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Mathematics: Post your doubts here!

Esme

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iKhaled said:
how to integrate the first part of question 10..10(i)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
Click to expand...

Posted by applepie1996

I = ⌡(tan^(n+2) x + tan^n x) dx
take tan^n out as a common factor
I = ⌡[tan^n x(tan^2 x + 1)] dx
u = tan x
dx = du/sec^2 x
I = ⌡[tan^n x(tan^2 x + 1)] du/sec^2 x
tan^2 x + 1=sec^x so you can cancel sec^2 x
I = ⌡u^n du
I = u^n+1/(n + 1)
put in the new limits
u = tan x
tan(π/4) = 1
tan(0) = 0
I = 1/(n + 1)
 
littlecloud11

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LMGD33

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LMGD33

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Esme

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farhan143 said:
CAN SOMEONE PLEASE EXPLAIN ME THE ANSWER TO THIS QUESTION

View attachment 26652


IT IS 2012 MAY-JUNE P61 STATISTICS - 9709
QP: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf

MS: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_61.pdf
Click to expand...

I suggest you post in this thread as everyone here is focusing on P3 which is tomorrow
https://www.xtremepapers.com/commun...tics-doubt-post-your-doubts-here.25911/page-9
 
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littlecloud11 said:
5) xy dy/dy = y^2 + 4
⌡(y/ y^2 +4) dy = ⌡1/x dx
Diffrentiating y^2+ 4 = 2y
so 1/2⌡2y/y^2 +4 dy = ln x + c
1/2 ln |y^2+4| = lnx +c
when y=0, x= 1
1/2 ln4 = ln1 +c
2*1/2 ln 2 = c
c= ln 2

1/2ln |y^2+4| = lnx +ln2
ln (y^2+4)^1/2 = ln (2x)
cancel ln from both sides
(y^2+4)^1/2 =2x
y^2 = 4x^2 - 4
Click to expand...
I didn't get the part where you diffrentiate y^2+4. PLEASE HELP ME! I'm panicking. I can't get anything through my brain! :(
 
Esme

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abbey789 said:
I didn't get the part where you diffrentiate y^2+4. PLEASE HELP ME! I'm panicking. I can't get anything through my brain! :(
Click to expand...

You don't have to differentiate. She shifted the terms so that you have y on one side and x on the other. Then you integrate them both.
 
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abbey789

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I G
littlecloud11 said:
5) xy dy/dy = y^2 + 4
⌡(y/ y^2 +4) dy = ⌡1/x dx
Diffrentiating y^2+ 4 = 2y
so 1/2⌡2y/y^2 +4 dy = ln x + c
1/2 ln |y^2+4| = lnx +c
when y=0, x= 1
1/2 ln4 = ln1 +c
2*1/2 ln 2 = c
c= ln 2

1/2ln |y^2+4| = lnx +ln2
ln (y^2+4)^1/2 = ln (2x)
cancel ln from both sides
(y^2+4)^1/2 =2x
y^2 = 4x^2 - 4
Click to expand...
I GOT IT! THANK YOU SO MUCH! You are my saviour!! :)
 
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abbey789

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Esme said:
You don't have to differentiate. She shifted the terms so that you have y on one side and x on the other. Then you integrate them both.
Click to expand...
We have to differentiate it so we have the numerator as the differentiation of the denominator. so, now we can write it as ln(y^2+4), as when this is differentiated it gives us numerator 2y. And (1/2) is there to balance it! :D
 
LMGD33

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abbey789 said:
I didn't get the part where you diffrentiate y^2+4. PLEASE HELP ME! I'm panicking. I can't get anything through my brain! :(
Click to expand...
The rule he used here is this one :
S is integral
S f ' (x)/ f(x) dx = ln |f(x)| + c
So he realized that if you diff y^2 + 4 you get 2y , but that's not the case in the numerator, to make it the case though, we can simply multiply inside the integral by two AND divide outside the integral by half , that way it's like we didn't change any value yet we were able to use the rule to get 1/2 ln (Y^2 + 4)
Yeah?
 
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