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Year 13 Maths Help Thread

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Looks like I won't be wasting my frees this year, timetable comes on wednesday and I start thursday
How do you integrate √(1−sin⁡x) \surd (1-\sin x) given that π2≤x≤5π2 \frac{\pi }{2} \leq x \leq \frac{5\pi }{2} ?
(edited 7 years ago)
Original post by Ano123
How do you integrate √(1−sin⁡x) \surd (1-\sin x) given that π2≤x≤5π2 \frac{\pi }{2} \leq x \leq \frac{5\pi }{2} ?


Use u=sin(x)u=sin(x)
Why isn't the 4 negative?

(Edexcel Jan 2006 C3 Q6a if you need to question)

image.jpg
Original post by jamestg
Why isn't the 4 negative?

(Edexcel Jan 2006 C3 Q6a if you need to question)



It should be.. as far as I can see.
(edited 7 years ago)
Original post by jamestg
Why isn't the 4 negative?

(Edexcel Jan 2006 C3 Q6a if you need to question)




Ah I see, easy mistake to make :s-smilie: you've expanded cos(x-a) rather than cos(x+a) given in the question.
Original post by RDKGames
Use u=sin(x)u=sin(x)


What did you get?
Original post by SeanFM
Ah I see, easy mistake to make :s-smilie: you've expanded cos(x-a) rather than cos(x+a) given in the question.


Ahh I see, thank you! I need to be more careful next time :angry:
Original post by Ano123
What did you get?


Is that restriction for x supposed to mean the limits of integration? I'm unsure how significant it is, but to integrate that expression;

∫1−sin(x).dx\displaystyle \int \sqrt{1-\text{sin}(x)} .dx

let u=sin(x)→du=cos(x).dx\displaystyle u=sin(x) \rightarrow du=cos(x) .dx

→∫1−ucos(x).du\displaystyle \rightarrow \int \frac{\sqrt{1-u}}{cos(x)} .du

Draw yourself a right angled triangle. Use the fact that u=sin(x) and you will find that cos(x)=1−u2cos(x)=\sqrt{1-u^2}

→∫1−u1−u2.du=∫1−u(1+u)(1−u).du\displaystyle \rightarrow \int \frac{\sqrt{1-u}}{\sqrt{1-u^2}} .du = \int \frac{\sqrt{1-u}}{\sqrt{(1+u)(1-u)}} .du

and I'll let you finish that off.
(edited 7 years ago)
Original post by jamestg
Ahh I see, thank you! I need to be more careful next time :angry:


No worries, you learn more from mistakes than getting things right, and eventually for each topic you build a mental checklist (trig is probably the worst for it) of things to look out for. Eg radians/degrees, cancelling solutions, right range of solutions etc. And if you have to, underlining key bits of information given in the question.
Original post by Ano123
How do you integrate √(1−sin⁡x) \surd (1-\sin x) given that π2≤x≤5π2 \frac{\pi }{2} \leq x \leq \frac{5\pi }{2} ?


Note that 1−sin⁡x≡sin2(x/2)−2sin⁡(x/2)cos⁡(x/2)+cos⁡2(x/2) \sqrt{1-\sin x}\equiv \sqrt{sin^2(x/2) -2\sin (x/2)\cos (x/2) + \cos^2 (x/2)}
≡(sin⁡(x/2)−cos⁡(x/2))2≡∣sin⁡(x/2)−cos⁡(x/2)∣ \equiv \sqrt{(\sin (x/2)-\cos (x/2) )^2 } \equiv |\sin (x/2) - \cos (x/2)| .
(edited 7 years ago)
Original post by B_9710
This integral is not expected is C4 but you can still apply normal substitution methods.


What cud i do
Original post by youreanutter
What cud i do


Exactly that. Use substitution x=tanθx=\text{tan}\theta . The method itself is C3.

If you're asking "How do I know to substitute that?" then just start with y=arctan(x)y=\text{arctan}(x) and find dy/dx. The substitution is based off this result.
(edited 7 years ago)
Original post by RDKGames
Exactly that. Use substitution x=tanθx=\text{tan}\theta . The method itself is C3.

If you're asking "How do I know to substitute that?" then just start with y=arctan(x)y=\text{arctan}(x) and find dy/dx. The substitution is based off this result.


The substitution isn't based off that result. You could integrate 1/(1+x^2) independently without knowing the derivative of arctan x.
Original post by B_9710
The substitution isn't based off that result. You could integrate 1/(1+x^2) independently without knowing the derivative of arctan x.


As far as C3/C4 goes? How?
Original post by RDKGames
As far as C3/C4 goes? How?


It's not in C3/C4. But that doesn't change the methods for finding the integral.
Reply 536
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Palette
OP
14
I've cooked STEP I 2007 Q3; STEP I 2007 Q1 has almost finished cooking in the oven too.

EDIT: STEP I 2007 Q1 ready too though I did some dodgy work when I ended up proving the general case in order to do show the special case.
(edited 7 years ago)
Just joining this thread in advance now...

I've dropped from Further Maths to just Maths, and I my applied unit will be M1...
I'm sad to no longer be doing FMaths, but it didn't match my career aspirations, and there was no way I was taking 4 subjects :lol:

I hate being in just a standard maths class already, though :lol:
Reply 538
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Palette
OP
14
STEP I 2007 Q8 in the bag too. Will try more MAT and IYGB papers now.
It's GCSE question but I'm in year 13 so, and I need to revise GCSEstyle maths for UKCAT.

"A ford focus holds 13 gallons of petrol, how many litres of petrol can the car hold if 1 gallon is equivalent to 4.5 L? "

I tried two methods of working out, both of which I have used for other questions like this and I can't figure out which one is right since there aren't accessible answers.

Method 1:
1 gallon = 4.5 L
so
13 = 4.5 x 13 L = 58.5

Method 2 :
(1/4.5) x 13 = 2.88

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