# hyperbolic calculus helpWatch

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#1
integral of root(x^2 - 4) using the substitution x = 2coshu
dx/du = 2sinhu so dx = 2sinhu du

now we have root(x^2-4)(2sinhu)du
factoring in u: ((2coshu)^2 -4)(2sinhu)du

the 2coshu squared expands to 4cosh^2 u
so we have root(4coshu^2 - 4)
for some reason the answer turns that into 2root(cosh^u -1), i dont understand where im going wrong?
https://imgur.com/a/o3R2zPj
thanks, also dont have a clue what they are doing to get 2sinhucoshu - 2u ?
0
2 weeks ago
#2
(Original post by Gent2324)
integral of root(x^2 - 4) using the substitution x = 2coshu
dx/du = 2sinhu so dx = 2sinhu du

now we have root(x^2-4)(2sinhu)du
factoring in u: ((2coshu)^2 -4)(2sinhu)du

the 2coshu squared expands to 4cosh^2 u
so we have root(4coshu^2 - 4)
for some reason the answer turns that into 2root(cosh^u -1), i dont understand where im going wrong?
https://imgur.com/a/o3R2zPj
thanks, also dont have a clue what they are doing to get 2sinhucoshu - 2u ?
Is the question just factoring the 4 out of the root sign to give 2?
0
2 weeks ago
#3
(Original post by Gent2324)
thanks, also dont have a clue what they are doing to get 2sinhucoshu - 2u ?
sinh 2x = 2 sinhx coshx is a well-known identity, the hyperbolic equivalent of sin 2x = 2 sinx cosx.

It's easily proven by going back to the definitions of sinh and cosh in terms of exp(x).
Last edited by MarkFromWales; 2 weeks ago
0
#4
(Original post by mqb2766)
Is the question just factoring the 4 out of the root sign to give 2?
(Original post by MarkFromWales)
sinh 2x = 2 sinhx coshx is a well-known identity, the hyperbolic equivalent of sin 2x = 2 sinx cosx.

It's easily proven by going back to the definitions of sinh and cosh in terms of exp(x).
ah ok im fine up to there now i see that they have just taken the 4 out and square rooted it. which identity are they using to convert 4 sinh^2u into 2 cosh2u -1 ?
0
2 weeks ago
#5
I remember these two...
cosh 2x = cosh²x + sinh²x
cosh²x - sinh²x = 1

Then I can work out others from those.
cosh 2x = 1 + sinh²x + sinh²x ... so ... cosh 2x = 1 + 2sinh²x <<< that's the one they're using.
0
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