1. Hyperbolic functions

2. FM Hyperbolic functions: Calculus
KUS objectives
BAT differentiate and integrate with hyperbolic functions
Starter: integrate
=− 1 3 c𝑜𝑠 3𝑥+2 +𝐶
sin (3𝑥+2)
= 1 4 sin 4𝑥 +𝐶
cos 4𝑥
=2 tan 𝑥 2 +𝐶
𝑠𝑒𝑐 2 𝑥 2

3. d dx sinh 𝑥 = cosh 𝑥 sinh 𝑥 𝑑𝑥= cosh 𝑥 +𝐶 𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥
Notes
d dx sinh 𝑥 = cosh 𝑥
𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥
d dx tanh 𝑥 = 𝑠𝑒𝑐ℎ 2 𝑥
sinh 𝑥 𝑑𝑥= cosh 𝑥 +𝐶
cosh 𝑥 𝑑𝑥= sinh 𝑥 +𝐶
d dx arsinh 𝑥 = 1 𝑥 2 +1
𝑑 𝑑𝑥 arcosh 𝑥 = 1 𝑥 2 −1 ,
d dx artanh 𝑥 = 1 1− 𝑥 2 ,
𝑥>1
𝑥 <1
1 𝑥 𝑑𝑥= 𝑎𝑟𝑠𝑖𝑛h 𝑥 +𝐶
1 𝑥 2 −1 , 𝑑𝑥= 𝑎𝑟𝑐𝑜𝑠h 𝑥 +𝐶 ,

4. WB E1 find these integrals
𝑎) cosh (4𝑥−1) 𝑑𝑥 b) 𝑥 𝑥 𝑑𝑥 by splitting to two integrals
- one a standard, the other using substitution 𝒖= 𝑥 /2
𝑎) = 1 4 sinh (4𝑥−1) +𝐶
d dx arsinh 𝑥 = 1 𝑥 2 +1
𝑑 𝑑𝑥 arcosh 𝑥 = 1 𝑥 2 −1 ,
d dx artanh 𝑥 = 1 1− 𝑥 2 ,
𝑏) = 𝑥 𝑑𝑥+ 5𝑥 𝑥 𝑑𝑥
Substitution
𝒖= 𝑥 /2
𝒅𝒖 𝒅𝒙 = 𝟏 𝟐 𝑥 −1/2 ×2𝑥
𝒅𝒖=𝒙 𝑥 −1/2 𝒅𝒙
= 𝑥 𝑑𝑥+5 𝑥 𝑥 −1/2 𝑑𝑥
= 𝑥 𝑑𝑥 𝑑𝑢
=2 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+5 𝑥 /2 +𝐶

5. WB E2 find these integrals
𝑎) cosh 5 2𝑥 sinh 2𝑥 𝑑𝑥 b) tanh 𝑥 𝑑𝑥 (a standard integral)
𝑎) = cosh 5 2𝑥 2sinh 2𝑥 𝑑𝑥
Substitution
𝒖= cosh 2𝑥
𝒅𝒖 𝒅𝒙 =𝟐𝒔𝒊𝒏𝒉 𝟐𝒙
𝒅𝒖=𝟐 𝒔𝒊𝒏𝒉 𝟐𝒙 𝒅𝒙
= u 5 𝑑𝑢
= u 6 +𝐶
= cosh 6 2𝑥+𝐶
Substitution
𝒖= cosh 𝑥
𝒅𝒖 𝒅𝒙 =𝒔𝒊𝒏𝒉 𝒙
𝒅𝒖= 𝒔𝒊𝒏𝒉 𝒙 𝒅𝒙
𝑎) = sinh 𝑥 cosh 𝑥 𝑑𝑥
= 𝑢 𝑑𝑢
= ln 𝑢 +𝐶
= ln cosh 𝑥 +𝐶
tanh 𝑥 𝑑𝑥= ln cosh 𝑥 +𝐶

