Integration question:

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Integration question:

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[-cosec(6x)\cot(6x)\]
\[-cosec(6x)\cot(6x)\] \[\frac{ -1 }{ \sin(6x) }*\frac{ \cos (6x) }{\sin (6x)}\] Integrate \[\frac{ -\cos(6x) }{ \sin^2(6x) }\]
HI!!

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Hi Misty1212 I need guidance from here plzz
do you know a function whose derivative is \[\csc(x)\cot(x)\]?
hint, you should !
if you do not, i can tell you, but i assure you it is in your book under the section of "derivatives of trig functions"
if y=cosec(x) then y'=-cosec(x)cot(x)
ok you got it
of course you have to adjust for the \(6x\) but that is a mental u - sub divide by 6 to take care of that
it it clear what i mean?
you need to change the sign, and divide by 6 to get the derivative you want, i.e. \[-\frac{1}{6}\csc(6x)\] will work
yes because when you integrate you tend to divide by the power multiplied by the coefficient of x?
yes it is the chain rule backwards
the check is easy right? take the derivative of \[-\frac{1}{6}\csc(6x)\] and see that you get the integrand \[\csc(6x)\cot(66x)\]
oops too many sixes, but you get the idea
this 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?
you still need the mental u - sub
and it would not be what you wrote, because of the square in the denominator
it would be more like \[\int\frac{f'(x)}{f^2(x)}dx\]
sorry i meant \[\frac{ -\cos(6x) }{ \sin^2(6x) }\]
Sorry I didnt think I made myself very clear Could I possibly use \[A \int\limits \frac{ f'(x) }{ f(x) } dx= A \ln \left| fx \right| +c\]
no, as i said, you are missing the square
\[\int\frac{f'(x)}{f^2(x)}dx\]j you will not get the log
oh 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
lol it is easy to do with integration look for the simple first \[\color\magenta\heartsuit\]

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