Zacken
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STEP I (paper here)
1: Solution by JeremyC
2: Solution by _gcx
3: Solution by Aditya33
4: Solution by Zacken
5: Solution by gnsbrg
6: Solution by 123Master321
7: Solution by JeremyC
8: Solution by Notnek
9: Solution by Notnek
10: Solution by mikelbird
11: Solution by mikelbird
12: Solution by etothepiplusone
13: Solution by Lordkpaj

STEP II (paper here)
1: Solution by Aditya33
2: Solution by I hate maths
3: Solution by JeremyC
4: Solution by roboliffe
5: Solution by I hate maths
6: Solution by IrrationalRoot
7: Solution by mikelbird
8: Solution by Zacken
9: Solution by mikelbird
10: Solution by FractalSteinway
11: Solution by ciberyad
12: Solution by Integer123
13: Solution by Gregorius


STEP III (paper here):
1: Solution by mikelbird
2: Solution by Zacken
3: Solution by jtSketchy
4: Solution by Garjun
5: Solution by Zacken
6: Solution by mikelbird
7: Solution by DFranklin
8: Solution by Zacken
9: Solution by jtSketchy
10:Solution by jtSketchy
11: Solution by jtSketchy
12: Solution by jtSketchy
13: Solution by Zacken
Last edited by Sir Cumference; 1 year ago
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Sir Cumference
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STEP I Q8 :

(i)

Spoiler:
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To show s^3 + c^3 is constant we can show that the derivative of the expression is 0.

(s^3+c^3)' = 3s^2s' + 3c^2c' = -3c's' + 3s'c' = 0

So we have shown that s^3+c^3 = k where k is constant.

Then plugging in the initial conditions:

s(0) + c(0) = k \Rightarrow 0 + 1 = k \Rightarrow k = 1

thus proving that

\displaystyle s(x)^3 + c(x)^3 = 1



(ii)

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Using the product rule:

\displaystyle \left(sc)' = s'c + c's = c^2 c - s^2 s = c^3-s^3 = c^3-(1-c^3) = 2c(x)^3-1

Using the quotient rule:

\displaystyle \left(\frac{s}{c}\right)' = \frac{cs' - sc'}{c^2} = \frac{c^3+s^3}{c^2} = \frac{1}{c(x)^2}



(iii)

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\displaystyle \int s^2 \ \mathrm{d}x = -\int c' \ \mathrm{d}x= -c(x) + \text{constant}

\displaystyle \begin{aligned}\int s^5 \ \mathrm{d}x&= \int (1-c^3)s^2 \ \mathrm{d}x \\ &= \int s^2 - c^3s^2\ \mathrm{d}x \\ &= \int -c' + c^3c' \ \mathrm{d}x \\ &= -c(x) + \frac{1}{4}c(x)^4 + \text{constant}\end{aligned}




(iv)

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\displaystyle u = s \Rightarrow u' = s'

\displaystyle \begin{aligned} \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \ \mathrm{d}u &= \int \frac{1}{\left(1-s^3\right)^{\frac{2}{3}}} \cdot s' \ \mathrm{d}x \\ &= \int   \frac{1}{c^2} \cdot c^2 \ \mathrm{d}x \\ &= \int 1 \ \mathrm{d}x \\ &= x +   \text{constant} \\& = s^{-1}(u) + \text{constant}\end{aligned}




(v)

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Using the same substitution as above:

\displaystyle \int \frac{1}{\left(1-u^3\right)^{\frac{4}{3}}} \ \mathrm{d}u = \int \frac{1}{c^4} \cdot s' \ \mathrm{d}x = \int \frac{c^2}{c^4} \ \mathrm{d}x = \int \frac{1}{c^2} \ \mathrm{d}x

Then using the second result in (ii):

\displaystyle \begin{aligned} \int \frac{1}{c^2} \ \mathrm{d}x &= \frac{s}{c} + \text{constant} \\ &= \frac{u}{\left(1-u^3\right)^\frac{1}{3}} + \text{constant}\end{aligned}


Again using the same substitution:

