Looking for a Casio graph....
I saw there were some connoisseurs here:
I have an offer for an original Casio 35 (not the 35+) for 30 Euro... Is is stupid buy this at this price?
PS - It needs to be a Casio, our course literature for material science is written for a 25+ or above.
PPS - I've been pulling my hair with this integral:
integrate (x/(1+3x)^4) dx
Solution -(9x+1)/54(1+3x)^3 +cte
I just need a pointer on how to do it... It's in a chapter about the substitution method, but it might be a mistake.
PPPS Sorry about being a mathspamcunt.
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Thu 12 Dec 2013, 11:10,
archived)
I have an offer for an original Casio 35 (not the 35+) for 30 Euro... Is is stupid buy this at this price?
PS - It needs to be a Casio, our course literature for material science is written for a 25+ or above.
PPS - I've been pulling my hair with this integral:
integrate (x/(1+3x)^4) dx
Solution -(9x+1)/54(1+3x)^3 +cte
I just need a pointer on how to do it... It's in a chapter about the substitution method, but it might be a mistake.
PPPS Sorry about being a mathspamcunt.
integrate (x/(1+3x)^4) dx
hmm seperate the x and (1+3x)^(-4) terms then integrate by parts.
There will be a trig substitution for 1/(1+3x) iirc
a^2+u^2 substitute u=a*tan(theta), where a=1 and u=sqrt(3x), that 4th power bit will probably come in handy but its a nasty problem (I hate integration by substitution).
Giving, 1/((1+3x)^4)=1/(a*tan(theta))^4,
I hope this helps :)
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Thu 12 Dec 2013, 11:42,
archived)
hmm seperate the x and (1+3x)^(-4) terms then integrate by parts.
There will be a trig substitution for 1/(1+3x) iirc
a^2+u^2 substitute u=a*tan(theta), where a=1 and u=sqrt(3x), that 4th power bit will probably come in handy but its a nasty problem (I hate integration by substitution).
Giving, 1/((1+3x)^4)=1/(a*tan(theta))^4,
I hope this helps :)
Oohh dear
Weird how the simple looking ones are a bitch to do...
Ta!
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Thu 12 Dec 2013, 12:03,
archived)
Ta!
Just realised that,
x=1/3(tan(theta))^2,
so there should be some nice cancelations to make!
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Thu 12 Dec 2013, 12:10,
archived)
x=1/3(tan(theta))^2,
so there should be some nice cancelations to make!
But but but
is that thing I did with the "t" to find the "x" legal? I was totally ad-libing there...
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Thu 12 Dec 2013, 17:17,
archived)
seems ok as you did define x, t, dx, dt, might be worth showing dx/dt or dt/dx to show how you got dx & dt though...
As far as I can see your equals signs hold everywhere which means it must be true! If the question didn't demand a specific substitution you should be fine. Merry Christams and good luck, also what is your field of study?
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Thu 12 Dec 2013, 18:02,
archived)
Engineering watch officer
Aka officer that's a mechanic on-board big ships (500 tons up).
Started last year. Passed all my exams except maths...
So I'm on the lookout for math trickery involving integrals and derivatives.
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Thu 12 Dec 2013, 18:06,
archived)
Started last year. Passed all my exams except maths...
So I'm on the lookout for math trickery involving integrals and derivatives.
And trigonemetry
But the calculus kind...
Cos(x)^3=cos(1-sin(x)^2) mumbo jumbo.
I'm so shit at trigo...
That's why I do thermodynamics and no map reading.
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Thu 12 Dec 2013, 18:17,
archived)
Cos(x)^3=cos(1-sin(x)^2) mumbo jumbo.
I'm so shit at trigo...
That's why I do thermodynamics and no map reading.
Trig is a bastard, it's worth knowing the complex exponential forms of sine and cosine as you can then avoid trying to remember the identities.
ie.
sin(x)=1/(2i)(e^(ix)-e^(-ix))
and,
cos(x)=1/2(e^(ix)+e^(-ix)
where,
i=sqrt(-1) (you might know 'i' as 'j')
Other than that I can only tell you the trick to learning maths is do lots of problems,
until your sick of em...
and then do some more.
Good luck with your studies :D
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Thu 12 Dec 2013, 20:34,
archived)
sin(x)=1/(2i)(e^(ix)-e^(-ix))
and,
cos(x)=1/2(e^(ix)+e^(-ix)
where,
i=sqrt(-1) (you might know 'i' as 'j')
Other than that I can only tell you the trick to learning maths is do lots of problems,
until your sick of em...
and then do some more.
Good luck with your studies :D