How to integrate (1/cos x) ,step by step?? not use formular?
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How to integrate (1/cos x) , or secx step by step?? not use formular?
the answer is ln (tan x + sec x) but how to do it step by step??
use winstan substitution?? but how? i cant seem to do it
i flipped all over the textbook but it say "memorize the following standard integrals" without showing the step by step how it's derived
can any maths pro please provide valuable guidance to a maths newbie like me?
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Integrate any 1/something = ln(something integrated) + C
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argghhh mathss!!!! miie hatess maths worrs... alwaes fail 0nes l0hs..
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Let's say you want to integrate 1/cos x.
After integration = ln sin x + C -
miieee teech you okies?Originally posted by jessie*:argghhh mathss!!!! miie hatess maths worrs... alwaes fail 0nes l0hs.. -
As for 1/sin x,
After integration = ln (tan x + sin x)
Why? Because the formula says so. Just memorise the formula if it is not given. -
at my level, i gotta derive it during exam, we are not allowed to simply "apply formular"Originally posted by aRcTiC_sTaR:As for 1/sin x,
After integration = ln (tan x + sin x)
Why? Because the formula says so. Just memorise the formula if it is not given.
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I'll use | for integrate symbol:
| 1/cos x dx
= | sec x dx
= | [sec x(sec x + tan x)]/(sec x + tan x) dx (ie, multiply (sec x + tan x) top and bottom)
= | (sec^2 x + sec x tan x)/(sec x + tan x) dx
Observe that d/dx (sec x + tan x) = (sec^2 x + sec x tan x), hence you can apply the logarithm derivative rule:
d/dx (ln [f(x)] = f'(x)/f(x)
Therefore,
| (sec^2 x + sec x tan x)/(sec x + tan x) dx
= ln (sec x + tan x) + c where c is an arbitrary constant. -
thnx man!
but the method is not as systematic way
muz remember the sec x + tanx factor to multiply.... might as well memorize
anyone tried using wistan substitution?
let tan x/2 = t
dx = (2 / 1 + t^2) dt
draw triangle blah blah, i cant seem to get it! always kanna stuck
tan(x/2) = t
cosx = (1-t²)/(1+t²) - (*)
Differentiating (*) w.r.t t
-sinx . dx/dt = 2tan-¹t
dx/dt = -2tan-¹t/[2t/(1+t²)]
dx/dt = [-2tan-¹t . (1+t²)]/2t
sinx = 2t/(1+t²)
INT [secx] .dx
= INT [1/(1-t²)/(1+t²) ]. dx/dt . dt
= INT [ [(1+t²)/(1-t²) . [-2tan-¹t . (1+t²)]/2t ].dt
= INT [ [-2tan-¹t(1+t²)²]/2t(1-t²) ].dt
expert please teach teach show me wadz wrong -
You know how to integrate 1/x? You know how to integrate cos x? Now you substitute the x in 1/x with cos x. Tada!
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It's a pity it won't work....Originally posted by ditzy:You know how to integrate 1/x? You know how to integrate cos x? Now you substitute the x in 1/x with cos x. Tada! -
whoa very smartOriginally posted by CoolMyth:I'll use | for integrate symbol:
| 1/cos x dx
= | sec x dx
= | [sec x(sec x + tan x)]/(sec x + tan x) dx (ie, multiply (sec x + tan x) top and bottom)
= | (sec^2 x + sec x tan x)/(sec x + tan x) dx
Observe that d/dx (sec x + tan x) = (sec^2 x + sec x tan x), hence you can apply the logarithm derivative rule:
d/dx (ln [f(x)] = f'(x)/f(x)
Therefore,
| (sec^2 x + sec x tan x)/(sec x + tan x) dx
= ln (sec x + tan x) + c where c is an arbitrary constant. -
then is there any sysmetic more way of solving it??
having to "guess" to multiply top and bottom tan x + sec x isnt sysmetic leh..
so is there?? -
there is no such thing as systematic way of doing integration, there are only strategies. and strategies are supposed to be flexible. if u see a certain question, dont right away use any strategy, try to simply it first.Originally posted by HanHao:then is there any sysmetic more way of solving it??
having to "guess" to multiply top and bottom tan x + sec x isnt sysmetic leh..
so is there?? -
Originally posted by HanHao:thnx man!
but the method is not as systematic way
muz remember the sec x + tanx factor to multiply.... might as well memorize
anyone tried using wistan substitution?
