Question

Simple calculus question

Answer 1

So simple it has no answer

Answer 2

Im stuck on a specific type of question, can someone help me?

Answer 3

do you know the integration of sec(t) ?

Answer 4

Why is that necessary?

Answer 5

its not really, but its a good check ...

Answer 6

Im not quite sure where sec come into this.

Answer 7

sec and csc have similar derivatives

i recall sec and then adjust for csc

Answer 8

the integration of sec was well known prior to the founding of integration :)

Answer 9

The derivative of csc i believe is -cscxcotx

Answer 10

the derivative yes, but the integration? its a sneaky little trick

Answer 11

\[\frac{csc}{1}*\frac{-csc-cot}{-csc-cot}\]
\[\frac{-csc^2-csc~cot}{-csc-cot}\implies~ln(csc+cot)\]

Answer 12

Im lost, simply how do you go from a derivative to an integral which is equal to the original function. Is it as simple as moving the derivative into the integral?

Answer 13

It's the fundamental theorem of calculus \[\frac{ d }{ dx } \int\limits_{a}^{x} f(t) dt = f(x)\]

Answer 14

we have 2 options, we can use the fundamental thrm:

\[\int_{a}^{x}f'(t)dt=F(x)-F(a)\] and take the derivative
\[\frac d{dx}\int_{a}^{x}f'(t)dt=f'(x)x'-f'(a)a'\]

Answer 15

So the derivative of the integral of a function in that format is equal to that function.. so since i have the derivative do i find the anti-derivative and plug that into the integral?

Answer 16

or we can work the integration, and then take the derivative, either way

Answer 17

My course has recently worked with the fundamental therom so I assume thats the way they want me to solve it.

Answer 18

you asked about which one it shouldbe, im just suggesting that in a pinch, you can work the long way if possible

Answer 19

So... does that make the answer b or am i on the wrong route?

Answer 20

It seems to me this problem is designed to be simple im just making it difficult.

Answer 21

it is designed to make you apply the fundamental thrm yes

Answer 22

Hey sorry, how did you get \[\frac{-\csc^2-\csc~\cot}{-\csc-\cot}\implies~\ln(\csc+\cot)\]

Answer 23

Hmm?

Answer 24

prolly an error in my head, thinkig to quick

-ln(csc + cot) derives to csc

like i said, im used to the sec form :)

Answer 25

x should be the high range, so its either the first or last option to me

Answer 26

Or b

Answer 27

the negation flips the limits, correct?

Answer 28

never mind

Answer 29

I know its not c (duh). I dont believe it is a. However i dont know whether i put the antiderivtive of the derivitive into the intergral or the derivative into the integral

Answer 30

IE, b or d

Answer 31

-ln(csc(x)+cot(x)) + C derives to csc(x)

since y = -ln(csc(x)+cot(x)) + C, i dont see why we would have a constant in the integration when dy/dx = csc(x)

Answer 32

a or d is my thought

Answer 33

the constant i believe has something to do with making the y value correct, they all have them so i think its correct.

Answer 34

A or D was what i started stuck between XD

Answer 35

im sorry b or d

Answer 36

\[y=\int y'(t)~dt +C\]
i see it now, these old eyes were placing it inside the dt

Answer 37

XD thats a correct therom? so that would make it d?

Answer 38

Essentially there youre using the intergral to find the antiderivative?

Answer 39

lets do the long way and check it out

\[y=\int_{a}^{x}csc(t)~dt+C=-ln(csc(x)+cot(x))+ln(csc(a)+cot(a))+C~\]
when x=4, y=-9
\[-9+ln(csc(4)+cot(4))=ln(csc(a)+cot(a))+C\]

Answer 40

im using the integral to determine that solution yes :)

Answer 41

let a=4, and C=-9

Answer 42

Thanks :) It really was simple XD

Answer 43

I figured

Answer 44

notice that by comparing like parts:

ln(csc(4)+cot(4)) = ln(csc(a)+cot(a)) , let a=4

-9 = C ...

Answer 45

another way to have viewed this, now that i have my bearings straight might be like this:

4 is in the domain element, it gets used as an input value ... it should be in the limit interval

-9 is a range element, its not necessarily a part of the integrals domain limits

Answer 46

but thats just a guess at what i see, i would not have determined that by just the FTC

Answer 47

Thanks: i have actually solved this equation before (and whtat you are saying sounds farmilar) but i was dead tired and couldnnt remember how i did it.

Answer 48

good luck :)

Answer 49

not to butt in, but why isn't the answer obviously 4?

Answer 50

Haha I was thinking the same thing, but anyway it was a nice question thanks @sccitestla and thanks @amistre64 :)