\displaystyle \int \left(1-u^3\right)^{\frac{1}{3}} \ \mathrm{d}u = \int c\cdot s' \ \mathrm{d}x = \int c^3 \ \mathrm{d}x

Rearranging the first result in (ii) gives

\displaystyle \begin{aligned}\int c^3 \ \mathrm{d}x &= \frac{1}{2}\left(sc + \int 1 \ \mathrm{d}x \right) \\ &= \frac{1}{2}\left( u(1-u^3)^\frac{1}{3} + x\right) + \text{constant} \\ &= \frac{1}{2}u(1-u^3)^\frac{1}{3} + \frac{1}{2}s^{-1}(u) + \text{constant}\end{aligned}

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_gcx
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Question 2

i
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x = \log_b (b^x) so c = b^x.

Choose x such that c = b^x then \frac{\log_a c} {\log_a b} = \frac{\log_a b^x} {\log_a b} = \frac{x \log_a b} {\log_a b} = x = \log_b b^x.


Note \log_2 2 = 1 and \log_5 5 = 1 so:

\displaystyle \frac 1 {\log_2 \pi} + \frac 1 {\log_5 \pi} = \log_\pi 2 + \log_\pi 5 = \log_\pi 10

We're given that \pi^2 < 10 so \log_\pi 10 > \log_\pi \pi^2 = 2. \blacksquare





ii

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We have that:

\displaystyle \frac {\ln \frac \pi e} {\ln 2} = \frac{\ln \pi - 1} {\ln 2} > \frac 1 5

So:

\displaystyle \ln \pi > \frac 1 5 \ln 2 + 1

From e^2 < 8 we have 2 < \ln 8 = 3\ln 2. So \frac 2 3 < \ln 2 and:

\displaystyle \ln \pi > \frac 1 5 \ln 2 + 1 > \frac 2 {15} + 1 = \frac {17} {15}. \blacksquare





iii

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We have:

\displaystyle \frac{\ln 2} {\ln 10} > \frac 3 {10}

So:

\displaystyle \ln 2 > \frac 3 {10} \ln 10 > \frac 3 {10} \ln \pi^2 = \frac 3 5 \ln \pi

From e^3 > 20 we have 3 > \ln 20 = \ln 2 + \ln 10.

So 3 - \ln 2 > \ln 10 > \ln \pi^2 = 2 \ln \pi.

So 3 - 2\ln \pi > \ln 2 and \displaystyle \frac 5 3 (3 - 2 \ln \pi) > \ln \pi. So 3 - 2\ln \pi > \frac 3 5 \ln \pi, then 3 > \frac {13} 5 \ln \pi, so \ln \pi < \frac{5 \cdot 3} {13} = \frac {15} {13}.



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(Original post by Notnek)
Zacken the link to the STEP III paper in your post is going to make a lot of people very excited
Evil
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Student1256
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(Original post by Notnek)
STEP I Q8 :

(i)

Spoiler:
Show


To show s^3 + c^3 is constant we can show that the derivative of the expression is 0.

(s^3+c^3)' = 3s^2s' + 3c^2c' = -3c's' + 3s'c' = 0

So we have shown that s^3+c^3 = k where k is constant.

Then plugging in the initial conditions:

s(0) + c(0) = k \Rightarrow 0 + 1 = k \Rightarrow k = 1

thus proving that

\displaystyle s(x)^3 + c(x)^3 = 1



(ii)

Spoiler:
Show


Using the product rule:

\displaystyle \left(sc)' = s'c + c's = c^2 c - s^2 s = c^3-s^3 = c^3-(1-c^3) = 2c(x)^3-1

Using the quotient rule:

\displaystyle \left(\frac{s}{c}\right)' = \frac{cs' - sc'}{c^2} = \frac{c^3+s^3}{c^2} = \frac{1}{c(x)^2}



(iii)

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\displaystyle \int s^2 \ \mathrm{d}x = -\int c' \ \mathrm{d}x= -c(x) + \text{constant}

