• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Integration Question (1 Viewer)

Ekman

Well-Known Member
Joined
Oct 23, 2014
Messages
1,615
Gender
Male
HSC
2015
Hey,
I recently did my extension 1 exam and there was one hard integration question that seemed a bit weird.





EDIT: I know there is the log way of solving it, but keep in mind that this exam was also done with 2u students who haven't learnt what logarithms are yet...
 
Last edited:

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
One of the standard integrals given on the back of the paper is the integral of secxtanx. Substituting u = pi/2 - x allows you to integrate cosecxcotx using this rule.

secx and cosec x can be integrated using the t-results.

But Ext 1 students are not expected to come up with either of these substutitions - you are supposed to be fed the substitution.
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
I solved I by converting both to sinx and cosx, then letting u = sinx
 

Ekman

Well-Known Member
Joined
Oct 23, 2014
Messages
1,615
Gender
Male
HSC
2015
I wasn't the issue, it was ii. As said by braintic, they should of built up the substitution for you in order to solve ii.
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
I wasn't the issue, it was ii. As said by braintic, they should of built up the substitution for you in order to solve ii.
They gave you nothing at all?
I'm pretty sure integrating cosecx is not required by the 3U syllabus
 

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
I wasn't the issue, it was ii. As said by braintic, they should of built up the substitution for you in order to solve ii.
I know this is not an English forum, but you just wrote something that irks me every time I see it.
There is no such thing as "should of". It is "should have".
 

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

Top