• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

Status
Not open for further replies.

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: MX2 Integration Marathon

int(f(x))=int((f(a-x))?
That will probably not work, this is beyond HSC difficulty though (but within the scope of the syllabus)

The answer is , it uses recursion in the solution, I can post my one on request

-----------------

 

Davo_01

Active Member
Joined
Apr 8, 2014
Messages
116
Gender
Male
HSC
N/A
Re: MX2 Integration Marathon

That will probably not work, this is beyond HSC difficulty though (but within the scope of the syllabus)

The answer is , it uses recursion in the solution, I can post my one on request

-----------------

Had a feeling there would be recursion, took me a while to find it:











 
Last edited:

Ikki

Active Member
Joined
Oct 29, 2012
Messages
130
Gender
Male
HSC
2014
Re: MX2 Integration Marathon

That will probably not work, this is beyond HSC difficulty though (but within the scope of the syllabus)

The answer is , it uses recursion in the solution, I can post my one on request

-----------------

Step back, davo's sol^n is like crazy. Sy did u have to mess around with identities as well or something else?

Alright, while i'm here...
Let t=tan(x/2)







Now casually using partial fractions...



At which point I basically ceebs typing the rest up, but should be integratable now easily.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: MX2 Integration Marathon

Step back, davo's sol^n is like crazy. Sy did u have to mess around with identities as well or something else?

Alright, while i'm here...
Let t=tan(x/2)







Now casually using partial fractions...



At which point I basically ceebs typing the rest up, but should be integratable now easily.
Yea his solution was pretty much exact same as mine

As for your solution to this, it is quite long and you will need to evaluate

Here is my solution:

 

Ikki

Active Member
Joined
Oct 29, 2012
Messages
130
Gender
Male
HSC
2014
Re: MX2 Integration Marathon

Ah k nice :)

Question: How would you determine value for tan (pi/8)?

----
Nvm. Figured it out. You can just split it into tan(pi/4-pi/8) and solve the quadratic.
 
Last edited:

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
Re: MX2 Integration Marathon








Can someone pose a new question, on my behalf, please.
 
Last edited:

Ikki

Active Member
Joined
Oct 29, 2012
Messages
130
Gender
Male
HSC
2014
Re: MX2 Integration Marathon








Can someone pose a new question, on my behalf, please.
Did not think of that, if you let u=arccot(3x^2) then you end up with a ln? Are both correct?

Also, in regards to the working, derivative of arccotx=negative * 1/1+x^2. So i believe there is a negative missing.
 
Last edited:

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
Re: MX2 Integration Marathon

Also, in regards to the working, derivative of arccotx=negative * 1/1+x^2. So i believe there is a negative missing.
You are correct. There should be a negative. When I tentatively started doing it, I thought there ought to be a negative. I then flipped thru 2 text books for the derivative of arccot x - looking out for the '-' sign. Both times I did not see the '-'sign. After seeing your feedback, I rechecked both and there was the '-'sign. How I missed it both times yesterday baffles me. It must mean my glasses are not working as well as I thought.



By the way, if you apply the Chain Rule to [arccot ( (3x^2)]^2 you will get a constant multiple of the integrand. That is you will get:

 
Last edited:

Ikki

Active Member
Joined
Oct 29, 2012
Messages
130
Gender
Male
HSC
2014
Re: MX2 Integration Marathon

Epic.

So, If i let u=arctan(3x^2) right, i get sqrt(2)/4 * ln(arctan(3x^2)) which if you derive also get a constant multiple of the integrand, so i assume both are correct.
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
Re: MX2 Integration Marathon

Epic.

So, If i let u=arctan(3x^2) right, i get sqrt(2)/4 * ln(arctan(3x^2)) which if you derive also get a constant multiple of the integrand, so i assume both are correct.
Where does this 'ln' come from ???
 

dunjaaa

Active Member
Joined
Oct 10, 2012
Messages
473
Gender
Male
HSC
2014
Re: MX2 Integration Marathon

Woops typo, was meant to be arctan(sqrt(2)tan(x)) instead of arctan(1+sqrt(2)tan(x)), typed it on the bus this morning
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top