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FP2 Revision Thread

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Reply 20

DFranklin

achilleas999

$\int^{\pi/2}_0 e^{sinx}sinxcosx\, \mathrm {d}x$

:confused:

TIA

Substitute $u = \sin x$ , then integrate by parts.

Reply 21

=gabriel=

I meant [arcosh(x)]^2.
By parts produces 2 repeated integrals, the first one being [arcosh(x)]^2 itself and then x arcosh(x)/sqrt(x^2-1). If you substitute variables for both, everything cancels out and you end up with 0 = 0.

I tried Mathematica but it gave me a result I don't understand:

x [arcosh(x)]^2 - 2sqrt[(x-1)/(x+1)](x+1)arcosh(x) + 2x

No idea where it came from...

Reply 22

nota bene

=gabriel=

By parts produces 2 repeated integrals, the first one being [arcosh(x)]^2 itself and then x arcosh(x)/sqrt(x^2-1). If you substitute variables for both, everything cancels out and you end up with 0 = 0.

Usually when you end up with that 0 you have chosen the wrong u and v' for the second 'by parts'... it means you get back to where you were and all cancels. I suggest you try changing u and v'... Sorry that I can't be of more help, but I have never done hyperbolic trig...

Reply 23

insparato

Okay i think ive got it out today.

$\displaystyle \int (arcoshx)^2$

$u = arcoshx$

$\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$

$\frac{dv}{dx} = arcoshx$

$v = \displaystyle \int arcoshx$

Consider arcoshx integral

$\displaystyle \int arcoshx$

$u = arcoshx$

$\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$

$\frac{dv}{dx} = 1$

$v = x$

$\displaystyle \int arcoshx = xarcoshx - \frac{x}{\sqrt{x^2-1}}$

Consider x/rt(x^2-1) integral
$\displaystyle \int \frac{x}{\sqrt{x^2-1}}$

$u^2 = x^2 -1$

$2u\frac{du}{dx} = 2x$

$\frac{u}{x} du = dx$

$\displaystyle \int \frac{x}{u}.\frac{u}{x} = \int du = u = \sqrt{x^2-1}$

So $\displaystyle \int arcoshx = xarcoshx - \sqrt{x^2-1}$

$\displaystyle \int (arcoshx)^2 = arcoshx(xarcoshx - \sqrt{x^2-1}) - \int \frac{xarcoshx - \sqrt{x^2-1}}{\sqrt{x^2-1}}$

$\int (arcoshx)^2 = x(arcoshx)^2 - arcoshx\sqrt{x^2-1} - \int \frac{xarcoshx}{\sqrt{x^2-1}} - 1$

Consider the integral xarcoshx/sqrt(x^2-1)

$\displaystyle \int \frac{xarcoshx}{\sqrt{x^2-1}}$

$u = arcoshx$

$\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$

$\frac{dv}{dx} = \frac{x}{\sqrt{x^2-1}}$

$v = \sqrt{x^2-1}$

$\displaystyle \int \frac{xarcoshx}{\sqrt{x^2-1}} = arcoshx\sqrt{x^2-1} - \int 1$

$\displaystyle \int \frac{xarcoshx}{\sqrt{x^2-1}} = arcoshx\sqrt{x^2-1} - x$

So!

$\displaystyle \int (arcoshx)^2 = x(arcoshx)^2 - arcoshx\sqrt{x^2-1} - \int \frac{xarcoshx}{\sqrt{x^2-1}} - 1$

$\displaystyle \int (arcoshx)^2 = x(arcoshx)^2 - arcoshx\sqrt{x^2-1} - [arcoshx\sqrt{x^2-1} - x - x]$

$\displaystyle \int (arcoshx)^2 = x(arcoshx)^2 - 2arcoshx\sqrt{x^2-1} + 2x + C$

Now anyone who says you forgot your du's and dx's ill get nasty :p:

(only kidding)

Anyone spot any mistakes?

If its right my god that is an AWFUL lot of work. I think thats STEP worthy.

Reply 24

DFranklin

Sorry, I think it's easier than that. Just integrate by parts twice

$\int 1.(arcosh x)^2 dx = x (arcosh x)^2 - 2 \int \frac{x}{\sqrt{x^2-1}} arcosh x dx$ (using $\frac{d}{dx} arcosh x = \frac{1}{\sqrt{x^2-1}}$ )

$= x (arcosh x)^2 - 2 \sqrt{x^2-1} arcosh x + 2 \int \sqrt{x^2-1} \frac{1}{\sqrt{x^2-1}}$ (using $\int \frac{x}{\sqrt{x^2-1}} = \sqrt{x^2-1}$ )

$= x (arcosh x)^2 - 2 \sqrt{x^2-1} arcosh x + 2x + C$

Reply 25

=gabriel=

Thanks, good stuff. I obviously missed the second bit i.e. $\int \frac{x}{\sqrt{x^2-1}} arcosh x dx$ ...

Reply 26

insparato

Yeah i would take the long way round :smile:

Reply 27

DFranklin

insparato

Yeah i would take the long way round :smile:

Though getting through all those integrations without a single error in terms of factors of 2 or -1 is pretty impressive in its own right!

