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# P3 integration AGAIN!!! watch

1. (Original post by imasillynarb)
Another P3 problem: A curve, C, is given by x = 2t + 3 y = t^3 - 4t

where t is a parameter. The point A has a parameter t = -1 and the line l is a tangent to C at A. The line l also interesects the curve at B
a) Show that an equation for l is 2y+x=7
b)Find the value of t at B

Ok, I subed in the values of x and y into the equation 2y+x = 7

I then used t+1 as a factor and did long division, to get t = 2

t=2 is the right answer, but was there an easier way to do it rather than doing long division?
dy/dt = 3t^2-4
dx/dt = 2

dy/dx = ((3t^2)/2) -2
sub in t=-1 to get the gradient, then work out the x,y co-ords then g on from there.
2. (Original post by Silly Sally)
But poor sally did that!!!
I=∫1/(xlnx)dx
so u=lnx => du=dx/x
I=∫(1/lnx)(dx/x)=∫(1/u)du
so I=ln|u|+c
I=ln|(lnx)|+c
3. (Original post by IntegralAnomaly)
I=∫1/(xlnx)dx
so u=lnx => du=dx/x
I=∫(1/lnx)(dx/x)=∫(1/u)du
so I=ln|u|+c
I=ln|(lnx)|+c
Oh - i did do that!!!

I just substituted incorrectly - the easiest part i can't see to do!!!
4. (Original post by Bhaal85)
dy/dt = 3t^2-4
dx/dt = 2

dy/dx = ((3t^2)/2) -2
sub in t=-1 to get the gradient, then work out the x,y co-ords then g on from there.
Ive done a, what about b? Is there an easier way?
5. (Original post by Bhaal85)
dy/dt = 3t^2-4
dx/dt = 2

dy/dx = ((3t^2)/2) -2
sub in t=-1 to get the gradient, then work out the x,y co-ords then g on from there.
X = 1
Y = 3

M = -1/2 at t=-1

Y-3=-1/2(X-1)

etc.

btw How on earth is the server busy at this time?
6. Ok then all you maths genius's out there!!! Some more integration questions to occupy your time!!!:

Ok - the integral of:

1. (cosx)^5

2. (sinx)^6

And i don't want any cheating - i want method as well!!!
7. (Original post by Silly Sally)
Where is the integration in that?!?
Didnt say it was integration :P
8. (Original post by imasillynarb)
Ive done a, what about b? Is there an easier way?
Another P3 problem: A curve, C, is given by x = 2t + 3 y = t^3 - 4t

where t is a parameter. The point A has a parameter t = -1 and the line l is a tangent to C at A. The line l also interesects the curve at B
a) Show that an equation for l is 2y+x=7
b)Find the value of t at B

Ok, I subed in the values of x and y into the equation 2y+x = 7

I then used t+1 as a factor and did long division, to get t = 2

t=2 is the right answer, but was there an easier way to do it rather than doing long division?

If 2y+x=7 then X=7-2y, then sub into one of the parametrics, and solve.
9. (Original post by Silly Sally)
Well thats the double angle formula - but they are both related anyway - but suppose the even power is more than 2 - suppose its 8!!!
sin^8 x = (sin²x)^4 = 1/16 (1-cos(2x))^4

= 1/16 (1 - 4cos2x + 6 cos²(2x) - 4 cos³(2x) + cos^4(2x))

= 1/16 (1-4cos(2x) + 6/2 (1-cos(4x)) - 4 cos(2x)(1-sin²(2x)) + 1/4 (1-cos(4x))²)

= 1/16 (1 - 4cos2x + 3 - 3cos(4x) - 4cos(2x) + 4cos(2x)sin²(2x) + 1/4 (1 - 2cos(4x) + cos²(4x)))

= 1/16 (17/4 - 3cos(2x) + 4cos(2x)sin²(2x) - 7/2 cos(4x) + 1/2 (1-cos(8x)))

= 1/16 (21/4 - 3cos(2x) + 4cos(2x)sin²(2x) - 7/2 cos(4x) -1/2 cos(8x))

= 21/64 - 3/16 cos(2x) + 1/4 cos(2x)sin²(2x) - 7/32 cos(4x) - 1/32 cos(8x)

=> integral = 21x/64 + 3/32 sin(2x) + 1/12 sin³(2x) + 7/128 sin(4x) + 1/256 cos(8x)

