Gent2324
Badges: 19
Rep:
?
#1
Report Thread starter 2 weeks ago
#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
reply
mqb2766
Badges: 17
Rep:
?
#2
Report 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
reply
MarkFromWales
Badges: 7
Rep:
?
#3
Report 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
reply
Gent2324
Badges: 19
Rep:
?
#4
Report Thread starter 2 weeks ago
#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
reply
MarkFromWales
Badges: 7
Rep:
?
#5
Report 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
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • De Montfort University
    Undergraduate Open Day Undergraduate
    Sat, 7 Dec '19
  • Buckinghamshire New University
    Undergraduate Undergraduate
    Sat, 7 Dec '19
  • Regent's University London
    Undergraduate Open Day Undergraduate
    Sat, 7 Dec '19

Which party will you be voting for in the General Election?

Conservatives (68)
19.48%
Labour (169)
48.42%
Liberal Democrats (49)
14.04%
Green Party (21)
6.02%
Brexit Party (4)
1.15%
Independent Group for Change (Change UK) (3)
0.86%
SNP (8)
2.29%
Plaid Cymru (3)
0.86%
Democratic Unionist Party (DUP) (0)
0%
Sinn Fein (5)
1.43%
SDLP (0)
0%
Ulster Unionist (1)
0.29%
UKIP (6)
1.72%
Other (1)
0.29%
None (11)
3.15%

Watched Threads

View All