\displaystyle \begin{aligned}\int s^5 \ \mathrm{d}x&= \int (1-c^3)s^2 \ \mathrm{d}x \\ &= \int s^2 - c^3s^2\ \mathrm{d}x \\ &= \int -c' + c^3c' \ \mathrm{d}x \\ &= -c(x) + \frac{1}{4}c(x)^4 + \text{constant}\end{aligned}




(iv)

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\displaystyle u = s \Rightarrow u' = s'

\displaystyle \begin{aligned} \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \ \mathrm{d}u &= \int \frac{1}{\left(1-s^3\right)^{\frac{2}{3}}} \cdot s' \ \mathrm{d}x \\ &= \int   \frac{1}{c^2} \cdot c^2 \ \mathrm{d}x \\ &= \int 1 \ \mathrm{d}x \\ &= x +   \text{constant} \\& = s^{-1}(u) + \text{constant}\end{aligned}




(v)

Spoiler:
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Using the same substitution as above:

\displaystyle \int \frac{1}{\left(1-u^3\right)^{\frac{4}{3}}} \ \mathrm{d}u = \int \frac{1}{c^4} \cdot s' \ \mathrm{d}x = \int \frac{c^2}{c^4} \ \mathrm{d}x = \int \frac{1}{c^2} \ \mathrm{d}x

Then using the second result in (ii):

\displaystyle \begin{aligned} \int \frac{1}{c^2} \ \mathrm{d}x &= \frac{s}{c} + \text{constant} \\ &= \frac{u}{\left(1-u^3\right)^\frac{1}{3}} + \text{constant}\end{aligned}


Again using the same substitution:

\displaystyle \int \left(1-u^3\right)^{\frac{1}{3}} \ \mathrm{d}u = \int c\cdot s' \ \mathrm{d}x = \int c^3 \ \mathrm{d}x

Rearranging the first result in (ii) gives

\displaystyle \begin{aligned}\int c^3 \ \mathrm{d}x &= \frac{1}{2}\left(sc + \int 1 \ \mathrm{d}x \right) \\ &= \frac{1}{2}\left( u(1-u^3)^\frac{1}{3} + x\right) + \text{constant} \\ &= \frac{1}{2}u(1-u^3)^\frac{1}{3} + \frac{1}{2}s^{-1}(u) + \text{constant}\end{aligned}

You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are. Also I’m not sure if it was *necessary* to give the second last integral in a form not involving inverse functions because the other part made you give the answer in s^-1 form
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Zacken
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Note that we have \displaystyle f'(x) = \frac{(1-\log^2 x)(\log^3 x - 3\log^2 x-\log x-1)}{x^2 \log^2 x} so that we can see that f,f' = 0 when \ln^2 x = 1.

Let u = \log t, then for x \in (0,1) we have

\displaystyle 

\begin{align*}F(x) = \int_{-1}^{\log x} \frac{(1-u^2)^2}{u} \, \mathrm{d}u &= \left[\log |u| - u^2 + \frac{u^4}{4}\right]_{-1}^{\log x} \\ & = \log (-\log x) - \log^2 x + \frac{1}{4}\log^4 x + 1 - 1/4\end{align*}

For any x > 0 with x \neq 1, we have F(x) = \log |\log x| - \log^2 x + \frac{1}{4} \log^4 x + 1 - 1/4

i) So F(x^{-1}) = \log |-\log x| - (-\log x)^2 + \frac{1}{4}(-\log x)^4 + 1 - 1/4 = F(x)

ii) Check on Desmos.
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Zacken
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(Original post by Student1256)
You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are.
It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.
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Student1256
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(Original post by Zacken)
It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.
Yes when I was doing the exam I was like wow this really feels like trig. I’m just saying we didn’t need to know what the actual function is to use it to aid us in finding the integral
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JeremyC
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(Original post by Zacken)
It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.
That's what I thought when I first saw the question, but it's slightly different.

In the question,
s'(x) = c(x)^2
c'(x) = -s(x)^2

vs

s'(x) = c(x)
c'(x) = -s(x)
for the normal sine and cosine functions.

