-
Looking for HSC notes and resources? Check out our Notes & Resources page
HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)
- Thread starter bleakarcher
- Start date
- Status
Sy123
This too shall pass
- Joined
- Nov 6, 2011
- Messages
- 3,730
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
) + \cos( \ln x) \ dx = x\sin \ln x + c )
 \ dx \\ u = e^{\pi/4} x \\ \\ = \frac{\sqrt{2}}{e^{\pi/4}} \int \sin \ln u \ du = x\sin \ln x + c )
 = \sqrt{2} u \sin \left(\ln u - \frac{\pi}{4}\right) )
Note that in an exam however you will not be marked if you do not use the previous part
I guess you can do that, though when I said Hence, I was thinking of:
Note that in an exam however you will not be marked if you do not use the previous part
Carrotsticks
Retired
- Joined
- Jun 29, 2009
- Messages
- 9,494
- Gender
- Undisclosed
- HSC
- N/A
Re: MX2 Integration Marathon
Haha I like this one, assuming you intended the method I have in mind. \ dx )
RealiseNothing
what is that?It is Cowpea
Re: MX2 Integration Marathon
 = \tan^{-1}(e^x))
 = \tan^{-1}(\frac{1}{e^x}))
+f(-x)=\frac{\pi}{2})
(just consider a right angled triangle)
So we can just make the curve
and change the limits from 0 to 1 and the answer is 
What I'm doing is "cutting" the section of the graph between -1 and 0, and "pasting" it on top of the graph between 0 and 1.
So we can just make the curve
What I'm doing is "cutting" the section of the graph between -1 and 0, and "pasting" it on top of the graph between 0 and 1.
Sy123
This too shall pass
- Joined
- Nov 6, 2011
- Messages
- 3,730
- Gender
- Male
- HSC
- 2013
Re: MX2 Integration Marathon
don't use t-sub plz
Yep, the more technical approach though would be to just set u=-x and apply the same formula.
So we can just make the curve
What I'm doing is "cutting" the section of the graph between -1 and 0, and "pasting" it on top of the graph between 0 and 1.
Bump
don't use t-sub plz
RealiseNothing
what is that?It is Cowpea
- Status