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FP2 Revision Thread

Reply 180

19

it's fine (as long as q does not equal p) but in this case $\frac{q^3-p^3}{q-p} \ne q^2-p^2$

$q^3 - p^3$ is actually $(q-p)(q^2+qp+p^2)$

Reply 181

jadc

As long as q - p is not equal to 0 it is mathematically correct to cancel the brackets, but note that q^3 - p^3 = (q - p)(q^2 + pq + p^2).

Reply 182

kamie

10

I'm going back to basics here, but I just want to check I've got the theory right :smile:

$\int \sin^4 \theta \cos \theta dx = \frac{\sin^5 \theta}{5}$

because when differentiating $\sin^5 \theta$ you knock one off the power of sin and multiply by it, then times by its differential. I missed a section of the integration topic and some of its only just clicking now! If someone could confirm I got this right or correct me, I'd really appreciate it!

Reply 183

.A.

kamie

I'm going back to basics here, but I just want to check I've got the theory right :smile:

$\int \sin^4 \theta \cos \theta dx = \frac{\sin^5 \theta}{5}$

because when differentiating $\sin^5 \theta$ you knock one off the power of sin and multiply by it, then times by its differential. I missed a section of the integration topic and some of its only just clicking now! If someone could confirm I got this right or correct me, I'd really appreciate it!

In cases like this you must watch out for signs. This is fine.

If you have cos^5 instead of sin the the differential of cos would have been
-sin. Thus a negative answer.

.A.

Reply 184

Aihara

Quick question that arises from the P5 textbook:

How can you tell that cot(t/2) = tan(pi/2-t/2)?
It's for an intrinsic one but I would never have got that step.

Reply 185

Wish

9

$\int \sqrt{(1+\cos t)^2 +(\sin t)^2} \ \mathrm{d}t$

Hints please?

Reply 186

.A.

Aihara

Quick question that arises from the P5 textbook:

How can you tell that cot(t/2) = tan(pi/2-t/2)?
It's for an intrinsic one but I would never have got that step.

This might not be such a clever method but here it is. Since sin (pi/2 -t/2) = cos t/2. And vice versa. Then do cos (pi/2 - t/2 ) divided by sin (pi /2 - t/2)

.A.

Reply 187

Daniel Freedman

7

Wish

$\int \sqrt{(1+\cos t)^2 +(\sin t)^2} \ \mathrm{d}t$

Hints please?

$\displaystyle \int \sqrt{(1+\cos t)^2 +(\sin t)^2} \ \mathrm{d}t \\[br]\\ = \int \sqrt{1 + 2\cos{t} + \cos^2{t} + \sin^2{t}} \ \mathrm{d}t \\[br]\\ = \int \sqrt{2+ 2\cos{t}} \ \mathrm{d}t$

Now write cos(t) in terms of cos(t/2), and you'll find things cancel nicely :smile:

Reply 188

.A.

Wish

$\int \sqrt{(1+\cos t)^2 +(\sin t)^2} \ \mathrm{d}t$

Hints please?

1 + cos^2t +2cost + sin^2t

2 + 2 cost

so take 2 out. Integral will become root 2 x root (1 + cost)

Then use cost = 2cos^2t/2 -1

And I am sure you will be fine.

.A.

Reply 189

Aihara

Wish

$\int \sqrt{(1+\cos t)^2 +(\sin t)^2} \ \mathrm{d}t$

Hints please?

Expand it all - you should get the integral of rt(2+2cost)
Then use the half angle identity for cost ( $cost=2cos^2(t/2) - 1$ )
Then it should be obvious? :smile:

Reply 190

nota bene

15

Wish

$\int \sqrt{(1+\cos t)^2 +(\sin t)^2} \ \mathrm{d}t$

Hints please?

expand then use c^2+s^2=1 and 1+cost=2cos^2(t/2)

Reply 191

Daniel Freedman

7

.A.

1 + cos^2t +2cost + sin^2t

2 + 2 cost

so take 2 out. Integral will become root 2 x root (1 + cost)

Then use cost = 2cos^2t/2 -1

And I am sure you will be fine.

.A.

No need to take the 2 out :smile:

Reply 192

.A.

Daniel Freedman

No need to take the 2 out :smile:

Well depends :mad:

Joking. My bad. :smile:

Reply 193

dannyboi18

err try letting y = arcosh x

change it into its e stuff then ln stuff n multiply stuff by e to the x

bearn in mind ive only just started learning the exam today

my teach hasnt had time to teach me so i sat fp1 with only being taught in the morning and im teaching my self this

the only prob im having is with reduction can someone give a systematic easy approach

... btw edexcel people cud try n consider some colourful books for fp1 n onwards how freakin boring!!!

wb much luv guys

Reply 194

Steffiewoo

12

Colourful books, for maths!! I am a little jealous!
With reduction formulae sometimes you can just see it. Sometimes by parts straight away. Sometimes split the power and multiply or by parts. Sometimes a trig identity. Sometimes by parts twice. Its usually intuitive because if you get an ${I_{n-1}}$ then you can see you'll either be doing by parts once or using a trig, although occasionally with multiplying out. Just try what seems intuitive, and then if that doesn't work see which other methods may be sensible. Good luck all. :smile:

Reply 195

bobbles_lass

7

Looking at the heinemann site it seems the books for the new FP's will be colourful. No fair eh?

Yeah, FP2 I am going to fail, I make millions of stupid mistakes and can't do basic stuff (like logs...) but I get the difficult stuff.Gyah....

:frown: