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Integration Q

I saw this question on a previous thread and nobody answered it.
- Integrate sqrt(x^2 - 16) dx

Can anyone do it?
Reply 1
Avatar for Simba
Simba
16
INT[(x² - 16)^0.5]dx

Let x = 4coshu.

dx/du = 4sinhu.

dx = 4sinhu.du

=> INT[(x² - 16)^0.5]dx

= INT[(16cosh²u - 16)^0.5.4sinhu]du

= INT[(16sinh²u)^0.5.4sinhu]du

= INT[16sinh²u]du

= INT[8(cosh2u - 1)]du

= 4sinh2u - 8u + c.

= 8sinhucoshu - 8arcosh(x/4) + c

= x(x^2 - 16)^0.5/2 - 8log[x/4 + (x^2/16 - 1)] + c.

Hope this helps,

~~Simba
Reply 2
Avatar for Notnek
Notnek
OP
21
Thanks
Its a bit too advanced for me though!
Reply 3
Avatar for Simba
Simba
16
Hehe, hyperbolic substitution, FP2 (on Edexcel anyway) :smile: ...
I've tried solving it for you, but it may or may not be 100% correct.....Havent touched integration since i finished A2....

I guess this soultion should be correct though....someone else can reconfirm it, and correct it should the need be.

Now where's my rep? I solved it on a sheet of paper and scanned it for you! lolz.....
IntegrationQ.jpg127.2KB
it's too complicated for a student of this level.....apparently, this doesnt appear as a product of two functions, and may not need integration by substitutuion.
Reply 6
Avatar for insparato
insparato
15
FP2 stuff mate :smile:.

Oh posted way too late lol.
Reply 7
Avatar for insparato
insparato
15
Red Thunder
I've tried solving it for you, but it may or may not be 100% correct.....Havent touched integration since i finished A2....

I guess this soultion should be correct though....someone else can reconfirm it, and correct it should the need be.

Now where's my rep? I solved it on a sheet of paper and scanned it for you! lolz.....


Need to look up your integration :smile:.
That's how u can integrate it if u use the methods in C3-C4....

And to the OP: add + C to each expression in the solution i provdided.
Reply 9
Avatar for Simba
Simba
16
Well, differentiating what you got:

Let y = uv, u = 4x/3, v = (x² - 16)^1.5

du/dx = 4/3, dv/dx = 3x(x^2 - 16)^0.5

dy/dx = v.du/dx + u.dv/dx = [4(x² - 16)^1.5]/3 + 4x²(x^2 - 16)^0.5.

Which, isn't what we wanted to integrate.

My method is the standard approach; that is the correct answer. Saying it's "too complicated" doesn't mean that there's an easier approach.

Red Thunder
That's how u can integrate it if u use the methods in C3-C4....


It's not possible using C3 and C4 techniques as far as I'm aware.
Reply 10
Avatar for Notnek
Notnek
OP
21
Red Thunder
I've tried solving it for you, but it may or may not be 100% correct.....Havent touched integration since i finished A2....

I guess this soultion should be correct though....someone else can reconfirm it, and correct it should the need be.

Now where's my rep? I solved it on a sheet of paper and scanned it for you! lolz.....


Im FP1 - almost there...
I dont know about this solution though
Sorry.....told u i have lost a bit of grip on it since I left maths at A2.....i only studied C1-C, M1 and S1...
Reply 12
Avatar for Notnek
Notnek
OP
21
That's Ok. Here's some + rep just for the effort :smile:
Reply 13
Avatar for Exile
Exile
1
Simba
INT[(x² - 16)^0.5]dx

Let x = 4coshu.

dx/du = 4sinhu.

dx = 4sinhu.du

=> INT[(x² - 16)^0.5]dx

= INT[(16cosh²u - 16)^0.5.4sinhu]du

= INT[(16sinh²u)^0.5.4sinhu]du

= INT[16sinh²u]du

= INT[8(cosh2u - 1)]du

= 4sinh2u - 8u + c.

= 8sinhucoshu - 8arcosh(x/4) + c

= x(x^2 - 16)^0.5/2 - 8log[x/4 + (x^2/16 - 1)^1/2] + c.

Hope this helps,

~~Simba


according to wolfram integrator its wrong,
they get

=[ x(x^2 - 16)^0.5] /2 - 8log[ x + (x^2-16)^0.5] + c

even by adding the missed ^1/2 in your answer its still different, I didnt point this out to say you're wrong but to say other then that I cant see why your answer is different can anyone spot a mistake?

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