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# P5 - hyperbolics watch

1. Hi,

I am unsure on the following questions, the first I have attempted with no success and the second I have an answer for:

1) using the substitution x = 3 + 3sinh(u) to show that:

∫Sqrt(x^2 - 6x + 18)dx = 9/2∫(1 + cosh(2u)) du

2) Hence find, in logarithm form, the value of:

7
∫Sqrt(x^2 - 6x + 18)dx
3

Many Thanks

Streety
2. 1) using the substitution x = 3 + 3sinh(u) to show that:

∫Sqrt(x^2 - 6x + 18)dx = 9/2∫(1 + cosh(2u)) du
x=3+3sinhu, (x^2-6x+18) = 9 + 9sinh^2 u + 18sinhucoshu - 18 - 18sinhu + 18
= 9 + 9sinh^2 u
= 9[1+sinh^2 u]
= 9[coshu]^2
SQR of the polynomial is hence 3coshu.

x=3+3sinhu
dx/du = 3coshu
dx = 3coshu du

Hence,
∫Sqrt(x^2 - 6x + 18)dx = ∫ 3coshu.3coshu du = ∫9(coshu)^2 du
cosh2u = 2cosh^2 u - 1
(coshu)^2 = 0.5(1+cosh2u)
∫9(coshu)^2 du = 9/2.(1+cosh2u)

2) Hence find, in logarithm form, the value of:

7
∫Sqrt(x^2 - 6x + 18)dx
3

Many Thanks

Streety
x=3+3sinhu
u = arsinh[[x-3]/3]
∫Sqrt(x^2 - 6x + 18)dx with limits x=7, x=3 becomes:
4.5∫(1+cosh2u) du with limits arsinh(4/3) and 0.
= 4.5[u + 0.5sinh2u] limits arsinh(4/3) and 0
= 4.5[arsinh(4/3) + 0.5sinh2(arsinh4/3)
3. 1)
x = 3 + 3sinh(u)
dx/du = 3cosh(u)

∫ sqrt[x^2 - 6x + 18] dx/du du = ∫ sqrt[9 + 18sinh(u) + 9sinh^2(u) - 18 - 18sinh(u) + 18] (3cosh(u)) du = ∫ sqrt[9 + 9sinh^2(u)] (2cosh(u)) du = ∫ 3sqrt[cosh^2(u)] 3cosh(u) du = 9 ∫ cosh^2(u) du
cosh(2u) = 2cosh^2(u) - 1 => cosh^2(u) = (cosh(2u) + 1)/2, so integral becomes:
(9/2) ∫ cosh(2u) + 1 du
as required.

b)
(9/2) ∫ cosh(2u) + 1 du = (9/4) ∫ [e^(2u) + e^(-2u)] du
The limits now become:
7 -> arsinh(4/3)
3 -> 0
4. Meh...
5. x = 3 + 3sinh(u)
dx = 3cosh(u) du

x² - 6x + 18
= 9 + 18sinh(u) + 9sinh²(u) - 18 - 18sinh(u) + 18
= 9 + 9sinh²(u)
= 9(1 + sinh²(u))
= 9cosh²(u)

I = int rt(x² - 6x + 18 ) dx
I = int rt(9cosh²(u)) 3cosh(u) du
I = int 9cosh²(u) du
I = int (9/2)(1 + cosh(2u)) du, using the identity cosh(2u) = 2cosh²(u) - 1
====================
6. Meh²...

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Updated: February 16, 2005
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