6. WB E3 find these integrals 𝑎) cosh 2 3𝑥 𝑑𝑥 b) sinh 3 3𝑥 𝑑𝑥
Identity
cosh 2𝑥 =2 𝑐𝑜𝑠ℎ 2 𝑥−𝟏
𝑐𝑜𝑠ℎ 2 𝑥= 𝟏 𝟐 𝒄𝒐𝒔𝒉 𝟐𝒙+𝟏
= sinh 6𝑥 +𝑥 +𝐶
= sinh 6𝑥 𝑥+𝐶
𝑏) = sinh 2 3𝑥 sinh 𝑥 𝑑𝑥
Substitution
𝒖= cosh 3𝑥
𝒅𝒖 𝒅𝒙 =𝟑𝒔𝒊𝒏𝒉 𝟑𝒙
𝟏 𝟑 𝒅𝒖= 𝒔𝒊𝒏𝒉 𝒙 𝒅𝒙
𝒖 𝟐 𝒅𝒖= 𝟏 𝟑 𝒖 𝟑
= 𝑐𝑜𝑠ℎ 2 3𝑥−𝟏 sinh 3𝑥 𝑑𝑥
= 𝑐𝑜𝑠ℎ 2 3𝑥 sinh 3𝑥 𝑑𝑥 − sinh 3𝑥 𝑑𝑥
=…= 1 9 𝑐𝑜𝑠ℎ 3 3𝑥− 1 3 cosh 3𝑥 +𝐶

7. WB E4 find these integrals 𝑎) e 2𝑥 sinh 𝑥 𝑑𝑥 b) −1/2 1/2 cosh 𝑥 𝑒 𝑥 𝑑𝑥
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
𝑎) = 𝑒 2𝑥 𝑒 𝑥 − 𝑒 −𝑥 2 𝑑𝑥
= 𝑒 3𝑥 − 𝑒 𝑥 𝑑𝑥
= 𝑒 3𝑥 3 − 𝑒 𝑥 +𝐶
= 𝑒 3𝑥 −3 𝑒 𝑥 +𝐶
𝑏) = −1/2 1/2 𝑒 𝑥 + 𝑒 −𝑥 2 𝑒 𝑥 𝑑𝑥
= −1/2 1/ 𝑒 −2𝑥 𝑑𝑥
= 𝑥 − 1 4 𝑒 −2𝑥 1/2 −1/2
= − 1 4 𝑒 −1 − − 1 4 − 1 4 𝑒 1
= 1 2 − 1 4𝑒 + 𝑒 4

8. WB E5a 𝑎) find 1 𝑥 2 − 𝑎 2 𝑑𝑥 using the substitution 𝒙= a cosh 𝑢
b) 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑥 2 −16 𝑑𝑥= ln
𝑎) = 𝑎 2 𝑐𝑜𝑠ℎ 2 𝑢 − 𝑎 𝑎 𝑠𝑖𝑛ℎ 𝑢 𝑑𝑢
Substitution
𝒙= a cosh 𝑢
𝒅𝒙 𝒅𝒖 =𝒂 𝒔𝒊𝒏𝒉 𝒖
𝒅𝒙=𝒂 𝒔𝒊𝒏𝒉 𝒖 𝒅𝒖
𝒙 𝟐 = 𝑎 2 𝑐𝑜𝑠ℎ 2 𝑢
= 𝑎 2 𝑠𝑖𝑛ℎ 2 𝑢 𝑎 𝑠𝑖𝑛ℎ 𝑢 𝑑𝑢
= 𝑑𝑢
= arcosh 𝑥 𝑎 +𝐶
=𝑢+𝐶
𝑢= arcosh 𝑥 𝑎
1 𝑥 2 − 𝑎 2 𝑑𝑥= arcosh 𝑥 𝑎 +𝐶
Also
1 𝑥 2 + 𝑎 2 𝑑𝑥= arsinℎ 𝑥 𝑎 +𝐶
Identity
𝒔𝒊𝒏𝒉 𝟐 𝒙= 𝒄𝒐𝒔𝒉 𝟐 𝒙−𝟏
𝒂 𝟐 𝒔𝒊𝒏𝒉 𝟐 𝒙= 𝒂 𝟐 𝒄𝒐𝒔𝒉 𝟐 𝒙− 𝒂 𝟐

