Integral of cot(x)^2 dx

  • Thread starter algebrist
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  • #1
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Main Question or Discussion Point

well, can somebody please tell me how do i get that:
Integral of cot(x)^2 dx == -x - cot(x)

what technique of integration should be used here?
 

Answers and Replies

  • #2
matt grime
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try differentiating -x-cot(x) and see what you get. but if you don't like that remember 1-sin^2=cos^2
 
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  • #3
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hmm

well, of course i get cot(x)^2 when i differentiate......
the question is how do i solve the integral without knowing that it's equal to -x - cot(x)

if i use cos(x)^2 + sin(x)^2 == 1 i get to solve
Integral of 1/sin(x)^2 dx
which i don't find easier
 
  • #4
matt grime
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you know your trig derivatives? that's how you solve these, you konw what differentiates to give what you want, and cot's derivative is cosec^2 or whatever.
 
  • #5
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why should i care about cot's derivative? i don't see your point
 
  • #6
matt grime
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err, because if F is such that F' =f, then the integral of f is F+k some constant k? fundamental theorem of calc? the same reason why the first suggestion of differentiating -x-cot was offered? if you know the derivatives of basic functions it might help you solve the questions that get set in the exams? any of those reasons interesting? most integrals can't be solved in any nice way, so it's good to know the ones that can.
 
  • #7
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ha-ha-ha do you imply that i just have to know the result by heart???

what i am asking is how to solve
Integral of cot(x)^2 dx

i know the result, i want to know how to obtain it

i have tried integration by parts and trigonometric identities but nothing seems to work
 
  • #8
matt grime
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the way you solve it is by knowing the antiderivative of cosec^2, because it is an elementary function. If you don't want to learn these things, fine, but don't get shirty.
 
  • #9
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ok the antiderivative of cosec^2 = -cot
so how do i use it?
 
  • #10
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ahhh i see.......
 
  • #11
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but i would like to know how this antiderivative is obtained
 
  • #12
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as i told you i don't find
Integral of 1/sin(x)^2 dx
an easy one
 
  • #13
matt grime
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Go back to the third post. You did the subs 1-sin^2=cos^2 and you yourself said that you needed to integrate 1/sin^2, so go from there. I'm at a slight loss as to how you would integrate cos(x) say under your prohibitive system of never using antiderivatives. I mean that's what you're doing in integration.

Edit: for all the follow up posts:

if you know something differentiates to give the integrand, you know it is the integral (up to a constant)

that's what integration is.

how did you do the integral of cos(x) first time? or 1 for that matter.
 
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  • #14
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but my problem is that i can't integrate
1/sin^2

well, it must be possible to derive it by just knowing the derivatives of sin and cos only.
i tried integration by parts but it would'n work
 
  • #15
matt grime
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yes, you can integrate it since 1/sin^2 is cosec^2. unless I've imagined the previous 14 posts in my fevered mind. Perhaps you'd just better accept that you ought to have learnt all the trig derivatives? why must it be possible only knowing the derivative of sin and cos? unless the integrand is nice then you cannot do an integral. the vast majority (almost all) integrals do not fall under that heading.
 
  • #16
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anyway......... i see your point i just wanted to do it in a more straight-forward fashion because this antiderivative business seems artificial to me........

with your logic you can just learn by heart the derivatives of infinitely many functions and never use any technique of integration except for antiderivatives......
 
  • #17
arildno
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algebrist said:
with your logic you can just learn by heart the derivatives of infinitely many functions and never use any technique of integration except for antiderivatives......
Basically, any "technique of integration" is a trick to let us find antiderivatives,
unless you want to abolish the fundamental theorem of calculus and resort to compute the limits of infinite sums in another way.
 
  • #18
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well, i'm always rediscovering the wheel you know... :smile:
 
  • #19
matt grime
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i am not asking you to learn all the derivatives of an infinite number of functions, just the common or garden ones, the elementary ones, and knowing cot's derivative should be one of them, (it's just lilke tan after all). otherwise you couldn't do any integrals. i think cot is a basic one to know, you don't, it appears.


even when you use these techniques, you're only reducing it to a case where you know the anti-derivative, so you ought to at least know some of them.


exercise: find the integral of x^n without usung the notion of an anti derivative.
 
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  • #20
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algebrist,

There's not always a turn-key method for finding the integral of a function, the way there is for finding a derivative. Sometimes you have to just feel your way along until something works. Experience and practice help a lot.

If I hadn't known the answer to this problem I think I could have gotten it once it became just the integral of 1/sin(x)^2. That sin(x)^2 in the denominator looks like the last thing you do when you differentiate a ratio (bottom times derivative of the top, minus top times derivative of the bottom, all divided by the bottom squared). So, the integral must be a ratio with sin(x) as the denominator. Then, when I differentiate this integral I'm going to get two terms in the top (bottom times derivative of the top minus....) and their difference has to be 1. Whenever you're working with sines and cosines and you see a 1, think sin^2 + cos^2. And there are my two terms! They add to 1 and since the sin and cos are eachother's derivative and integral, things are looking good. Then I just have to fiddle around with the signs.
 
  • #21
pig
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algebrist, this is how you could calculate it without knowing it by heart:

integral of dx/sin(x)^2

I don't know what else to do here, so:

t=sinx, dt=cosxdx, dx=dt/cosx, dx=dt/sqrt(1-t^2)

= integral of dt/(t^2*sqrt(1-t^2))

This can be simplified by substituting 1/t:

u=1/t, du=-t^(-2)dt, dt=-t^2du

= - integral of du/sqrt(1-1/u^2)

By simple algebraic manipulation we can turn this into:

= - integral of udu/sqrt(u^2-1)

v=u^2-1, dv=2udu, du=dv/2u

= -1/2 integral of dv/sqrt(v)
= -sqrt(v) + C
= -sqrt(u^2-1) + C
= -sqrt(1/t^2 - 1) + C
= -sqrt((1-t^2)/t^2) + C
= -sqrt(cos(x)^2/sin(x)^2) + C
= -ctg(x) + C

The problem with integrals is that there isn't a smart way to directly calculate them. You can only transform them until you get something for what you know what it is a derivative of (up to a constant). :)
 
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  • #22
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word of advice..

hey algebrist,
just a word of advice, if your in physics, eng, or anything requiring high level calc your going to want to learn the trig derivatives and integrals. I don't want to sound harsh, but the truth is I can garentee you will fail when (or before) they start teaching vector functions and dimension n integrals since this **** is the basis of further assumed knowledge (hyperbolic trig integrals and whatnot) for those courses. Please for the love of your university (or college) carreer learn the trig...

Cheers
 
  • #23
arildno
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This thread being 2 years old, I'm not too sure OP is going to read your reply, since he hasn't logged onto PF during that period, either.
 

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