Reply 28

insparato

DFranklin

Though getting through all those integrations without a single error in terms of factors of 2 or -1 is pretty impressive in its own right!

Thanks

. I just gotta see the simpler things. I am pleased the answer was the same.

Reply 29

ad absurdum

insparato

Okay i think ive got it out today.

$\displaystyle \int (arcoshx)^2$

$u = arcoshx$

$\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$

$\frac{dv}{dx} = arcoshx$

$v = \displaystyle \int arcoshx$

Consider arcoshx integral

$\displaystyle \int arcoshx$

$u = arcoshx$

$\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$

$\frac{dv}{dx} = 1$

$v = x$

$\displaystyle \int arcoshx = xarcoshx - \frac{x}{\sqrt{x^2-1}}$

Consider x/rt(x^2-1) integral
$\displaystyle \int \frac{x}{\sqrt{x^2-1}}$

$u^2 = x^2 -1$

$2u\frac{du}{dx} = 2x$

$\frac{u}{x} du = dx$

$\displaystyle \int \frac{x}{u}.\frac{u}{x} = \int du = u = \sqrt{x^2-1}$

So $\displaystyle \int arcoshx = xarcoshx - \sqrt{x^2-1}$

$\displaystyle \int (arcoshx)^2 = arcoshx(xarcoshx - \sqrt{x^2-1}) - \int \frac{xarcoshx - \sqrt{x^2-1}}{\sqrt{x^2-1}}$

$\int (arcoshx)^2 = x(arcoshx)^2 - arcoshx\sqrt{x^2-1} - \int \frac{xarcoshx}{\sqrt{x^2-1}} - 1$

Consider the integral xarcoshx/sqrt(x^2-1)

$\displaystyle \int \frac{xarcoshx}{\sqrt{x^2-1}}$

$u = arcoshx$

$\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$

$\frac{dv}{dx} = \frac{x}{\sqrt{x^2-1}}$

$v = \sqrt{x^2-1}$

$\displaystyle \int \frac{xarcoshx}{\sqrt{x^2-1}} = arcoshx\sqrt{x^2-1} - \int 1$

$\displaystyle \int \frac{xarcoshx}{\sqrt{x^2-1}} = arcoshx\sqrt{x^2-1} - x$

So!

$\displaystyle \int (arcoshx)^2 = x(arcoshx)^2 - arcoshx\sqrt{x^2-1} - \int \frac{xarcoshx}{\sqrt{x^2-1}} - 1$

$\displaystyle \int (arcoshx)^2 = x(arcoshx)^2 - arcoshx\sqrt{x^2-1} - [arcoshx\sqrt{x^2-1} - x - x]$

$\displaystyle \int (arcoshx)^2 = x(arcoshx)^2 - 2arcoshx\sqrt{x^2-1} + 2x$

Now anyone who says you forgot your du's and dx's ill get nasty :p:

(only kidding)

Anyone spot any mistakes?

If its right my god that is an AWFUL lot of work. I think thats STEP worthy.

Sadly, yes. You've done all the difficult integration perfectly, and forgot +C :p:

Would you lose a mark for that in an A-Level exam? I've done this so many times and we lose marks for it :frown:

Reply 30

insparato

Yes you would. Im usually alright with the +C, this was quite involved so i was absorbed into finding the integral.

Reply 31

achilleas999

$given \, that \quad I_n= \int^{\pi/4}_0 \sec^n\theta \, \mathrm{d}\theta[br][br]\quad show by differentiating \quad \sec^{n-2}\theta\tan\theta [br][br]that \, for \, n\geq2 \,[br][br](n-1)I_n=(n-2)I_{n-2}+2^{(n-1)/2}[br][br][br]$

ffs it took me 20 minutes to write that in latex :eek:

Reply 32

DFranklin

It's easier to only use LaTeX for the maths bits.

So, did you differentiate the function they gave you? What did you get?

Reply 33

insparato

Yep differentiate it. Then you might want to think about integrating both sides :smile:

Reply 34

Eau

I suck at integration/differentiation. Coordinate is okay.

Reply 35

generalebriety

I hate this module.

That is all.

Reply 36

zrancis

I think ocr fp2 must be quite nice compared to what you all do!

Everyone in my class seems to be enjoying it more than anything.

Is there a topic on it that makes it horrible or does everyone just dislike it in its entirety?

Reply 37

Eau

generalebriety

I hate this module.

That is all.

Me too - most stressful math module ever. Seemingly easy - yet so hard.

Reply 38

generalebriety

Mauve

Me too - most stressful math module ever. Seemingly easy - yet so hard.

It's just ridiculous. There's always a question on "find the radius of curvature of this", which really just means "differentiate it twice, look in the formula book and don't make a mistake". And of course everyone does make a mistake. I don't think I need to mention hyperbolic equations, or reduction formulae...

Reply 39

Eau

generalebriety

Well, yeah - they are okay - for exam's sake - because they are easy marks. But I mean the huge amount of integration/differentiation is so annoying sometimes.

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