this reduction formula to evaluate 𝑥+1 𝑥+2 1−𝑥 𝑑𝑥
Given that 𝐼 𝑛 = 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥 , where 𝑛 is a positive integer,
Show that 𝐼 𝑛 =− 𝑥 𝑛 𝑒 −𝑥 +𝑛 𝐼 𝑛−1 , 𝑛≥1
Find 𝑥 3 𝑒 −𝑥 𝑑𝑥
Evaluate 𝑥 4 𝑒 −𝑥 𝑑𝑥 , giving your answer in terms of 𝑒. 1 2 3 4

36. Exercise 4F 5 𝐼 𝑛 = tanh 𝑛 𝑥 𝑑𝑥 ,
By writing tanh 𝑛 𝑥 = tanh 𝑛−2 𝑥 tanh 2 𝑥 show that for 𝑛≥2, 𝐼 𝑛 = 𝐼 𝑛−2 − 1 𝑛−1 tanh 𝑛−1 𝑥
Find tanh 5 𝑥 𝑑𝑥
Show that 0 ln tanh 4 𝑥 𝑑𝑥 = ln 2 −
Given that tan 𝑛 𝑥 𝑑𝑥 = 1 𝑛−1 tan 𝑛−1 𝑥 − tan 𝑛−2 𝑥 𝑑𝑥
Find tan 4 𝑥 𝑑𝑥
Evaluate 0 𝜋 4 tan 5 𝑥 𝑑𝑥
Show that 0 𝜋 3 tan 6 𝑥 𝑑𝑥= − 𝜋 3
Given that 𝐼 𝑛 = 1 𝑎 ln 𝑥 𝑛 𝑑𝑥 , where 𝑎>1 is a constant,
Show that, for 𝑛≥1, 𝐼 𝑛 =𝑎 ln 𝑎 𝑛 −𝑛 𝐼 𝑛−1
Find the exact value of ln 𝑥 3 𝑑𝑥
Show that 1 𝑒 ln 𝑥 𝑑𝑥=5 53𝑒−144 6 7

37. Exercise 4F 8 9 10

38. Section C: Arc lengths and surface area Section B: Reduction Formulae
Lengths of a curve.
A technique for dealing with large powers in integration.
Surface area of volumes of revolution.
Using substitutions
Use 𝑥=𝑎 tan 𝜃 to find
1 9 𝑥 2 +6𝑥+5 𝑑𝑥
Completing the square
1 9 𝑥 2 +6𝑥+5 𝑑𝑥
Standard results
sinh 𝑥 𝑑𝑥
Section A: General Skills
By parts
0 1 𝑒 2𝑥 𝑠𝑖𝑛ℎ 𝑥 𝑑𝑥

39. Lengths of curves
You’re used to adding lots of infinitely small things to get an area or volume. We use 𝛿𝑥 to denote a small change, which becomes 𝑑𝑥 as 𝛿𝑥→0. 𝛿𝑥 𝑦 𝑎 𝑏 𝑥
𝛿𝑥 𝛿𝑥 𝛿𝑥 𝛿𝑥 𝛿𝑥
𝑦 1
𝑦 2 𝑎 𝑏 𝑥 𝑦
Area of each rectangle:
𝑦×𝛿𝑥
All rectangles:
Σ𝑦 𝛿𝑥
In limit as 𝛿𝑥→0:
𝑏 𝑎 𝑦 𝑑𝑥 ?
Each cylinder volume:
𝜋 𝑦 2 𝛿𝑥
In limit as 𝛿𝑥→0:
𝑏 𝑎 𝜋 𝑦 2 𝑑𝑥 ? ?
You also used a similar strategy in FP2 to get the area under a polar curve. ? ?

40. Lengths of curves
So what infinitely small things should we add this time for the length of a curve? 𝑦
We add together infinitely small straight lines/chords. We can use Pythagoras to get the length of each line. 𝛿𝑥 𝛿𝑦 𝑥
𝑥 𝑎
𝑥 𝑏
Each chord has length 𝛿 𝑥 2 +𝛿 𝑦 2 , thus summing these up and letting 𝛿𝑥→0:
𝑥 𝑎 𝑥 𝑏 𝑑 𝑥 2 +𝑑 𝑦 = 𝑥 𝑎 𝑥 𝑏 𝑑𝑦 𝑑𝑥 𝑑𝑥
We factored 𝑑𝑥 out so on the end.
The above is a tiny bit dodgy because you’d always integrate with respect to something, so the 𝑑𝑥 should have been on the end to start with. Another way of thinking about it is that integration is the opposite of differentiation. If 𝑠 is the length of the curve so that 𝑑𝑠= 𝑑 𝑥 2 +𝑑 𝑦 2 , then we’re integrating 𝑑𝑠 𝑑𝑥 with respect to 𝑥 to get to 𝑠, i.e. 𝑑𝑠 𝑑𝑥 𝑑𝑥=∫𝑑𝑠 , as we did above.

41. Lengths of curves ! The length 𝑠 of an curve is:
𝑠= 𝑥 𝑎 𝑥 𝑏 𝑑𝑦 𝑑𝑥 𝑑𝑥
Equivalently:
𝑠= 𝑦 𝑎 𝑦 𝑏 𝑑𝑥 𝑑𝑦 𝑑𝑦
For parametric equations:
𝑠= 𝑡 𝑎 𝑡 𝑏 𝑑𝑥 𝑑𝑡 + 𝑑𝑦 𝑑𝑡 𝑑𝑡
This is because the 𝑑𝑦 and 𝑑𝑥 are interchangeable in 𝑑 𝑠 2 =𝑑 𝑥 2 +𝑑 𝑦 2
Informally you can see that 𝑑𝑥 𝑑𝑡 + 𝑑𝑦 𝑑𝑡 𝑑𝑡 simplifies to 𝑑 𝑥 2 +𝑑 𝑦 2 , but more formally, we should use 𝑑𝑠 𝑑𝑡 𝑑𝑡 to prove.

42. Example
Find the exact length of the arc on the parabola with equation 𝑦= 1 2 𝑥 2 , from the origin to the point 𝑃 4,8 . ?
𝑦= 1 2 𝑥 2 → 𝑑𝑦 𝑑𝑥 =𝑥
𝑠= 𝑑𝑦 𝑑𝑥 𝑑𝑥= 𝑥 2 𝑑𝑥
As we saw earlier in the chapter, for 1+ 𝑥 2 we use the substitution 𝑥= sinh 𝑢
𝑑𝑥 𝑑𝑢 = cosh 𝑢 → 𝑑𝑥= cosh 𝑢 𝑑𝑢
𝑥 2 𝑑𝑥= 0 𝑎𝑟𝑠𝑖𝑛ℎ sinh 2 𝑢 cosh 𝑢 𝑑𝑢 = 0 𝑎𝑟𝑠𝑖𝑛ℎ sinh 2 𝑢 cosh 𝑢 𝑑𝑢 = 0 𝑎𝑟𝑠𝑖𝑛ℎ 4 cosh 2 𝑢 𝑑𝑢 = 0 𝑎𝑟𝑠𝑖𝑛ℎ cosh 2𝑢 2 𝑑𝑢
Continued on next slide… ?

43. Example
Find the exact length of the arc on the parabola with equation 𝑦= 1 2 𝑥 2 , from the origin to the point 𝑃 4,8 .
= 0 𝑎𝑟𝑠𝑖𝑛ℎ 4 cosh 2 𝑢 𝑑𝑢 = 0 𝑎𝑟𝑠𝑖𝑛ℎ cosh 2𝑢 2 𝑑𝑢 = 𝑢 sinh 2𝑢 0 𝑎𝑟𝑠𝑖𝑛ℎ 4 = 𝑢 sinh 𝑢 cosh 𝑢 0 𝑎𝑟𝑠𝑖𝑛ℎ 4 = 𝑢 sinh 𝑢 sinh 2 𝑢 0 𝑎𝑟𝑠𝑖𝑛ℎ 4 =…= 1 2 ln ?
The point of these two steps is so we can end up with sinh arcsinh … , e.g. sinh arcsinh 4 =4

44. Test Your Understanding? ?

45. Exercise 4G

46. Fun Fact
From Year 7 you know how to find the length of the curve if you have the arc of a circle. We expect 𝜋 2 for the diagram below, but can we obtain this result using the technique we’ve just learned?
𝑥 2 + 𝑦 2 =1 𝑦= 1− 𝑥 2 𝑑𝑦 𝑑𝑥 =− 𝑥 1− 𝑥 2 𝑠= 𝑥 2 1− 𝑥 𝑑𝑥 = − 𝑥 𝑑𝑥 = arcsin 𝑥 = 𝜋 2 −0= 𝜋 2 = ?
Only considering positive 𝑦 𝑦 1 𝑥 1
Standard results: (in formula booklet) 𝑎 2 − 𝑥 2 𝑑𝑥 = arcsin 𝑥 𝑎 +𝐶, 𝑥 <𝑎

47. Fun Fact
Surprisingly, it is not possible to find the exact length of a general ellipse.
𝑥 2 𝑎 𝑦 2 𝑏 2 =1 𝑦= 𝑎 2 𝑏 2 − 𝑏 2 𝑥 2 =𝑏 𝑎 2 − 𝑥 2 𝑑𝑦 𝑑𝑥 = −𝑏𝑥 𝑎 2 − 𝑥 2 𝑠= 𝑏 2 𝑥 2 𝑎 2 − 𝑥 𝑑𝑥
And sadly this does not integrate unless 𝑎=𝑏…  ?

