Question

can someone show me how to integral this question

Answer 1

\[\int\limits \frac{ \cos^5x }{ \sin^7x } dx\]

Answer 2

i split it into \[\int\limits \frac{ \cos^5x }{ \sin^5x }dx +\int\limits \frac{ 1 }{ \sin^2x }dx \]

Answer 3

\[\int\limits \tan5xdx + \int\limits \csc^2xdx\]

Answer 4

i mean cot^5x

Answer 5

\(\Large \int\limits \dfrac{ \cos^5x }{ \sin^7x } dx\)


\(\Large = \int\limits \cot^5x \csc^2 x dx\)

Answer 6

oh yea i messed up i cannot split it

Answer 7

youu have a natural differentiation pattern in there :-)

Answer 8

it make sense now thank you

Answer 9

you're thnking \((cot x)' = - csc^2 x\)?!?!

Answer 10

yea

Answer 11

brill!

Answer 12

the answer i got is \[-\frac{ \cot^6x }{ 6 }+C\]

Answer 13

\[\int\limits \cot^5x(1+\cot^2x)dx\]

Answer 14

u=cotx
du=-csc^2xdx

Answer 15

\(\Large \dfrac{d}{dx} (-\frac{ \cot^6x }{ 6 }+C) \)

\(\Large = \dfrac{d}{dx} (-\frac{ \cot^6x }{ 6 }) \)

\(\Large = (-6 \frac{ \cot^5x }{ 6 }) *(- csc^2 x)\)

\(\Large = \cot^5x ~ csc^2 x\)

Answer 16

so your answer was right and then ..... ????

Answer 17

what does it mean by Rewrite the given integral using this change of variables.?

Answer 18

\[\int\limits \frac{ \cos^5x }{ \sin^7x }dx =\int\limits du\]

Answer 19

you'll need a pro to answer that. i just do it on sight

Answer 20

i tried \[\int\limits u^5 du\]

Answer 21

and it says its wrong

Answer 22

so \( u = \cot x\) type thing?

Answer 23

yea

Answer 24

it said rewrite it so i changed everything in U form

Answer 25

work it through then to see how it plays

Answer 26

du = ??

Answer 27

oh i left out the negative sign -

Answer 28

personal preference: substitution takes your eye of the ball, it is far better to look for patterns if you can see them

Answer 29

so it's \(u = \cot x~ du = - \csc^2 x\)??

that's how you did it?

Answer 30

with a dx at the end!

Answer 31

ah! is that the bit you missed?

Answer 32

yea need to pay attention more

Answer 33

i've just done it myself on paper

the csc cancels out because you do that stuff with the dx-du

that's witchcraft!!!

Answer 34

but its also the chain rule :-))

Answer 35

Irish is a witch! Burn him at the stake!

Answer 36

nah mate!

i don't like the witch craft!!

Answer 37

That's exactly what a witch would say!

Answer 38

lol!!!!

Answer 39

Grand Inquisitor

Answer 40

i don't get subs though, most of the time, Torquemada!

Answer 41

That sounds like Grand Wizard... you racist bastard.

Answer 42

I get subs. They're delicious.

Answer 43

Cheap, yes

delicious?

Answer 44

Not subway ones. Real ones from other places.

Answer 45

Right

Answer 46

I don't get subs. I just don't.

Answer 47

Even the real ones from other places

Answer 48

that was 2 back, agent!

Answer 49

and then a load more!!

Answer 50

\[\int\limits \cot^5xcosec^2xdx\]

Use the substitution u = cotx
du = -cosec^2x dx

You have
\[-\int\limits u^5 du = -\frac{ u^6 }{ 6 } +c =-\frac{ 1 }{ 6}\cot^6x+c\]

Answer 51

Good meatball ones. Or chicken pamigiana.

Answer 52

sure pythagoreas, but then you are in a process and maybe not seeing the pattern. just a thought.

Answer 53

how do you choose the sub?

Answer 54

what is pamigiana?

Answer 55

Parmigiana.

Answer 56

Parmesan?!

In Old Blighty

Answer 57

that British comment did not go down well here :-)

Answer 58

this thread started so well

Answer 59

lol!

Answer 60

Parmesan is not parmigiana

Answer 61

Educate me

Answer 62

We done?

Answer 63

https://i.ytimg.com/vi/xjQDCrNdsVU/maxresdefault.jpg

Answer 64

that's just cheese!

Answer 65

Choose the substitution that when integrated gives the other product in the original integrand. E.g. in this case we choose u=cotx because when differentiated cotx yields -cosec^2x

Answer 66

awesome. so you're picking the thing deliberately. my bad.

Answer 67

but then why bother switching letters about? and you have to deal with the bit at the end, you know the dx -> du switch.

Answer 68

and how do you explain that bit?

Answer 69

It's breaded chicken, with marinara sauce, and cheese.

Answer 70

real Italian!

Answer 71

you only get pizza here, these days. and processed garbage.

Answer 72

\[u=cotx\]
\[\frac{ du }{ dx }=-cosec^2x\]
\[du=-cosec^2x dx\]
\[-du=cosec^2x dx\]
Original integrand is rewritten as
\[\int\limits \cot^5xcosec^2xdx\]

Substitute cosec^2x dx with -du
And cot^5x with u^5

Answer 73

yep. my question is how do you do the switch, the du for the dx

not trying to be an a/hole, jus interested.

Answer 74

in fact, what do the du and the dx mean?

Answer 75

i put it down to the chain rule. FWIW

Answer 76

@IrishBoy123 That I'm not really sure

Answer 77

yes, that's the thing i wonder about. because it is a brilliant process for doing stuff but what is really going on. it can be chain-ruled.

Answer 78

like

\(I = \int f(x) \ dx\)

\(\dfrac{dI}{dx} = f(x)\)

\(x = x(u)\)

\(\dfrac{dI}{du} = f(x) \dfrac {dx}{du}\)

\(I = \int f(x) \dfrac {dx}{du} \ du\)

or words to that effect

Answer 79

so it's that simple?

Answer 80

well, looks like everyone's died of boredom so here's a challenge

prove that \(\dfrac{dy}{dx} = \dfrac{1}{\dfrac{dx}{dy}}\)

without just flipping Liebnitz notation

Answer 81

You can use u = sin x as well.

\[\int\limits \frac{ \cos^5x }{ \sin^7x } dx = \int\limits \frac{ \cos^4x \cos x }{ \sin^7x }dx = \int\limits \frac{ (1-\sin^2x)^2\cos x }{ \sin^7x }dx = \int\limits \frac{ (1-u^2)^2 }{ u^7 }du\]

Answer 82

@IrishBoy123 isn't that like proving the inverse function derivative?

Answer 83

yes agent

when i found it, i had to look that up :-(