# Indefinite Integrals

• Jun 22nd 2011, 02:45 AM
nav92
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.
• Jun 22nd 2011, 02:50 AM
mr fantastic
Re: Indefinite Integrals
Quote:

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.
• Jun 22nd 2011, 04:01 AM
nav92
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.
• Jun 22nd 2011, 04:24 AM
CaptainBlack
Re: Indefinite Integrals
Quote:

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
• Jun 22nd 2011, 04:27 AM
CaptainBlack
Re: Indefinite Integrals
Quote:

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
• Jun 22nd 2011, 04:32 AM
tom@ballooncalculus
Re: Indefinite Integrals
Quote:

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...

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

Balloon Calculus: standard integrals, derivatives and methods
• Jun 22nd 2011, 04:45 AM
CaptainBlack
Re: Indefinite Integrals
Quote:

Originally Posted by tom@ballooncalculus
Typo - rather :

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

fixed

CB
• Jun 22nd 2011, 07:10 AM
nav92
Re: Indefinite Integrals
Quote:

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...

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

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?
• Jun 22nd 2011, 07:29 AM
tom@ballooncalculus
Re: Indefinite Integrals
Quote:

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?)
• Jun 22nd 2011, 07:51 AM
nav92
Re: Indefinite Integrals
Quote:

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.
• Jun 22nd 2011, 08:01 AM
tom@ballooncalculus
Re: Indefinite Integrals
sin(2x) = 2 sinx cosx
• Jun 23rd 2011, 04:28 AM
nav92
Re: Indefinite Integrals
Quote:

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.
• Jun 23rd 2011, 06:35 AM
tom@ballooncalculus
Re: Indefinite Integrals
Hi there,

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

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

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

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

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

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

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 AM
nav92
Re: Indefinite Integrals
Roger that, thank you.