I agree with Student1256 - it's an awesome question.
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Sir Cumference
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(Original post by Student1256)
You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are. Also I’m not sure if it was *necessary* to give the second last integral in a form not involving inverse functions because the other part made you give the answer in s^-1 form
I agree, it is a well put together question.

I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?
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etothepiiplusone
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I can do Q12, it'll be 30 minutes, watch this space!

EDIT: Q12 parts i) and ii) : (. I need to comne back to this tomorrow to add parts iii) and iv), because bizarrely I can't do the algebra without the time pressure, when I was able to do it with pressure in the exam ...

EDIT EDIT: Done Zacken ! Although I will need to proofread this.

i)
Spoiler:
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\frac{p_1 + p_2 + p_3}{3}

ii)
Spoiler:
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E(N_1) = \sum_{x=0}^2 xP(N_1 = x) = P(N_1 = 1) + 2P(N_2 = 2) = 2p(1-p) + 2p^2 = 2p.

Recall  Var(N_1) = E(N_1 ^2) - E(N_1)^2. E(N_1 ^2) = P(N_1 = 1) + 4P(N_2 = 2) = 2p^2 + 2p = 2p(p+1). Thus Var(N_1) = 2p(1+p) - (2p)^2 = 2p - 2p^2 = 2p(1-p) as required.


iii)
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E(N_2) = P(N_2 = 1) + 2P(N_2=2) = \frac13 (p_1(1-p_2) + p_1(1-p_3) + p_2(1-p_1) + p_2(1-p_3) + p_3(1-p_1) + p_3(1-p_2)) + \frac23(p_1p_2 + p_2p_3 + p_3p_1) = 2p.

 E(N_2^2) = P(N_2 = 1) + 4P(N_2=2) = \frac13 (p_1(1-p_2) + p_1(1-p_3) + p_2(1-p_1) + p_2(1-p_3) + p_3(1-p_1) + p_3(1-p_2)) + \frac43(p_1p_2 + p_2p_3 + p_3p_1) = 2p + \frac23(p_1p_2 + p_2p_3 + p_3p_1).

Thus Var(N_2) = 2p - 4p^2 + \frac23(p_1p_2 + p_2p_3 + p_3p_1) . (Simplest form of that?)


iv)
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 Var(N_1) \geq Var(N_2)

 \Leftrightarrow 2p - 2p^2 \geq 2p - 4p^2 + \frac23(p_1p_2 + p_2p_3 + p_3p_1)

 \Leftrightarrow 2p^2 \geq \frac23(p_1p_2 + p_2p_3 + p_3p_1)

 \Leftrightarrow \frac29 (p_1^2 + p_2^2 + p_3^2 + 2p_1p_2 + 2p_2p_3 + 2p_3p_1) \geq \frac23(p_1p_2 + p_2p_3 + p_3p_1)

 \Leftrightarrow \frac29(p_1^2 + p_2^2 + p_3^2) \geq \frac29(p_1p_2 + p_2p_3 + p_3p_1)

 \Leftrightarrow \frac29(p_1^2 + p_2^2 + p_3^2) \geq \frac29(p_1p_2 + p_2p_3 + p_3p_1)

And this is true since (p_1-p_2)^2 + (p_2-p_3)^2 + (p_3-p_1)^2 \geq 0, where clearly equality holds iff the probabilities are equal.
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Zacken
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(Original post by JeremyC)
That's what I thought when I first saw the question, but it's slightly different.

In the question,
s'(x) = c(x)^2
c'(x) = -s(x)^2

vs

s'(x) = c(x)
c'(x) = -s(x)
for the normal sign and cosine functions.
Ah right yeah, juggling typing and looking on a phone so missed that
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Rohan77642
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(Original post by Zacken)
Note that we have \displaystyle f'(x) = \frac{(1-\log^2 x)(\log^3 x - 3\log^2 x-\log x-1)}{x^2 \log^2 x} so that we can see that f,f' = 0 when \ln^2 x = 1.

