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.

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- Jun 22nd 2011, 02:45 AMnav92Indefinite 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. - Jun 22nd 2011, 02:50 AMmr fantasticRe: Indefinite Integrals
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. - Jun 22nd 2011, 04:01 AMnav92Re: 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. - Jun 22nd 2011, 04:24 AMCaptainBlackRe: Indefinite Integrals
- Jun 22nd 2011, 04:27 AMCaptainBlackRe: Indefinite Integrals
- Jun 22nd 2011, 04:32 AMtom@ballooncalculusRe: Indefinite Integrals
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...

http://www.ballooncalculus.org/asy/cotDiff.png

Balloon Calculus: standard integrals, derivatives and methods - Jun 22nd 2011, 04:45 AMCaptainBlackRe: Indefinite Integrals
- Jun 22nd 2011, 07:10 AMnav92Re: Indefinite Integrals
- Jun 22nd 2011, 07:29 AMtom@ballooncalculusRe: Indefinite Integrals
- Jun 22nd 2011, 07:51 AMnav92Re: Indefinite Integrals
- Jun 22nd 2011, 08:01 AMtom@ballooncalculusRe: Indefinite Integrals
sin(2x) = 2 sinx cosx

- Jun 23rd 2011, 04:28 AMnav92Re: Indefinite Integrals
- Jun 23rd 2011, 06:35 AMtom@ballooncalculusRe: Indefinite Integrals
Hi there,

PS you have a typo in your latest (i.e. should be -2 not 2-, maybe that was the bug). - Jun 23rd 2011, 07:34 AMnav92Re: Indefinite Integrals
Roger that, thank you.