48. Surface Area of Revolution
We saw we could use chords to get the area of a curve.
What if we want a slice of the surface area formed by revolving the curve 2𝜋 around the 𝑥-axis? 𝑦 𝛿𝑠 𝑦
𝑦+𝛿𝑦 𝑥 𝑎 𝑏
It’s the curved surface area of a frustum!
𝑙 2 𝛿𝑠 𝛿𝑥
𝑙 1
𝑦+𝛿𝑦 𝑦 𝛿𝑥

49. Surface Area of Revolution
Recall that 𝜋𝑟𝑙 is the curved surface area of a cone, where 𝑙 is the slant height.
If 𝑆 is the total surface area we have:
𝛿𝑆=𝜋 𝑦+𝛿𝑦 𝑙 2 −𝜋𝑦 𝑙 =𝜋𝑦 𝑙 2 − 𝑙 1 +𝜋 𝑙 2 𝛿𝑦 =𝜋𝑦𝛿𝑠+𝜋 𝑙 2 𝛿𝑦
But we don’t want the 𝑙 2 in our expression; we can eliminate using similar triangles:
𝑙 2 𝑙 1 = 𝑦+𝛿𝑦 𝑦 𝑙 2 𝑙 2 −𝛿𝑠 = 𝑦+𝛿𝑦 𝑦 𝑙 2 = 𝛿𝑠 𝑦+𝛿𝑦 𝛿𝑦
∴𝛿𝑆=𝜋𝑦𝛿𝑠+𝜋 𝛿𝑠 𝑦+𝛿𝑦 𝛿𝑦 𝛿𝑦 𝛿𝑆=𝜋 2𝑦+𝛿𝑦 𝛿𝑠
𝛿𝑆 𝛿𝑥 =𝜋 2𝑦+𝛿𝑦 𝛿𝑠 𝛿𝑥
𝑙 2 𝛿𝑠
𝑙 1
𝑦+𝛿𝑦 𝑦 𝛿𝑥
As 𝛿𝑥→0, 𝛿𝑦→0
We get 𝛿𝑆 𝛿𝑥 → 𝑑𝑆 𝑑𝑥 and 𝛿𝑠 𝛿𝑥 → 𝑑𝑠 𝑑𝑥
∴ 𝑑𝑆 𝑑𝑥 =2𝜋𝑦 𝑑𝑠 𝑑𝑥
Integrating both sides w.r.t 𝑥:
𝑆=∫2𝜋𝑦 𝑑𝑠 𝑑𝑥 𝑑𝑥 =∫2𝜋𝑦 𝑑𝑠
We want 𝛿𝑆 𝛿𝑥 because we’re later finding 𝑑𝑆 𝑑𝑥 𝑑𝑥

50. Surface Area of Revolution
𝑆=∫2𝜋𝑦 𝑑𝑠 𝑑𝑥 𝑑𝑥
We found earlier that 𝑑𝑠 𝑑𝑥 = 𝑑𝑦 𝑑𝑥 or 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 2
We therefore get the following formulae (from the formula booklet):

51. Example
The curve 𝐶 has equation 𝑦= 𝑥 3−𝑥 . The arc of the curve between the points with 𝑥-coordinates 1 and 3 is completely rotated about the 𝑥-axis. Find the area of the surface generated.
𝑦= 𝑥 − 1 3 𝑥 ⇒ 𝑑𝑦 𝑑𝑥 = 𝑥 − 1 2 − 𝑥 1 2
So 1+ 𝑑𝑦 𝑑𝑥 2 =…= 𝑥 − 𝑥
Therefore 𝑆=2𝜋 𝑥 − 1 3 𝑥 𝑥 − 𝑥 𝑑𝑥 =… = 16 9 𝜋
! Key strategy:
Since 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 2 , we wish to end ?

