marigirl
  • marigirl
Integration question:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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marigirl
  • marigirl
\[-cosec(6x)\cot(6x)\]
marigirl
  • marigirl
\[-cosec(6x)\cot(6x)\] \[\frac{ -1 }{ \sin(6x) }*\frac{ \cos (6x) }{\sin (6x)}\] Integrate \[\frac{ -\cos(6x) }{ \sin^2(6x) }\]
misty1212
  • misty1212
HI!!

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marigirl
  • marigirl
Hi Misty1212 I need guidance from here plzz
misty1212
  • misty1212
do you know a function whose derivative is \[\csc(x)\cot(x)\]?
misty1212
  • misty1212
hint, you should !
misty1212
  • misty1212
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"
marigirl
  • marigirl
if y=cosec(x) then y'=-cosec(x)cot(x)
misty1212
  • misty1212
ok you got it
misty1212
  • misty1212
of course you have to adjust for the \(6x\) but that is a mental u - sub divide by 6 to take care of that
misty1212
  • misty1212
it it clear what i mean?
misty1212
  • misty1212
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
marigirl
  • marigirl
yes because when you integrate you tend to divide by the power multiplied by the coefficient of x?
misty1212
  • misty1212
yes it is the chain rule backwards
misty1212
  • misty1212
the check is easy right? take the derivative of \[-\frac{1}{6}\csc(6x)\] and see that you get the integrand \[\csc(6x)\cot(66x)\]
misty1212
  • misty1212
oops too many sixes, but you get the idea
marigirl
  • marigirl
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?
misty1212
  • misty1212
you still need the mental u - sub
misty1212
  • misty1212
and it would not be what you wrote, because of the square in the denominator
misty1212
  • misty1212
it would be more like \[\int\frac{f'(x)}{f^2(x)}dx\]
marigirl
  • marigirl
sorry i meant \[\frac{ -\cos(6x) }{ \sin^2(6x) }\]
marigirl
  • marigirl
Sorry I didnt think I made myself very clear Could I possibly use \[A \int\limits \frac{ f'(x) }{ f(x) } dx= A \ln \left| fx \right| +c\]
misty1212
  • misty1212
no, as i said, you are missing the square
misty1212
  • misty1212
\[\int\frac{f'(x)}{f^2(x)}dx\]j you will not get the log
marigirl
  • marigirl
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
misty1212
  • misty1212
lol it is easy to do with integration look for the simple first \[\color\magenta\heartsuit\]

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