ashokkumar
Mr
integrate cosx/(sinx)^2
How'd you know the integral of (Cosec x Cot x) = - Cosec x? Is that just a general rule? (It isn't on the standard integrals )Cos x/(Sin x)^2 = Cosec x Cot x.
Integration(Cosec x Cot x) = - Cosec x.
Let u=sin(x): du=cos(x) dxreverse chain rule what?
learn your mafs! look in a 2unit textbook.reverse chain rule what?
Differentiation of Sec x = Sec x Tan x.How'd you know the integral of (Cosec x Cot x) = - Cosec x? Is that just a general rule? (It isn't on the standard integrals )
Basically what the 'reverse' chain rule states is :reverse chain rule what?
You shouldn't write your cosecs that way. Too easily mistaken with arcsin.Basically what the 'reverse' chain rule states is :
Thus for this example:
( csc x = cosec x )
This rule can be very helpful in Ext. 1 and Ext. 2.
Everywhere else in the world uses csc for Cosec. It is only the NSW syllabus which uses Cosec, which is kinda crap. (Much like they use -mCAT for chemistry)You shouldn't write your cosecs that way. Too easily mistaken with arcsin.
what he meant was if you write csc x = sin-1x it can be mistaken as arcsin which is generally written as sin-1xwhich way shouldnt i write cosecs? if you mean the 'csc' then i cant help that because thats how the function is loaded in the latex equation editor.
Yes the confusion between inverse trig with powers is stupid.what he meant was if you write csc x = sin-1x it can be mistaken as arcsin which is generally written as sin-1x
i.e csc x = sin-1x can be mistaken to mean csc x = arcsin x
That's the way to do it. By the way it's "with respect to sin x"integrate: cosx/sinx^2 dx
d/dx sinx = cosx
therefore dsinx = cosxdx
so we are now integrating
(1/sinx^2)d(sinx) ---> in respect to sinx.
and so:
=sinx^-2 d(sinx)
=(-1)sinx^-1
=-cosecx+c
i love integration. you have to be flexible <3