Hyperbolic differentiation questionWatch

#1
13. i) Show d/dy (0.5sinh(4y) + 4 sinh (2y) + 6y) = 16cosh^4 (y)
ii) given x = 2 sinh y, show sinh (2y) = 0.5x(x^2 + 4)^0.5
and sinh(4y) = 0.5x(x^2 + 2) (x^2 + 4)^0.5

I have done both of these

Part iii)
Use i) and ii) to show the integral of (x^2 + 4)^0.5 dx = 0.25x(x^2 + 10)(x^2 + 4)^0.5 + 6 arcsinh(0.5x) + C

I'm not really sure what to do, I've tried substituting 2sinh(2y)/x = (x^2 + 4)^0.5 but this doesn't get anywhere near the right answer. Any help?
0
4 months ago
#2
Is the question definitely integral of (x^2 + 4)^0.5?
Because looking at it, I think it should be (x^2 +4)^1.5
1
4 months ago
#3
(Original post by psycholeo)
13. i) Show d/dy (0.5sinh(4y) + 4 sinh (2y) + 6y) = 16cosh^4 (y)
ii) given x = 2 sinh y, show sinh (2y) = 0.5x(x^2 + 4)^0.5
and sinh(4y) = 0.5x(x^2 + 2) (x^2 + 4)^0.5

I have done both of these

Part iii)
Use i) and ii) to show the integral of (x^2 + 4)^0.5 dx = 0.25x(x^2 + 10)(x^2 + 4)^0.5 + 6 arcsinh(0.5x) + C

I'm not really sure what to do, I've tried substituting 2sinh(2y)/x = (x^2 + 4)^0.5 but this doesn't get anywhere near the right answer. Any help?
Something about part (iii) looks odd. Both my back of the envelope scrawl and Wolfram Alpha give the integral of (x^2 + 4)^0.5 as (x/2)(x^2 + 4)^(1/2) + 2arcsinh(x/2).
0
4 months ago
#4
(Original post by psycholeo)
13. i) Show d/dy (0.5sinh(4y) + 4 sinh (2y) + 6y) = 16cosh^4 (y)
ii) given x = 2 sinh y, show sinh (2y) = 0.5x(x^2 + 4)^0.5
and sinh(4y) = 0.5x(x^2 + 2) (x^2 + 4)^0.5

I have done both of these

Part iii)
Use i) and ii) to show the integral of (x^2 + 4)^0.5 dx = 0.25x(x^2 + 10)(x^2 + 4)^0.5 + 6 arcsinh(0.5x) + C

I'm not really sure what to do, I've tried substituting 2sinh(2y)/x = (x^2 + 4)^0.5 but this doesn't get anywhere near the right answer. Any help?
I think the trick is to use i) first for the integration then use ii) for the substitution

So you want to try and map
integral (x^2 + 4)^0.5 dx
onto cosh^4
Use the simple
cosh^2 - sinh^2 = 1
to get it in terms of cosh^2 = 1 + sinh^2
So
(x^2 + 4)^0.5 = 2cosh(y)
Also
dx = 2 cosh(y) dy
So
integral (x^2 + 4)^0.5 dx = integral 4cosh^2(y) dy.

We want the integrand to be 16 cosh^4 (square of what I've got), so I may be missing something? Possibly use a half angle formula ?

Not evaluated fully ... and the previous post seems to suggest the expression isn't right?
Last edited by mqb2766; 4 months ago
0
#5
Sorry it is to the power of 3/2. Just really looked like 1/2. How would i do it if it was ^1.5
0
4 months ago
#6
Start by substituting in x = 2 sinh y. Then simplify this using cosh^2 - sinh^2 = 1 and try working through from there

Hint:
Spoiler:
Show
you are aiming to integrate 16cos^4(y)
0
#7
Okay so I could only get (4cos^2(y))^1.5

this is my method:

integrate (x^2 + 4)^1.5

sub in x = 2 sinh y

therefore, integrate (4sinh^2(y) + 4)^1.5
Sub in cosh^2 - sinh^2 = 1
so sinh^2 = cosh^2 - 1
therefore integrate (4(cosh^2(y) -1) + 4)^1.5
therefore integrate (4cosh^2(y))^1.5
therefore integrate 8cosh^3(y)

How are you supposed to get 16cosh^4(y)???
0
4 months ago
#8
(Original post by psycholeo)
Okay so I could only get (4cos^2(y))^1.5

this is my method:

integrate (x^2 + 4)^1.5

sub in x = 2 sinh y

therefore, integrate (4sinh^2(y) + 4)^1.5
Sub in cosh^2 - sinh^2 = 1
so sinh^2 = cosh^2 - 1
therefore integrate (4(cosh^2(y) -1) + 4)^1.5
therefore integrate (4cosh^2(y))^1.5
therefore integrate 8cosh^3(y)

How are you supposed to get 16cosh^4(y)???
You haven't accounted for the substitution ().
0
4 months ago
#9
(Original post by psycholeo)
Sorry it is to the power of 3/2. Just really looked like 1/2. How would i do it if it was ^1.5
Think I got it just about there apart in post 4 apart from the wrong power. You should be able to fiddle it.
0
#10
(Original post by DFranklin)
You haven't accounted for the substitution ().
I don't get what you mean, when should I multiply by that?
0
4 months ago
#11
When changing the variable you are integrating with respect to (from x to y in this case), you need to multiply the integrand by dx/dy to integrate with respect to y instead of x.

Integral (x2 + 4)^1.5 dx = Integral (4sinh^2(y) + 4)^1.5 dx/dy dy

Clearly, dx/dy = 2cosh(y)
0
4 months ago
#12
(Original post by psycholeo)
I don't get what you mean, when should I multiply by that?
Not to be unkind, but if you don't know, you need to review integration by substitution.
0
#13
(Original post by DFranklin)
Not to be unkind, but if you don't know, you need to review integration by substitution.
I haven't learnt it yet. That's single and we were given this exercise in Further
0
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