It's far easier to use DeMoivre's theorem, but you dont come accross that till P6
10. (Original post by imasillynarb)
Didnt say it was integration :P
Look at the thread name in future!!!
11. (Original post by Silly Sally)
Ok then all you maths genius's out there!!! Some more integration questions to occupy your time!!!:

Ok - the integral of:

1. (cosx)^5

2. (sinx)^6

And i don't want any cheating - i want method as well!!!
u try them,use exactly the same method!
12. *******s.
13. (Original post by elpaw)
sin^8 x = (sin²x)^4 = 1/16 (1-cos(2x))^4

= 1/16 (1 - 4cos2x + 6 cos²(2x) - 4 cos³(2x) + cos^4(2x))

= 1/16 (1-4cos(2x) + 6/2 (1-cos(4x)) - 4 cos(2x)(1-sin²(2x)) + 1/4 (1-cos(4x))²)

= 1/16 (1 - 4cos2x + 3 - 3cos(4x) - 4cos(2x) + 4cos(2x)sin²(2x) + 1/4 (1 - 2cos(4x) + cos²(4x)))

= 1/16 (17/4 - 3cos(2x) + 4cos(2x)sin²(2x) - 7/2 cos(4x) + 1/2 (1-cos(8x)))

= 1/16 (21/4 - 3cos(2x) + 4cos(2x)sin²(2x) - 7/2 cos(4x) -1/2 cos(8x))

= 21/64 - 3/16 cos(2x) + 1/4 cos(2x)sin²(2x) - 7/32 cos(4x) - 1/32 cos(8x)

=> integral = 21x/64 + 3/32 sin(2x) + 1/12 sin³(2x) + 7/128 sin(4x) + 1/256 cos(8x)
Do you think that i would get anything this hard for Edexcel P3 integration?
14. (Original post by Silly Sally)
Ok then all you maths genius's out there!!! Some more integration questions to occupy your time!!!:

Ok - the integral of:

1. (cosx)^5

2. (sinx)^6

And i don't want any cheating - i want method as well!!!
(cosx)^5 = cosx(cos^2x)(cos^2x) = cosx(1-sin^2x)(1-sin^2x)
=cosx(1-2sin^2x + sin^4x)

u=sinx
du=cosx dx
(1-2u^2 + u^4)cosx x du/cosx

=(1-2u^2 + u^4) du

= u - 2/3u^3 + u^5/5

= sinx-2/3sin^3x + sin^5x/5 + c

Is that right?
15. (Original post by IntegralAnomaly)
u try them,use exactly the same method!
Well i would, but you have put me off integration for good integralanomaly - if i get another question wrong, i will be too depressed to take the final exam - let it be on your head if i lose my uni place!!! (only teasing!!! )
16. (Original post by Silly Sally)
Do you think that i would get anything this hard for Edexcel P3 integration?
theoretically, yes, because its in the syllabus. but theoretically they could also ask you (sin x)^128934758656 - it's probably a bit too much work, will give you millions of UMS and would take you till your forty, so no.
17. (Original post by Silly Sally)
Well i would, but you have put me off integration for good integralanomaly - if i get another question wrong, i will be too depressed to take the final exam - let it be on your head if i lose my uni place!!! (only teasing!!! )
When I first started doing the integration questions, I would get stuck on one, post it here, get help, then thinking the next one I would breeze it with my newfound techniques/knowledge, get stuck again, and then again, and again, you have no idea how depressing it was! Well you do, as youre probably going through it right now, but it does get easier
18. (Original post by imasillynarb)
(cosx)^5 = cosx(cos^2x)(cos^2x) = cosx(1-sin^2x)(1-sin^2x)
=cosx(1-2sin^2x + sin^4x)

u=sinx
du=cosx dx
(1-2u^2 + u^4)cosx x du/cosx

=(1-2u^2 + u^4) du

= u - 2/3u^3 + u^5/5

= sinx-2/3sin^3x + sin^5x/5 + c

Is that right?
YAY thats what i got for the first one!!! Thanks
19. (Original post by imasillynarb)
When I first started doing the integration questions, I would get stuck on one, post it here, get help, then thinking the next one I would breeze it with my newfound techniques/knowledge, get stuck again, and then again, and again, you have no idea how depressing it was! Well you do, as youre probably going through it right now, but it does get easier
I don't think that there is an easy way of doing your question. As I got Y=(X^3-9X^2+27X-27)/8 and couldn't be arsed doing the rest, so I think you would do what you did, which was to divide by (X+1).
20. here u go:
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