## marigirl one year ago Integration question:

1. marigirl

$-cosec(6x)\cot(6x)$

2. marigirl

$-cosec(6x)\cot(6x)$ $\frac{ -1 }{ \sin(6x) }*\frac{ \cos (6x) }{\sin (6x)}$ Integrate $\frac{ -\cos(6x) }{ \sin^2(6x) }$

3. misty1212

HI!!

4. marigirl

Hi Misty1212 I need guidance from here plzz

5. misty1212

do you know a function whose derivative is $\csc(x)\cot(x)$?

6. misty1212

hint, you should !

7. 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"

8. marigirl

if y=cosec(x) then y'=-cosec(x)cot(x)

9. misty1212

ok you got it

10. 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

11. misty1212

it it clear what i mean?

12. 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

13. marigirl

yes because when you integrate you tend to divide by the power multiplied by the coefficient of x?

14. misty1212

yes it is the chain rule backwards

15. misty1212

the check is easy right? take the derivative of $-\frac{1}{6}\csc(6x)$ and see that you get the integrand $\csc(6x)\cot(66x)$

16. misty1212

oops too many sixes, but you get the idea

17. 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?

18. misty1212

you still need the mental u - sub

19. misty1212

and it would not be what you wrote, because of the square in the denominator

20. misty1212

it would be more like $\int\frac{f'(x)}{f^2(x)}dx$

21. marigirl

sorry i meant $\frac{ -\cos(6x) }{ \sin^2(6x) }$

22. marigirl

Sorry I didnt think I made myself very clear Could I possibly use $A \int\limits \frac{ f'(x) }{ f(x) } dx= A \ln \left| fx \right| +c$

23. misty1212

no, as i said, you are missing the square

24. misty1212

$\int\frac{f'(x)}{f^2(x)}dx$j you will not get the log

25. 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

26. misty1212

lol it is easy to do with integration look for the simple first $\color\magenta\heartsuit$