Let u = \log t, then for x \in (0,1) we have

\displaystyle</b>

<b>\begin{align*}F(x) = \int_{-1}^{\log x} \frac{(1-u^2)^2}{u} \, \mathrm{d}u &= \left[\log |u| - u^2 + \frac{u^4}{4}\right]_{-1}^{\log x} \\ & = \log (-\log x) - \log^2 x + \frac{1}{4}\log^4 x + 1 - 1/4\end{align*}

For any x &gt; 0 with x \neq 1, we have F(x) = \log |\log x| - \log^2 x + \frac{1}{4} \log^4 x + 1 - 1/4

i) So F(x^{-1}) = \log |-\log x| - (-\log x)^2 + \frac{1}{4}(-\log x)^4 + 1 - 1/4 = F(x)

ii) Check on Desmos.
How much do you (and other people) think I would get for the question if I did the parts highlighted in the solution?
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Zacken
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(Original post by GcseLad-_-)
Zacken Opinions on difficulty of the paper?
Haven’t had a chance to look properly, but I’ve seen a few very easy/short questions on a short glance (Q4/8/etc...) but my opinion might be wrong nowadays, after two years of uni maths. Think it’s just a standard paper. Not too hard, not too easy.
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gnsbrg
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STEP I Q5
I)
f(x)=k(x-1)(x-2)(x-3)(x-4)+1 where k is a non-zero real constant.

II)
From I) we know

P(x)=k(x-1)\ldots (x-N)+1 again non-zero k

P(N+1)=k(N)\ldots 1+1=kN!+1

If P(N+1)=1,then kN!=0,which is impossible.

Given P(N+1)=2, then kN!=1,that is

k=\frac{1}{N!}

Therefore,
P(N+r)=\frac{1}{N!}(N+r-1)\ldots (r)+1=\frac{(N+r-1)!}{N!(r-1)!}+1={N+r-1\choose N}+1

Using this, it is plain that r=2 verifies P(N+r)=N+r

III)

Since the coefficient of x^4 is 1, k=1

S(x)=(x-a)(x-b)(x-c)(x-d)+2001



a) If there were one, we would have that 

[latex](e-a)(e-b)(e-c)(e-d)=17

This is impossible if a,b,c,d are distinct and integers as 17 is prime and three factors should equal 1.

b)
S(0)=2017 implies

abcd=16=2^4

Taking into account a&amp;lt;b&amp;lt;c&amp;lt;d, we find that there are 5 sets of integers (a,b,c,d)=(-2,-1,2,4),(-4,-1,1,4),(-4,-2,1,2),(-2,-1,1,8),(-8,-1,1,2)

Thanks to @DFranklin for some corrections in this last bit.
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Student1256
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(Original post by Notnek)
I agree, it is a well put together question.

So you're saying you could leave the penultimate integral as s/c? I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?
Previously it said show that the integral is equal to s^-1(u) +c and that’s it. So I’m pretty sure if you left the other integrals in this form it’d be okay. I’m not saying to leave it in x. You still have to substitute back u
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Sir Cumference
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(Original post by Student1256)
Previously it said show that the integral is equal to s^-1(u) +c and that’s it. So I’m pretty sure if you left the other integrals in this form it’d be okay. I’m not saying to leave it in x. You still have to substitute back u
Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?
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(Original post by Notnek)
Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?
If you write x = s^{-1}(u) then you could have:
\frac{s(x)}{c(x)} = \frac{u}{c(s^{-1}(u))}
This is what I answered (plus the constant) but I think your answer is probably nicer as it avoids the inverse function - although I think both should hopefully get the marks.

A similar approach can be used for the second part of (v).
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Student1256
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(Original post by Notnek)
Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?
So (for the second last integral) I = s(x)/c(x) +constant. We used the substitution s(x)= u so x=s^-1(u)
So I= u/c(s^-1(u)) + constant.
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(Original post by Notnek)
I agree, it is a well put together question.

I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?
I think you'd lose a mark or 2 TBH. It's like writing cos(arcsin x) instead of sqrt(1-x^2).

But it wouldn't coat more than that, for sure.
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