Question

what is the integral of csc^5 6x cot 6x dx???

Answer 1

strip out a csc\[\int\csc^5(6x)\cot(6x)dx=\int\csc^4(6x)\cdot\csc(6x)\cot(6x)dx\]now it's an easy u-sub\[u=\csc(6x)\to du=-6\csc(6x)\cot(6x)\]\[-\frac16\int u^4du\]

Answer 2

thanks for ur help!!

Answer 3

welcome :D

Answer 4

if u dont mind, can u help me with one more??

Answer 5

sure

Answer 6

thanks

Answer 7

what is the integral of (x dx) / sin^2 4x^2

Answer 8

\[\int\frac{xdx}{\sin^2(4x^2)}\]?

Answer 9

yes

Answer 10

\[\int\frac{xdx}{\sin^2(4x^2)}\]\[u=4x^2\]do you have any idea what to do next?

Answer 11

in u substitutions we need u and du...
we can't integrate without a differential

Answer 12

let me see, then du= 8x dx

Answer 13

right

Answer 14

1/8 du = x dx

Answer 15

wait I'm sorry that won't work!

Answer 16

that would be the right answer though...

Answer 17

so now what do we do next?

Answer 18

integration by parts

Answer 19

what do u mean by that?

Answer 20

\[\int udv=uv-\int vdu\]

Answer 21

how do u come up with that?

Answer 22

it's an integration technique you learn in calc II
what level calculus are you in?

Answer 23

calc I

Answer 24

are there any way we can move sin from denominator to numerator??

Answer 25

sort of :D\[\frac{x}{\sin^2(4x^2)}=x\csc^2(4x^2)\]now use the same substitution
good idea ;)

Answer 26

oh really??? it was just my random thought

Answer 27

think you can do it now?

Answer 28

let me try, im not sure though, just give me a minute

Answer 29

oh wait a minute, what do i do with cot

Answer 30

what do you mean ?

Answer 31

when i did the u substitution, i got du=-8x csc 4x^2 cot 4x^2 dx

Answer 32

that's not the sub I showed you...

Answer 33

u=4x^2
du=8xdx
1/8csc^2udu

Answer 34

but then we can't take antiderivative with csc ^2

Answer 35

\[u=4x^2\]\[du=8xdx\]\[\frac18\int\csc^2udu\]oh yes we can, think back

somethings derivative is -csc^2

Answer 36

oh yeah so the antiderivaative would be cot, right??

Answer 37

yeah, but we're missing the negative sign, so we have to add it\[\frac18\int\csc^2udu=-\frac18\int-\csc^2 udu=-\frac18\cot u\]

Answer 38

...+C

Answer 39

so the final answer is -1/8 cot 4x^2

Answer 40

yep

Answer 41

oh yes, i always forget the +C

Answer 42

thanks so much!! that was such a big help

Answer 43

anytime !

Answer 44

i think i did this wrong

Answer 45

when i did the u sub du= -8x csc 4x^2 cot 4x^2, so i substitute all those into the integral x csc^2 4x^2, so the x and csc^2 cancelled out but then we have u cot 4 x^2 left, and we still can't integrate it like that

Answer 46

how did you get that du ?
u=4x^2
du=8xdx
you don't include the csc in that substitution

Answer 47

oh never mind, my bad i made this harder than it should be

Answer 48

\[\int\frac{xdx}{\sin^2(4x^2)}=\int x\csc^2(4x^2)dx\]\[u=4x^2\to du=8xdx\]\[\frac18\int\csc^2udu=-\frac18\cot u+C=-\frac18\cot(4x^2)+C\]ta dah!

Answer 49

oh yeah i just got it now, sorry!!

Answer 50

no prob :D

Answer 51

im sorry to bother u this much, but i have one more problem, if y+x = sin y cos x, what is dy/dx

Answer 52

implicit differentiation!

Answer 53

yes

Answer 54

taking the derivative of both sides with respect to x we need to remember that we need to use the chain rule on functions of y\[\frac d{dx}f(y)=f'(y)y'\]we also need the product rule on the right part\[y'+1=\cos x\cdot\cos y\cdot y'+\sin y\cdot(-\sin x)\]solve for y'

Answer 55

but dont we need to separate all y to one side and all x to other side?

Answer 56

you need to get y' by itself, however you want to go about it

Answer 57

is that the same thing?

Answer 58

yeah, I guess so
you said something about getting x on one side and y on the other though, so it's not quite that
just solve for the variable y', it's algebra

Answer 59

sorry i got it mixed up with integration

Answer 60

this one ask for derivative, not antiderivative, well i guess cuz i been doing the the antideriv

Answer 61

I gave you the result of taking the derivative, you just have to solve for the variable

Answer 62

oh ok thanks :D

Answer 63

for final answer i got, (-1-sin x sin y) / (1-cos x cos y)

Answer 64

thanks for helping me today!!

Answer 65

you're right!
you're also welcome :)