- Messages
- 16
- Reaction score
- 0
- Points
- 1
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
liuweijian94 said:oh no~~~i am using the text book which is written by Braian and Mark Gaulter~~~Oxford press ~~~~am i using the wrong text book? :sorry:
usman said:liuweijian94 said:oh no~~~i am using the text book which is written by Braian and Mark Gaulter~~~Oxford press ~~~~am i using the wrong text book? :sorry:
Assalam-o-alaikum
Hi there 'liuweijian94', according to my experience, the book "Further Pure Mathematics by Brian and Mark Gaulter, Oxford University press" is infact a very good book covering, in a very nice way, many topics of CIE Further Maths (and I use it frequently).
So don't worry, you may well use this book, along with "Further Pure Mathematics by L. Bostock, S. Chandler, C. Rourke; Stanley Thomes Publishers," which is also a similar, but a bit better choice (though it also contains a considerable amount of irrelevant material for CIE Further Maths, but covers many more topics than the book by B.M Gaulter as a whole) + Regularly go through the past-papers of each topic when you are done with it e.g. differential equations in this case (though there are questions relating only to Second Order Differential Equations in the CIE -papers, but practising First Order Differential Equations as well, will atleast give you a nice background for 2nd O.D.E).
As far as your problem is concerned, i worked out the following solution for it--->
Continuing on from your last step.... (Don’t lose your patience, cos it’s gonna get xtremely messy)
....y*sin(x) = int [ exp(2x)*sin(x) ] dx
Using integration by parts on int [ exp(2x)*sin(x) ] dx, setting:-
du/dx = exp(2x) while v = sin(x)
Then, u = 0.5*exp(2x) while dv/dx = cos(x)
---> y*sin(x) = 0.5*[ exp(2x)*sin(x) ] - 0.5*int [ exp(2x)*cos(x) ] dx
Let's use integration by parts again, on int [ exp(2x)*cos(x) ] dx, setting:-
dp/dx = exp(2x) while q = cos(x)
Then, p = 0.5*exp(2x) while dq/dx = -sin(x)
So, int [ exp(2x)*cos(x) ] dx = 0.5*exp(2x)*cos(x) + 0.5*int [ exp(2x)*sin(x) ] dx
We can observe that the integral we started with, i.e. int [ exp(2x)*sin(x) ] dx, has now appeared again as the last term in the expression for “int [ exp(2x)*cos(x) ] dx.” Bingo!
I know it’s gettin’ messy as hell…;p Bt anyways, let’s bear with it.
So, upto now, we’ve found that
int [ exp(2x)*sin(x) ] dx = 0.5*[ exp(2x)*sin(x) ] – 0.5*[0.5*exp(2x)*cos(x) + 0.5*int { exp(2x)*sin(x) } dx ]
int [ exp(2x)*sin(x) ] dx = 0.5*[ exp(2x)*sin(x) ] – 0.25*[ exp(2x)*cos(x) ] - 0.25*int { exp(2x)*sin(x) } dx ]
(1 + 0.25)* int [ exp(2x)*sin(x) ] dx = 0.5*exp(2x)*sin(x) – 0.25*[ exp(2x)*cos(x) ]
(5/4)* int [ exp(2x)*sin(x) ] dx = [0.25*exp(2x)] *[2sin(x) – cos(x)]
int [ exp(2x)*sin(x) ] dx = [0.25*(4/5)*exp(2x)] *[2sin(x) – cos(x)]
int [ exp(2x)*sin(x) ] dx = 0.2*exp(2x)*[2sin(x) – cos(x)]
--->int [ exp(2x)*sin(x) ] dx = (1/5)*exp(2x)*[2sin(x) – cos(x)] + constant of integration (let it be ‘C’)
Therefore, y*sin(x) = (1/5)*exp(2x)*[2sin(x) – cos(x)] + C
NB: The constant of integration is shown only in the last step, to avoid repetition of constants of integration, over and over again.
P.S.
The question you asked, as I know, is from Further Maths by B.M. Gaulter --> Exercise 4A --> Q.7, which is very similar to the Q.15 of Exercise 7a of Further Maths by L.Bostock [the difference being that there is an exp(x) in that question rather than an exp(2x) as in this case]…;p
Anyways, hope that helps (rather than confuses yew further….;p)….!
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now