The Student Room Group

integration help

will post in a min

Scroll to see replies

Reply 1
Capture.PNG

4/2x+1
we express this function as
4/2 X(multiply) 1 /x+1
then it should be
2ln (x+1)

then in solutions why there is
(2x+1)???
Original post by Qer
Capture.PNG

4/2x+1
we express this function as
4/2 X(multiply) 1 /x+1
then it should be
2ln (x+1)

then in solutions why there is
(2x+1)???


421x+1=42x+242x+1 \frac{4}{2} \cdot \frac{1}{x+1} = \frac{4}{2x+2} \neq \frac{4}{2x+1}
Reply 3
Original post by Qer
Capture.PNG

4/2x+1
we express this function as
4/2 X(multiply) 1 /x+1
then it should be
2ln (x+1)

then in solutions why there is
(2x+1)???


its not 4/2 multiply 1/x+1. anyway its better off thinking about chain rule the other way round, find the derivative of the bottom function and do 1/derivative multiplied by standard integral.
here that is derivative of 2x+1 which is 2 divided by 4 multiplied by the standard integral of 1/f(x) which is lnf(x)
Original post by Qer
Capture.PNG

4/2x+1
we express this function as
4/2 X(multiply) 1 /x+1
then it should be
2ln (x+1)

then in solutions why there is
(2x+1)???


42x+12×22x+1\dfrac{4}{2x+1} \equiv 2 \times \dfrac{2}{2x+1}, which can be integrated by inspection.

What you have written isn't a correct simplification :smile:
Reply 5
thanks everyone

another example
integrate (e^x +1)^2


The solution that i got in which they expand the bracket by using formula a^2 +2ab+b^2 and then integrate
each term separately.But why i cant do like that

(e^x +1)^3 /3??
Original post by Qer
thanks everyone

another example
integrate (e^x +1)^2


The solution that i got in which they expand the bracket by using formula a^2 +2ab+b^2 and then integrate
each term separately.But why i cant do like that

(e^x +1)^3 /3??


If it were (x+1)2(x+1)^2, you could integrate it to get (x+1)3/3(x+1)^3/3 (as the derivative of x+1x+1 is 1), but since it is (ex+1)2(e^x+1)^2, that method is invalid, and expanding the brackets is the best way to do this.
Reply 7
Original post by Integer123
If it were (x+1)2(x+1)^2, you could integrate it to get (x+1)3/3(x+1)^3/3 (as the derivative of x+1x+1 is 1), but since it is (ex+1)2(e^x+1)^2, that method is invalid, and expanding the brackets is the best way to do this.


thanks
Reply 8
Capture.PNG


I actually got the right answer in this question but i am thinking that if we replace 1 + tan^2x with sec^2x it gives a different answer(-cotx . tanx +c) ? so how I know that in which form i have to give answer unless they not stated in question
Original post by Qer
Capture.PNG


I actually got the right answer in this question but i am thinking that if we replace 1 + tan^2x with sec^2x it gives a different answer(-cotx . tanx +c) ? so how I know that in which form i have to give answer unless they not stated in question


you should get equivalent answers. the examiners are experts as this & will know if you done good.
Original post by Qer

I actually got the right answer in this question but i am thinking that if we replace 1 + tan^2x with sec^2x it gives a different answer(-cotx . tanx +c) ? so how I know that in which form i have to give answer unless they not stated in question


2cot(2x)=2cos(2x)sin(2x)=2cos2(x)+2sin2(x)2sin(x)cos(x)-2 \cot(2x) = \dfrac{-2\cos(2x)}{\sin(2x)}=\dfrac{-2\cos^2(x)+2\sin^2(x)}{2\sin(x) \cos(x)}

=2cos2(x)2sin(x)cos(x)+2sin2(x)2sin(x)cos(x)=\dfrac{-2\cos^2(x)}{2\sin(x)\cos(x)}+ \dfrac{2\sin^2(x)}{2\sin(x)\cos(x)}

=cot(x)+tan(x)=- \cot(x) + \tan(x)

So it's NOT a different answer!!


Both forms would be accepted in an exam unless specified
(edited 6 years ago)
Reply 11
Original post by RDKGames
2cot(2x)=2cos(2x)sin(2x)=2cos2(x)+2sin2(x)2sin(x)cos(x)-2 \cot(2x) = \dfrac{-2\cos(2x)}{\sin(2x)}=\dfrac{-2\cos^2(x)+2\sin^2(x)}{2\sin(x) \cos(x)}

=2cos2(x)2sin(x)cos(x)+2sin2(x)2sin(x)cos(x)=\dfrac{-2\cos^2(x)}{2\sin(x)\cos(x)}+ \dfrac{2\sin^2(x)}{2\sin(x)\cos(x)}

=cot(x)+tan(x)=- \cot(x) + \tan(x)

So it's NOT a different answer!!


Both forms would be accepted in an exam unless specified


Is this solution to a example that i posted?
Capture.PNG Is that is correct?
Original post by Qer
Is this solution to a example that i posted?
Is that is correct?


Huh? cot(x)tan(x)1-\cot(x) \cdot \tan(x) \equiv -1 is not an answer... it's a constant, and you cannot even manipulate it into the given answer!

In prev. post, I thought you had ended up with 2cot(2x)-2\cot(2x) - guess not.
(edited 6 years ago)
Reply 13
Original post by RDKGames
Huh? cot(x)tan(x)1-\cot(x) \cdot \tan(x) \equiv -1 is not an answer... it's a constant, and you cannot even manipulate it into the given answer!

In prev. post, I thought you had ended up with 2cot(2x)-2\cot(2x) - guess not.


:confused::confused::creep:
Reply 14
Capture.PNG

where this 1/2 come from?
Original post by Qer
Is this solution to a example that i posted?
Capture.PNG Is that is correct?


Original post by RDKGames
Huh? cot(x)tan(x)1-\cot(x) \cdot \tan(x) \equiv -1 is not an answer... it's a constant, and you cannot even manipulate it into the given answer!

In prev. post, I thought you had ended up with 2cot(2x)-2\cot(2x) - guess not.
I think the OP is thinking you can integrate cosec^2 sec^2 by integrating cosec^2 and sec^2 separately and multiplying the results together.

To OP: you can't do that, it's not a valid step.
Original post by Qer


where this 1/2 come from?


Identity relating cos(2x)\cos(2x) and cos2(x)\cos^2(x)
Reply 17
Original post by DFranklin
I think the OP is thinking you can integrate cosec^2 sec^2 by integrating cosec^2 and sec^2 separately and multiplying the results together.

To OP: you can't do that, it's not a valid step.


yeah, i was thinking what you said above .....so any two trig identities that are multiplying i cant integrate them separately?
Original post by DFranklin
I think the OP is thinking you can integrate cosec^2 sec^2 by integrating cosec^2 and sec^2 separately and multiplying the results together.

To OP: you can't do that, it's not a valid step.


That would explain it.

There are a few obvious red alerts going on with that approach which should be obvious to OP at this level IMO.
Reply 19
Original post by RDKGames
Identity relating cos(2x)\cos(2x) and cos2(x)\cos^2(x)


is this possible for you to show me working? i don't understand it

Quick Reply