RealiseNothing
what is that?It is Cowpea
Re: HSC 2013 2U Marathon
is not a constant, you can't just take it outside the integral.
2)![](https://latex.codecogs.com/png.latex?\bg_white \frac{d}{dx} x \ln(x) \neq \frac{1}{x})
1)I think this is right?
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2)
1)I think this is right?
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this may help you see the answer more easilyI think this is right?
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hint:I'm not sure, all I know is 1/x integrated is ln(x), not sure about 1/ln(x).
Oh dear. Four days to get this out of your system.I think this is right?
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The area under a cosecx graph between 0 and pi/2?
The area under a cosecx graph between 0 and pi/2?
I'm getting infinity :/ Which is wrong.
Too hard lol
You would integrate it by writing 1/sinx as sinx/sinx^2 ie. sinx/1-cosx^2
so many optionsToo hard lol
But i do remember something...when integrating a cosecx, you multiply top and bottom by (csc^2 - cot^2) or (csc^2 + cot^2) or (csc + cot) or (csc - cot) then you use![]()
so say for instance we change the limits to pi/3 to p/2 the area wouldnt be infinite but is there anyway the function can be differentiated using 2U or 3U even?You would integrate it by writing 1/sinx as sinx/sinx^2 ie. sinx/1-cosx^2
Then perform substitution u=cosx , partial fractions etc. (too hard for 2U). Its just that the limits in your question stuffs things up.
ohh theres the answer how gay is that maths just turned into a guessing game
<------change csc and cot into sine and cosine, apply quotient rule to differentiate
In the end, this is just a memory exercise (unless somehow you knew to multiply top and bottom by that)
nah mate just stick to yourohh theres the answer how gay is that maths just turned into a guessing game
wow your questions are so amazing, i cant solve that teach me please, teach me your ways synah mate just stick to your
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what's wrong with thatnah mate just stick to your
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do it yourselfwow your questions are so amazing, i cant solve that teach me please, teach me your ways sy
but i cant because sy's questions are harddo it yourself