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# Hit me with a difficult P3 or P1 question watch

1. (Original post by *dave*)
I have NEVER answered a question about arctanx or arccosx or all of those. I swear to god if those come up i am buggered.
It won't come up, but if you are interested the differentiation of such functions is fairly simple. For example, if you want to find the derivive of arcsin(x) you would do this:

y = arcsin(x)
sin(y) = x

Then, differentiating w.r.t. x

cos(y).dy/dx = 1
dy/dx = 1/cos(y)

Now, we have

sin^2(y) + cos^2(y) = 1
cos^2(y) = 1 - sin^2(y)
cos(y) = sqrt[ 1 - sin^2(y) ]

and since sin(y) = x we have

cos(y) = sqrt( 1 - x^2 )

therefore

dy/dx = d( arcsin(x) )/dx = 1/sqrt(1-x^2)

If you are interested you can use similar methods to find the derivitives of arccos(x) and arctan(x).
2. (Original post by fishpaste)
Much respect to you for getting that far. Well done.
BTW, I answered the question you posted here: http://www.uk-learning.net/t46254.html.
3. Here is another nasty integral if anybody wants to spend their time churning out the algebra:

INT 1/(1+x^3) dx
4. (Original post by mikesgt2)
Here is another nasty integral if anybody wants to spend their time churning out the algebra:

INT 1/(1+x^3) dx
Any hints?
5. (Original post by Bhaal85)
Any hints?
Well, you will need to factorise the denominator. However, you should know that this integral is a bit beyond P3 since you will need arctan integrals.
6. (Original post by mikesgt2)
Well, you will need to factorise the denominator. However, you should know that this integral is a bit beyond P3 since you will need arctan integrals.
I only went up to P3, what is x^3+1 factorised? Is there a rule for factorising cubics quickly? I had to use some P2 stuff, and equate like terms:

1/[(x+1)(x^2-x+1)] ?
7. (Original post by mikesgt2)
Well, you will need to factorise the denominator. However, you should know that this integral is a bit beyond P3 since you will need arctan integrals.

I got as far as 1/3ln(x+1) + 1/3int.(2-x)/(x^2-x+1)

(although its probably littered with mistakes as I'm too tired to work effectively at this hour).
8. (Original post by Bhaal85)
I only went up to P3, what is x^3+1 factorised? Is there a rule for factorising cubics quickly? I had to use some P2 stuff, and equate like terms:

1/[(x+1)(x^2-x+1)] ?
Then use partial fractions.
9. (Original post by mikesgt2)
Then use partial fractions.
1/3(x+1)^-1 + 2/3(x^2-x+1)^-1

Is it assumed that x=sinx? I have never been told that before, the only reason I mention this is that you said something about arcsin?
10. (Original post by Bhaal85)
1/3(x+1)^-1 + 2/3(x^2-x+1)^-1

Is it assumed that x=sinx? I have never been told that before, the only reason I mention this is that you said something about arcsin?
Im not sure your partial fractions are correct. The numerator of the second fraction should have an x-coefficient.
11. (Original post by Ralfskini)
I got as far as 1/3ln(x+1) + int.(2-x)/(x^2-x+1)

(although its probably littered with mistakes as I'm too tired to work effectively at this hour).
This is correct so far, but I think the integral should also be multipled by 1/3.

(Original post by Bhaal85)
1/3(x+1)^-1 + 2/3(x^2-x+1)^-1

Is it assumed that x=sinx? I have never been told that before, the only reason I mention this is that you said something about arcsin?
Your partial fractions are not correct, you need to find constants A,B,C such that

1/[(x+1)(x^2-x+1)] = A/(x+1) + (Bx+C)/(x^2-x+1)
12. (Original post by mikesgt2)
This is correct so far, but I think the integral should also be multipled by 1/3.

Your partial fractions are not correct, you need to find constants A,B,C such that

1/[(x+1)(x^2-x+1)] = A/(x+1) + (Bx+C)/(x^2-x+1)
I think this must be more advanced BE. I just assumed that I could do:

A/(x+1) + B/(x^2-x+1) alas, not that easy then. Oh well.
13. Peter, non linear partial fractions DO come up, make sure you're familiar with them, along with things like quadratic denominators.

E.g.

1/(x-1)(x+3)^2 into partial fractions.
14. (Original post by fishpaste)
Peter, non linear partial fractions DO come up, make sure you're familiar with them, along with things like quadratic denominators.

E.g.

1/(x-1)(x+3)^2 into partial fractions.
I am aware that that would be in the form:

1/(x-1)(x+3)^2 = A/(x-1) + B/(x+3) + C/(x+3)^2

However:

1/[(x+1)(x^2-x+1)] looked dissimiliar as the quadratic is inside the bracket, which is why I treated it the same, this is somewhat confusing.
15. (Original post by mik1a)
Some difficult P1, as I can't ask questions I don't understand!

1. A curve has the equation:

x = 4y^2 - 9
Find the area between the curve and the line x = 1

Find the minimum value of x, and the value of y that gives this.

When y = 17, and c is set to 9, find the equations of the normal to the curve and find where it meets the x axis.
Total area is twice area under curve from x=-9 to x=1.

A=∫{-9 -> 1}y dx
= ½ ∫ (x+9)^½ dx
= (1/2)(2/3)[(x+9)^(3/2)]{-9 -> 1}
= (1/3){(10)^(3/2) - (0)^(3/2)}
= (1/3)(10)^(3/2)
= 10.54
2A = 21.08
=======

From the graph,
min x = -9 @ y=0
======= ===

I'm not sure what you mean by 'c', but

at y=17, x = 4*17² - 9 = 1147

gives P(1147,17), a point on the curve.

dy/dx = 1/(dx/dy) = 1/(8y)
=> m = -8y, for normal
========

(y-y1) = m(x-x1)
y - 17 = -8*17(x - 1147)
y + 136x = 156,009, eqn of normal at P
=============

This normal meets x-axis, when y=0, giving
x = 156,009/17
x = 9,177
======
Attached Images

16. (Original post by mikesgt2)
I get this nasty thing:

[ ln(x^2 + 2x + 2) - ln(x^2 - 2x + 2) + 2arctan(x+1) + 2arctan(x-1) ]/16 + C
It seems that it is necessary to use the substitutions (x+1)^2 = tan(beta) and (x-1)^2 = tan(alpha), right?
But does it appear in P3?
17. (Original post by keisiuho)
It seems that it is necessary to use the substitutions (x+1)^2 = tan(beta) and (x-1)^2 = tan(alpha), right?
But does it appear in P3?
No.
18. (Original post by Bhaal85)
No.
It doesnt appear in P3 or the substitutions arent needed?
19. (Original post by keisiuho)
It doesnt appear in P3 or the substitutions arent needed?
Yes.
20. (Original post by Bhaal85)
Yes.
Postcount + 1

no you wont need to do anything with inverse trig functions in p3 (edexcel)

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