9. 𝑎) find 1 𝑥 2 − 𝑎 2 𝑑𝑥 b) 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 5 8 1 𝑥 2 −16 𝑑𝑥= ln 2+ 3 2
WB E5b
𝑎) find 𝑥 2 − 𝑎 2 𝑑𝑥 b) 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑥 2 −16 𝑑𝑥= ln
𝑏) = 𝑥 2 − 4 2
sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2
cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2
tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1
arsinℎ 𝑥 = ln 𝑥+ 𝑥 2 +1
arcosh 𝑥 = ln 𝑥+ 𝑥 2 −1
= arcosh 𝑥
=𝑎𝑟𝑐𝑜𝑠ℎ 2−arcosh 5 4
= ln −1 − ln −1
= ln − ln 2
= ln QED

10. WB E6 use the substitution 𝒙= sinh 𝑢 to
𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑥 2 𝑑𝑥 = 1 2 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+ 1 2 𝑥 1+ 𝑥 2 +𝐶
Substitution
𝒙= sinh 𝑢
𝒅𝒙 𝒅𝒖 =𝒄𝒐𝒔𝒉 𝒖
𝒅𝒙=𝒄𝒐𝒔𝒉 𝒖 𝒅𝒖
1+ 𝑥 2 𝑑𝑥= 𝑠𝑖𝑛ℎ 2 𝑥 cosh 𝑢 𝑑𝑢c
= 𝑐𝑜𝑠ℎ 2 𝑢 𝑑𝑢
= cosh 2𝑢 𝑑𝑢
𝒖= 𝑎𝑟sinh 𝑥
= 1 2 𝑢 sinh 2𝑢 +C
𝒄𝒐𝒔𝒉 𝒖= 𝟏+ 𝒔𝒊𝒏𝒉 𝟐 𝒖
= 𝟏+ 𝒙 𝟐
= 1 2 𝑢 sinh 𝑢 cosh 𝑢 +C
Identities
𝒔𝒊𝒏𝒉 𝟐 𝒙= 𝒄𝒐𝒔𝒉 𝟐 𝒙−𝟏
𝒄𝒐𝒔𝒉 𝟐𝒙= 𝟐 𝒄𝒐𝒔𝒉 𝟐 𝒙−𝟏
𝒔𝒊𝒏𝒉 𝟐𝒙=𝟐𝒔𝒊𝒏𝒉 𝒙 𝒄𝒐𝒔𝒉 𝒙
= 1 2 𝑎𝑟𝑠𝑖𝑛ℎ 𝑥+ 1 2 𝑥 𝟏+ 𝒙 𝟐 +C

11. WB E7 use the hyperbolic substitution 𝑥=3 sinh 𝑢 to
evaluate 𝑥 𝑥 𝑑𝑥
𝑥 𝑥 𝑑𝑥= 𝑠𝑖𝑛ℎ 3 𝑢 𝑠𝑖𝑛ℎ 2 𝑥 cosh 𝑢 𝑑𝑢
Substitution
𝒙= 3sinh 𝑢
𝒅𝒙 𝒅𝒖 =𝟑𝒄𝒐𝒔𝒉 𝒖
𝒅𝒙=𝟑𝒄𝒐𝒔𝒉 𝒖 𝒅𝒖
𝒙 𝟑 =𝟐𝟕 𝑠𝑖𝑛ℎ 3 𝑢
= 𝑠𝑖𝑛ℎ 3 𝑢 𝑐𝑜𝑠ℎ 2 𝑥 cosh 𝑢 𝑑𝑢
= 𝑠𝑖𝑛ℎ 3 𝑢 𝑑𝑢
= 𝑐𝑜𝑠ℎ 2 𝐮−𝟏 𝑠𝑖𝑛ℎ 𝑢 𝑑𝑥
= 𝑐𝑜𝑠ℎ 2 𝑢 sinh 𝑢 𝑑𝑢 − sinh 𝑢 𝑑𝑥
= 𝑐𝑜𝑠ℎ 3 𝑢− cosh 𝑢 +𝐶

