1. ## 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.

2. ## Re: Indefinite Integrals

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.

3. ## Re: Indefinite Integrals

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.

4. ## Re: Indefinite Integrals

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

5. ## Re: Indefinite Integrals

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

6. ## Re: Indefinite Integrals

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

7. ## Re: Indefinite Integrals

Originally Posted by tom@ballooncalculus
Typo - rather :

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

CB

8. ## Re: Indefinite Integrals

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?

9. ## Re: Indefinite Integrals

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?)

10. ## Re: Indefinite Integrals

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.

11. ## Re: Indefinite Integrals

sin(2x) = 2 sinx cosx

12. ## Re: Indefinite Integrals

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.

13. ## Re: Indefinite Integrals

Hi there,

$\displaystyle \frac{1}{\cot(x)}\ (-\csc^2(x))$

$\displaystyle =\ \tan(x)\ \frac{-1}{\sin^2(x)}$

$\displaystyle =\ \frac{- \sin(x)}{\cos(x) \sin^2(x)}$

$\displaystyle =\ \frac{-1}{\sin(x) \cos(x)}$

$\displaystyle =\ \frac{-1}{\frac{1}{2} \sin(2x)}$

$\displaystyle =\ \frac{-2}{\sin(2x)}$

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

14. ## Re: Indefinite Integrals

Roger that, thank you.