Hyperbolic functions

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

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 2 +1 𝑑𝑥= 𝑎𝑟𝑠𝑖𝑛h 𝑥 +𝐶 1 𝑥 2 −1 , 𝑑𝑥= 𝑎𝑟𝑐𝑜𝑠h 𝑥 +𝐶 ,

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

WB E2 find these integrals 𝑎) cosh 5 2𝑥 sinh 2𝑥 𝑑𝑥 b) tanh 𝑥 𝑑𝑥 (a standard integral) 𝑎) = 1 2 cosh 5 2𝑥 2sinh 2𝑥 𝑑𝑥 Substitution 𝒖= cosh 2𝑥 𝒅𝒖 𝒅𝒙 =𝟐𝒔𝒊𝒏𝒉 𝟐𝒙 𝒅𝒖=𝟐 𝒔𝒊𝒏𝒉 𝟐𝒙 𝒅𝒙 = 1 2 u 5 𝑑𝑢 = 1 12 u 6 +𝐶 = 1 12 cosh 6 2𝑥+𝐶 Substitution 𝒖= cosh 𝑥 𝒅𝒖 𝒅𝒙 =𝒔𝒊𝒏𝒉 𝒙 𝒅𝒖= 𝒔𝒊𝒏𝒉 𝒙 𝒅𝒙 𝑎) = sinh 𝑥 cosh 𝑥 𝑑𝑥 = 1 𝑢 𝑑𝑢 = ln 𝑢 +𝐶 = ln cosh 𝑥 +𝐶 tanh 𝑥 𝑑𝑥= ln cosh 𝑥 +𝐶

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 = 1 2 1 6 sinh 6𝑥 +𝑥 +𝐶 = 1 12 sinh 6𝑥 + 1 2 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

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 = 1 2 𝑒 3𝑥 − 𝑒 𝑥 𝑑𝑥 = 1 2 𝑒 3𝑥 3 − 𝑒 𝑥 +𝐶 = 1 6 𝑒 3𝑥 −3 𝑒 𝑥 +𝐶 𝑏) = −1/2 1/2 𝑒 𝑥 + 𝑒 −𝑥 2 𝑒 𝑥 𝑑𝑥 = −1/2 1/2 1 2 + 1 2 𝑒 −2𝑥 𝑑𝑥 = 1 2 𝑥 − 1 4 𝑒 −2𝑥 1/2 −1/2 = 1 4 − 1 4 𝑒 −1 − − 1 4 − 1 4 𝑒 1 = 1 2 − 1 4𝑒 + 𝑒 4

WB E5a 𝑎) find 1 𝑥 2 − 𝑎 2 𝑑𝑥 using the substitution 𝒙= a cosh 𝑢 b) 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 5 8 1 𝑥 2 −16 𝑑𝑥= ln 2+ 3 2 𝑎) = 1 𝑎 2 𝑐𝑜𝑠ℎ 2 𝑢 − 𝑎 2 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 = 1 𝑎 2 𝑠𝑖𝑛ℎ 2 𝑢 𝑎 𝑠𝑖𝑛ℎ 𝑢 𝑑𝑢 = 1 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

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 1 𝑥 2 − 𝑎 2 𝑑𝑥 b) 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 5 8 1 𝑥 2 −16 𝑑𝑥= ln 2+ 3 2 𝑏) = 5 8 1 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 4 8 5 =𝑎𝑟𝑐𝑜𝑠ℎ 2−arcosh 5 4 = ln 2+ 2 2 −1 − ln 5 4 + 5 4 2 −1 = ln 2+ 3 − ln 2 = ln 2+ 3 2 QED

WB E6 use the substitution 𝒙= sinh 𝑢 to 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 1+ 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= 1+ 𝑠𝑖𝑛ℎ 2 𝑥 cosh 𝑢 𝑑𝑢c = 𝑐𝑜𝑠ℎ 2 𝑢 𝑑𝑢 = 1 2 + 1 2 cosh 2𝑢 𝑑𝑢 𝒖= 𝑎𝑟sinh 𝑥 = 1 2 𝑢+ 1 4 sinh 2𝑢 +C 𝒄𝒐𝒔𝒉 𝒖= 𝟏+ 𝒔𝒊𝒏𝒉 𝟐 𝒖 = 𝟏+ 𝒙 𝟐 = 1 2 𝑢+ 1 4 2 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

WB E7 use the hyperbolic substitution 𝑥=3 sinh 𝑢 to evaluate 0 6 𝑥 3 𝑥 2 +9 𝑑𝑥 𝑥 3 𝑥 2 +9 𝑑𝑥= 27 𝑠𝑖𝑛ℎ 3 𝑢 9 𝑠𝑖𝑛ℎ 2 𝑥+9 3cosh 𝑢 𝑑𝑢 Substitution 𝒙= 3sinh 𝑢 𝒅𝒙 𝒅𝒖 =𝟑𝒄𝒐𝒔𝒉 𝒖 𝒅𝒙=𝟑𝒄𝒐𝒔𝒉 𝒖 𝒅𝒖 𝒙 𝟑 =𝟐𝟕 𝑠𝑖𝑛ℎ 3 𝑢 = 27 𝑠𝑖𝑛ℎ 3 𝑢 3 𝑐𝑜𝑠ℎ 2 𝑥 3cosh 𝑢 𝑑𝑢 =27 𝑠𝑖𝑛ℎ 3 𝑢 𝑑𝑢 =27 𝑐𝑜𝑠ℎ 2 𝐮−𝟏 𝑠𝑖𝑛ℎ 𝑢 𝑑𝑥 =27 𝑐𝑜𝑠ℎ 2 𝑢 sinh 𝑢 𝑑𝑢 −27 sinh 𝑢 𝑑𝑥 =27 1 3 𝑐𝑜𝑠ℎ 3 𝑢− cosh 𝑢 +𝐶

WB E7 (cont) use the hyperbolic substitution 𝑥=3 sinh 𝑢 to evaluate 0 6 𝑥 3 𝑥 2 +9 𝑑𝑥 𝑥 3 𝑥 2 +9 𝑑𝑥=27 1 3 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 3 𝑥 2 +9 𝑑𝑥 =27 1 3 𝑐𝑜𝑠ℎ 3 𝑢− cosh 𝑢 𝑎𝑟𝑠𝑖𝑛ℎ 2 0 =27 5 5 3 − 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

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

WB E9 Use the substitution 𝑥= 1 2 3+4 cosh 𝑢 to find 1 4 𝑥 2 −12𝑥−7 𝑑𝑥 4 𝑥 2 −12𝑥−7 = 4 1 2 3+4 cosh 𝑢 2 − 12 2 3+4 cosh 𝑢 −7 Substitution 𝒙= 𝟏 𝟐 𝟑+𝟒𝒄𝒐𝒔𝒉 𝒖 𝒅𝒙 𝒅𝒖 =𝟐 𝒔𝒊𝒏𝒉 𝒖 𝒅𝒙=𝟐𝒔𝒊𝒏𝒉 𝒖 𝒅𝒖 = 9+24 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= 1 4 sinh 𝑢 2𝑠𝑖𝑛ℎ 𝑢 𝑑𝑢 𝒖=𝒂𝒓𝒄𝒐𝒔 𝟐𝒙−𝟑 𝟒 = 1 2 𝑑𝑢 = 1 2 𝑢+𝐶 = 1 2 𝑑𝑢 = 1 2 𝑢+𝐶 = 1 2 𝑢+𝐶 = 1 2 𝒂𝒓𝒄𝒐𝒔 𝟐𝒙−𝟑 𝟒 +𝐶 NOW DO EX 6E

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 –

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