12. WB E7 (cont) use the hyperbolic substitution 𝑥=3 sinh 𝑢 to
evaluate 𝑥 𝑥 𝑑𝑥
𝑥 𝑥 𝑑𝑥= 𝑐𝑜𝑠ℎ 3 𝑢− cosh 𝑢 +𝐶
Substitution
𝒙= 3sinh 𝑢
𝒅𝒙 𝒅𝒖 =𝟑𝒄𝒐𝒔𝒉 𝒖
𝒅𝒙=𝟑𝒄𝒐𝒔𝒉 𝒖 𝒅𝒖
𝒙 𝟑 =𝟐𝟕 𝑠𝑖𝑛ℎ 3 𝑢
0 6 𝑥 𝑥 𝑑𝑥
= 𝑐𝑜𝑠ℎ 3 𝑢− cosh 𝑢 𝑎𝑟𝑠𝑖𝑛ℎ 2 0
= − 5 − 1 3 −1
Bounds
𝒙= 3sinh 𝑢
𝒊𝒇 𝒙=𝟔 =𝟑𝒔𝒊𝒏𝒉 𝒖
𝒖=𝒂𝒓𝒔𝒊𝒏𝒉 𝟐
𝒊𝒇 𝒙=𝟎 =𝟑𝒔𝒊𝒏𝒉 𝒖
𝒖=𝒂𝒓𝒔𝒊𝒏𝒉 𝟎=𝟎
=𝟏𝟖 𝟓 +𝟏𝟖

13. WB E8 a) show that 1 12𝑥+2 𝑥 2 = 1 2 𝑢 2 −9 using a suitable substitution
b) Hence find −𝜋 𝑥+2 𝑥 𝑑𝑥
1 12𝑥+2 𝑥 2 = 𝑥 2 +6𝑥
= (𝑥+3) 2 −9
= 𝑢 2 −9
a) First
Use 𝑢=𝑥+3
−𝜋 𝑥+2 𝑥 𝑑𝑥= 𝑢(−𝜋) 𝑢(0) 𝑢 2 −9 𝑑𝑥 b)
= 𝑎𝑟𝑐𝑜𝑠ℎ 𝑢 3
= 𝑎𝑟𝑐𝑜𝑠ℎ 𝑥 −𝜋
= −1.618 =1.144

14. WB E9
Use the substitution 𝑥= cosh 𝑢 to find 𝑥 2 −12𝑥−7 𝑑𝑥
4 𝑥 2 −12𝑥−7 = cosh 𝑢 2 − cosh 𝑢 −7
Substitution
𝒙= 𝟏 𝟐 𝟑+𝟒𝒄𝒐𝒔𝒉 𝒖
𝒅𝒙 𝒅𝒖 =𝟐 𝒔𝒊𝒏𝒉 𝒖
𝒅𝒙=𝟐𝒔𝒊𝒏𝒉 𝒖 𝒅𝒖
= cosh 𝑢+16 𝑐𝑜𝑠ℎ 2 𝑢 −18−24𝑐𝑜𝑠ℎ𝑢−7
= 16 𝑐𝑜𝑠ℎ 2 𝑢−16
=4 𝑐𝑜𝑠ℎ 2 𝑢−1 =𝟒 𝐬𝐢𝐧𝐡 𝐮
1 4 𝑥 2 −12𝑥−7 𝑑𝑥= sinh 𝑢 2𝑠𝑖𝑛ℎ 𝑢 𝑑𝑢
𝒖=𝒂𝒓𝒄𝒐𝒔 𝟐𝒙−𝟑 𝟒
= 𝑑𝑢 = 1 2 𝑢+𝐶
= 𝑑𝑢 = 1 2 𝑢+𝐶
= 1 2 𝑢+𝐶
= 1 2 𝒂𝒓𝒄𝒐𝒔 𝟐𝒙−𝟑 𝟒 +𝐶
NOW DO EX 6E

15. One thing to improve is –
KUS objectives
BAT differentiate and integrate with hyperbolic functions
self-assess
One thing learned is –
One thing to improve is –

16. END