let tan x/2 = t
dx = (2 / 1 + t^2) dt
draw triangle blah blah, i cant seem to get it! always kanna stuck
tan(x/2) = t
cosx = (1-t²)/(1+t²) - (*)
Differentiating (*) w.r.t t
-sinx . dx/dt = 2tan-¹t
dx/dt = -2tan-¹t/[2t/(1+t²)]
dx/dt = [-2tan-¹t . (1+t²)]/2t
sinx = 2t/(1+t²)
INT [secx] .dx
= INT [1/(1-t²)/(1+t²) ]. dx/dt . dt
= INT [ [(1+t²)/(1-t²) . [-2tan-¹t . (1+t²)]/2t ].dt
= INT [ [-2tan-¹t(1+t²)²]/2t(1-t²) ].dt
expert please teach teach show me wadz wrong
Very "CHIM" lea..
Er...this type learn in A Level? or Poly or Uni??(Jus to widen my knowledge..i really dunno...)
I never even see b4...
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he is doing some kind of substitution, I didnt bother to figure out what hes doing because its wrong.Originally posted by bo liao: Very "CHIM" lea..
Er...this type learn in A Level? or Poly or Uni??(Jus to widen my knowledge..i really dunno...)
I never even see b4... -
I don't even do such things in uni. Such things are left for scientists, not engineers.Originally posted by bo liao: Very "CHIM" lea..
Er...this type learn in A Level? or Poly or Uni??(Jus to widen my knowledge..i really dunno...)
I never even see b4... -
Just a simple parameterization....you already know it, so no need to explain....Originally posted by HanHao:thnx man!
but the method is not as systematic way
muz remember the sec x + tanx factor to multiply.... might as well memorize
anyone tried using wistan substitution?
let tan x/2 = t
dx = (2 / 1 + t^2) dt
draw triangle blah blah, i cant seem to get it! always kanna stuck
tan(x/2) = t
cosx = (1-t²)/(1+t²) - (*)
Differentiating (*) w.r.t t
-sinx . dx/dt = 2tan-¹t
dx/dt = -2tan-¹t/[2t/(1+t²)]
dx/dt = [-2tan-¹t . (1+t²)]/2t
sinx = 2t/(1+t²)
INT [secx] .dx
= INT [1/(1-t²)/(1+t²) ]. dx/dt . dt
= INT [ [(1+t²)/(1-t²) . [-2tan-¹t . (1+t²)]/2t ].dt
= INT [ [-2tan-¹t(1+t²)²]/2t(1-t²) ].dt
expert please teach teach show me wadz wrong -
Haha I agree man, in uni, engineers-to-be oni apply the derived formulae, Actual derivation is left to the maths geeks haha (no offence to the Mathematicians)Originally posted by ditzy:I don't even do such things in uni. Such things are left for scientists, not engineers. -
haha..ic..cos i sch drop out..tot how come now teach till this kind of equation...lik maths fomular..haha..Originally posted by ditzy:I don't even do such things in uni. Such things are left for scientists, not engineers.
i see till blur
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I study engineering science, I have to derive and prove all the tough mathematical SHit, prove 0+0=0, what the hellOriginally posted by w2k_83:Haha I agree man, in uni, engineers-to-be oni apply the derived formulae, Actual derivation is left to the maths geeks haha (no offence to the Mathematicians) -
Why 0 + 0 = 0?Originally posted by the.raven:I study engineering science, I have to derive and prove all the tough mathematical SHit, prove 0+0=0, what the hell
By axiom of fields, there is an element 0 such that 0 + x = x for every element x in the field.
Since 0 is an element in the field, therefore 0 + 0 = 0.
Did I miss something? -
lol I was just joking abut the 0+0=0, just trying to show how weird the things I need to prove can get.Originally posted by CoolMyth:Why 0 + 0 = 0?
By axiom of fields, there is an element 0 such that 0 + x = x for every element x in the field.
Since 0 is an element in the field, therefore 0 + 0 = 0.
Did I miss something?
btw, what do u study, u seem to be maths expert 
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Currently majoring in Maths.Originally posted by the.raven:lol I was just joking abut the 0+0=0, just trying to show how weird the things I need to prove can get.
btw, what do u study, u seem to be maths expert
And yes, I'm surprised at how weird Maths can be in the advanced level, big headache.... when "0" is not zero and when "1" is not one.... *shakes head*