The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

WHERE AM I GOING WRONG ?!?! :'(:'( Please help me with this integration!

Problem solved :smile:
(edited 9 years ago)
Reply 1
Avatar for Sena5
Sena5
20
Original post by cl1.
This is how I attempted the following integration:

∫(1+x^2)^(3/2) dx

let x = sinh u, so dx = cosh u du

subbing in:


∫(1+sinh^(2)u)^(3/2) * cosh u du

Using, cosh^(2)u - sin^(2)u = 1, so cosh^(2)u = 1+ sin^(2)u

subbing this in:



∫(cosh^(2)u)^(3/2) * cosh u du = cosh^(4) u du = ∫cosh^(2) u * cosh^(2) u


Taking, cosh 2u = 2cosh^(2)u + 1, so cosh^(2)u = (1/2)(cosh(2u) - 1):



∫cosh^(2) u * cosh^(2) u du = ∫(1/2)^(2)*(cosh(2u) - 1)^(2) du

= (1/4) ∫ (cosh^(2) 2u - 2cosh 2u + 1) du


... using substitution cosh so cosh^(2) 2u = (1/2)(cosh(4u) - 1)



= (1/4) ∫ ( (1/2)cosh(4u) - 2cosh(2u) + 1/2 ) du , then take out the half:

(1/8) ∫ ( cosh(4u) - cosh(2u) + 1 )

= (1/8) [ (1/4)*sinh(4u) - (1/2)*sinh(2u) + u ] + C , taking out the quarter:


= (1/32) [ sinh(4u) - 2sinh(2u) + 4u ] + C <= and this is wrong?



This is a past paper and the answers say the above line should be:

= (1/32) [ sinh(4u) + 8sin(2u) + 12u ] + C

Do you have any idea where I went wrong?

I'm sorry about the length of this, I just wanted to make it clear what i did so the error is clearer.

Thank you if you have read all of this, and any help would be greatly appreciated!

I will help you. Tell me which past paper year and month. Is this edexcel C1 maths?
I can surely help
Reply 2
Avatar for Sena5
Sena5
20
could you tell me which past papers so i will see the question.
This ain't C1.

This is FP3 :rofl:


Posted from TSR Mobile
(edited 9 years ago)
Reply 4
Avatar for Sena5
Sena5
20
BETTER ATTACH THE PAST PAPER ALSO. Your method is not clear
Reply 5
Avatar for cl1.
cl1.
OP
8
Original post by kers123
BETTER ATTACH THE PAST PAPER ALSO. Your method is not clear


I can't, it's a uni past paper, we're not allowed to post them online!

The question is simply:

solve ∫(1+x^2)^(3/2) dx
Reply 6
Avatar for Sena5
Sena5
20
Original post by cl1.
I can't, it's a uni past paper, we're not allowed to post them online!

The question is simply:

solve ∫(1+x^2)^(3/2) dx


it's not from edexcel Mathematics then?
Reply 7
Avatar for cl1.
cl1.
OP
8
Original post by kers123
it's not from edexcel Mathematics then?


nope
Hmm
Posted from TSR Mobile
(edited 9 years ago)
Reply 9
Avatar for cl1.
cl1.
OP
8
Original post by TheKingOfTSR


supposed to be sinh, just a typo. Unless you knew that and mean mathematically, then yeh it's correct, according to my textbook anyway.
(edited 9 years ago)
Reply 10
Avatar for Sena5
Sena5
20
Original post by cl1.
I can't, it's a uni past paper, we're not allowed to post them online!

The question is simply:

solve ∫(1+x^2)^(3/2) dx


I think you book a tutorial in this website which i just saw. They will teach free for one hour anything. So you can ask this too. even university questions are taught. try it

https://www.tutorme.co.uk/student-home
I just did it. The answer given (not yours) is correct. Something is wrong with your working.

Posted from TSR Mobile
Reply 12
Avatar for cl1.
cl1.
OP
8
Original post by kers123
I think you book a tutorial in this website which i just saw. They will teach free for one hour anything. So you can ask this too. even university questions are taught. try it

https://www.tutorme.co.uk/student-home


The next available time is 30 minutes before my exam haha,

but thanks anyway, I'm going to book one for Chemistry. How helpful did you find them?
Reply 13
Avatar for cl1.
cl1.
OP
8
Original post by TheKingOfTSR
I just did it. The answer given (not yours) is correct. Something is wrong with your working.

Posted from TSR Mobile


Did you use the same substitution as me? If so where did we differ?
Reply 14
Avatar for Sena5
Sena5
20
Original post by cl1.
The next available time is 30 minutes before my exam haha,

but thanks anyway, I'm going to book one for Chemistry. How helpful did you find them?


i see. no i just typed right now on google and found it
Original post by cl1.
Did you use the same substitution as me? If so where did we differ?


I used the same subs (after looking at your subs) but slightly different method. :yy:

Original post by cl1.



= (1/4) ∫ ( (1/2)cosh(4u) - 2cosh(2u) + 1/2 ) du , then take out the half:

(1/8) ∫ ( cosh(4u) - cosh(2u) + 1 )


Taking that half out...it leads to -4cosh(2u) not -1cosh(2u)

This (1/8) ∫ ( cosh(4u) - cosh(2u) + 1 ) should have been (1/8) ∫ ( cosh(4u) - 4cosh(2u) + 1 )


Posted from TSR Mobile
(edited 9 years ago)
Reply 16
Avatar for cl1.
cl1.
OP
8
Original post by TheKingOfTSR
I used the same subs (after looking at your subs) but slightly different method. :yy:



Taking that half out...it leads to -4cosh(2u) not -1cosh(2u)

Posted from TSR Mobile


Aw yeh that was a stupid mistake, but that still doesn't get me the correct answer.

So what identities did you use afterwards?
Original post by cl1.
Aw yeh that was a stupid mistake, but that still doesn't get me the correct answer.

So what identities did you use afterwards?


No..some thing is wrong before this.

Is this right?

Taking, cosh 2u = 2cosh^(2)u + 1, so cosh^(2)u = (1/2)(cosh(2u) - 1):


Should have been:


Taking, cosh 2u = 2cosh^(2)u - 1 , so cosh^(2)u = (1/2)(cosh(2u) + 1):


Posted from TSR Mobile
(edited 9 years ago)
Reply 18
Avatar for cl1.
cl1.
OP
8
Original post by TheKingOfTSR
No..some thing is wrong before this.

Is this right?

Taking, cosh 2u = 2cosh^(2)u + 1, so cosh^(2)u = (1/2)(cosh(2u) - 1):


Should have been:


Taking, cosh 2u = 2cosh^(2)u - 1 , so cosh^(2)u = (1/2)(cosh(2u) + 1):


Posted from TSR Mobile


****, yeh that's it. I read Osborn's rule completely wrong! Thank you! At least I won't make this mistake in the exam :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you