52. Further Example
The curve with parametric equations 𝑥=𝑡− sin 𝑡 , 𝑦=1− cos 𝑡 , from 𝑡=0 to 𝑡=2𝜋, is rotated through 360° about the 𝑥-axis. Find the area of the curve generated.
Because we’re about to square root this expression, we’d ideally like it as a single term. The trick for 1− cos 𝑡 is to use a half-angle formula:
cos 2𝑡 =1−2 sin 2 𝑡 ∴ cos 𝑡 =1−2 sin 2 𝑡 2
𝑑𝑥 𝑑𝑡 =1− cos 𝑡 , 𝑑𝑦 𝑑𝑡 = sin 𝑡 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 2 =…=2−2 cos 𝑡 =4 sin 2 𝑡 2 𝑆=2𝜋 0 2𝜋 1− cos 𝑡 2 sin 𝑡 2 𝑑𝑡 =2𝜋 0 2𝜋 2 sin 2 𝑡 sin 𝑡 𝑑𝑡 =2𝜋 0 2𝜋 4 sin 3 𝑡 𝑑𝑡 =8𝜋 0 2𝜋 sin 𝑡 − cos 2 𝑡 𝑑𝑡 =8𝜋 0 2𝜋 sin 𝑡 2 − sin 𝑡 cos 2 𝑡 2 𝑑𝑡 =8𝜋 −2 cos 𝑡 2 − cos 3 𝑡 𝜋 =…= 64 3 𝜋 ? ? ? ?
In general you can integrate sin 3 𝑥 and cos 3 𝑥 by splitting and using sin 2 𝑥 + cos 2 𝑥 ≡1. We saw this earlier in the chapter. ?
sin 2 𝑡 2 = − cos 𝑡
Use to get 1− cos 𝑡 as a single term. ? ? ? ? ? ?

53. Test Your Understanding? ?

54. Exercise 4H

55. Summary cheat sheet ? ? ? ? ? ? ? ? ? ? ? ? ? 𝒇(𝒙) 𝒇 𝒙 𝒅𝒙
Strategy/Notes
FmBk?
sinh 𝑥
cosh 𝑥 +𝐶
Use definition of sinh 𝑥
Yes
cosh 𝑥
sinh 𝑥 +𝐶
tanh 𝑥
ln cosh 𝑥 +𝐶
sech 2 𝑥
tanh 𝑥 +𝐶
(Same as non-hyp) No
𝑐𝑜𝑠𝑒𝑐 ℎ 2 𝑥
− coth 𝑥 +𝐶
sech 𝑥 tanh 𝑥
− sech 𝑥 +𝐶
(NOT same as non-hyp)
𝑐𝑜𝑠𝑒𝑐ℎ 𝑥 coth 𝑥
−𝑐𝑜𝑠𝑒𝑐ℎ 𝑥+𝐶
1 𝑎 2 − 𝑥 2
arcsin 𝑥 𝑎
Substitution 𝑥=𝑎 sin 𝜃 or 𝑥=𝑎 tanh 𝑢
1 𝑎 2 + 𝑥 2
1 𝑎 arctan 𝑥 𝑎
𝑥=𝑎 sinh 𝑢
1 𝑎 2 + 𝑥 2
𝑎𝑟𝑠𝑖𝑛ℎ 𝑥 𝑎
1 𝑥 2 − 𝑎 2
𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 𝑎 +𝐶
𝑥=𝑎 cosh 𝑢
1 𝑎 2 − 𝑥 2
1 2𝑎 ln 𝑎+𝑥 𝑎−𝑥 +𝐶
Partial Fractions
1 𝑥 2 − 𝑎 2
1 2𝑎 ln 𝑥−𝑎 𝑥+𝑎 +𝐶 ? ? ? ? ? ? ? ? ? ? ? ? ?

56. Summary cheat sheet ? ? ? ? ? ? ? 𝒇(𝒙) 𝒇 𝒙 𝒅𝒙 Strategy/Notes FmBk?
sinℎ 2 𝑥
1 4 𝑠𝑖𝑛ℎ 2𝑥− 1 2 𝑥+𝐶
Double angle formula for cosh 2𝑥 . No
tanh 2 𝑥
𝑥− tanh 𝑥 +𝐶
1+ tanh 2 𝑥 ≡ sech 2 𝑥
sech 𝑥
2 arctan 𝑒 𝑥 +𝐶
sech 𝑥 ≡ 1 cosh 𝑥 ≡ 2 𝑒 𝑥 𝑒 2𝑥 +1
Then use 𝑢= 𝑒 𝑥 .
𝑒 𝑥 sinh 𝑥
1 4 𝑒 2𝑥 −2𝑥 +𝐶
Do NOT use parts. Use def of sinh 𝑥 .
1− cos 𝑥
−2 2 cos 𝑥 +𝐶
cos 𝑥 ≡1−2 sin 𝑥
sin 3 𝑥
− cos 𝑥 cos 3 𝑥 +𝐶
sin 3 𝑥 ≡ sin 𝑥 sin 2 𝑥 ≡ sin 𝑥 1− cos 2 𝑥 ≡ sin 𝑥 − sin 𝑥 cos 2 𝑥
1+ 𝑥 2
1 2 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+ 1 2 𝑥 1+ 𝑥 2 +𝐶
Substitution 𝑥= sinh 𝑢 ? ? ? ? ? ? ?