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marigirl
 one year ago
Integration question:
marigirl
 one year ago
Integration question:

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marigirl
 one year ago
Best ResponseYou've already chosen the best response.0\[cosec(6x)\cot(6x)\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0\[cosec(6x)\cot(6x)\] \[\frac{ 1 }{ \sin(6x) }*\frac{ \cos (6x) }{\sin (6x)}\] Integrate \[\frac{ \cos(6x) }{ \sin^2(6x) }\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0Hi Misty1212 I need guidance from here plzz

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3do you know a function whose derivative is \[\csc(x)\cot(x)\]?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3if you do not, i can tell you, but i assure you it is in your book under the section of "derivatives of trig functions"

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0if y=cosec(x) then y'=cosec(x)cot(x)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3of course you have to adjust for the \(6x\) but that is a mental u  sub divide by 6 to take care of that

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3it it clear what i mean?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3you need to change the sign, and divide by 6 to get the derivative you want, i.e. \[\frac{1}{6}\csc(6x)\] will work

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0yes because when you integrate you tend to divide by the power multiplied by the coefficient of x?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3yes it is the chain rule backwards

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3the check is easy right? take the derivative of \[\frac{1}{6}\csc(6x)\] and see that you get the integrand \[\csc(6x)\cot(66x)\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3oops too many sixes, but you get the idea

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0this is a once step thing which is fantastic thank you! but if for some reason I decided to make things complicated and go \[\frac{ \cos(6x) }{ \sin^2(x) }\] can i use \[\int\limits \frac{ f'(x) }{ f(x) } \] rule somehow?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3you still need the mental u  sub

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3and it would not be what you wrote, because of the square in the denominator

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3it would be more like \[\int\frac{f'(x)}{f^2(x)}dx\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0sorry i meant \[\frac{ \cos(6x) }{ \sin^2(6x) }\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I didnt think I made myself very clear Could I possibly use \[A \int\limits \frac{ f'(x) }{ f(x) } dx= A \ln \left fx \right +c\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3no, as i said, you are missing the square

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3\[\int\frac{f'(x)}{f^2(x)}dx\]j you will not get the log

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0oh yea i get it now .. best method is to use the basic rules outlined . Thanks Heaps! I cant believe i made it so complicated lol

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3lol it is easy to do with integration look for the simple first \[\color\magenta\heartsuit\]
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