Thread: Indefinite Integrals

Given that y= ln cot x, find dy/dx. Hence, find ∫(2-cosec 2x)dx.

Okay basically, I'm stuck on how to differentiate y=ln cot x & then inserting it into the integration equation.

Originally Posted by nav92

Given that y= ln cot x, find dy/dx. Hence, find ∫(2-cosec 2x)dx.

Okay basically, I'm stuck on how to differentiate y=ln cot x & then inserting it into the integration equation.

To get dy/dx, use the chain rule (you will also need to use the quotient rule during this process).

I suggest you review some examples from your class notes or textbook (integration by recognition is the name this 'technique' is often given).

If you need more help, please show what you've done and say where you get stuck.

Alright, I know the how the chain rule works and here goes....

y= ln u ; u=cotx

dy/du= 1/u ; du/dx= -cosec^2 x

but why dy/dx cotx = -cosec^2 x . I don't get it.

Originally Posted by nav92

Alright, I know the how the chain rule works and here goes....

y= ln u ; u=cotx

dy/du= 1/u ; du/dx= -cosec^2 x

but why dy/dx cotx = -cosec^2 x . I don't get it.

Trivial, either it is a standard derivative and so you should know it, or write:

(d/dx)[cot(x)]=(d/dx)[cos(x)/sin(x)]=-1/sin^2(x)


CB

Originally Posted by nav92

Given that y= ln cot x, find dy/dx. Hence, find ∫(2-cosec 2x)dx.

Okay basically, I'm stuck on how to differentiate y=ln cot x & then inserting it into the integration equation.

You might like to consider using brackets around function arguments, and "^" to denote raising to a power.

CB

Originally Posted by CaptainBlack

cot(x)=cos(x)/sin(x)=(d/dx)[-1/sin^2(x)]

Typo - rather :

(d/dx)cot(x)=(d/dx)[cos(x)/sin(x)]=[-1/sin^2(x)]

And, just in case a picture helps with this part...



Balloon Calculus: standard integrals, derivatives and methods

Originally Posted by tom@ballooncalculus

Typo - rather :

(d/dx)cot(x)=(d/dx)[cos(x)/sin(x)]=[-1/sin^2(x)]

fixed

CB

Originally Posted by tom@ballooncalculus

Typo - rather :

(d/dx)cot(x)=(d/dx)[cos(x)/sin(x)]=[-1/sin^2(x)]

And, just in case a picture helps with this part...



Balloon Calculus: standard integrals, derivatives and methods

Perfect. That diagram really sparked my understanding in the derivation of cosec^2 (x).

Well now how to get dy/dx. I know it should be;
dy/dx= (1/cot x)*( -cosec^2 (x) )
= ??
The answer suppose to be dy/dx= -2 cosec 2x but I don't know how to get it?

Originally Posted by nav92

dy/dx= (1/cot x)*( -cosec^2 (x) )
= ??

... is correct. Then just say all of that in terms of sin and cos, and then apply the trig identity that gives 2 sin x cos x. (Know the one?)

Originally Posted by tom@ballooncalculus

... is correct. Then just say all of that in terms of sin and cos, and then apply the trig identity that gives 2 sin x cos x. (Know the one?)

Which trig identity are you referring to? I'm not sure.

sin(2x) = 2 sinx cosx

Originally Posted by tom@ballooncalculus

sin(2x) = 2 sinx cosx

Okay, now I'm not sure where should I put the sin(2x)= 2sinxcosx to get 2-cosec(2x). Pls elaborate on this.

dy/dx= (1/cot x)*( -cosec^2 (x) )
= ??

Thank you.

Hi there,













PS you have a typo in your latest (i.e. should be -2 not 2-, maybe that was the bug).

Roger that, thank you.