Simple calculus question

- anonymous

Simple calculus question

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- schrodinger

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- nincompoop

So simple it has no answer

- anonymous

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

- anonymous

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## More answers

- amistre64

do you know the integration of sec(t) ?

- anonymous

Why is that necessary?

- amistre64

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

- anonymous

Im not quite sure where sec come into this.

- amistre64

sec and csc have similar derivatives
i recall sec and then adjust for csc

- amistre64

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

- anonymous

The derivative of csc i believe is -cscxcotx

- amistre64

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

- amistre64

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

- anonymous

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?

- anonymous

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

- amistre64

we have 2 options, we can use the fundamental thrm:
\[\int_{a}^{x}f'(t)dt=F(x)-F(a)\] and take the derivative
\[\frac d{dx}\int_{a}^{x}f'(t)dt=f'(x)x'-f'(a)a'\]

- anonymous

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?

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

it is designed to make you apply the fundamental thrm yes

- anonymous

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

- anonymous

Hmm?

- amistre64

prolly an error in my head, thinkig to quick
-ln(csc + cot) derives to csc
like i said, im used to the sec form :)

- amistre64

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

- anonymous

Or b

- anonymous

the negation flips the limits, correct?

- anonymous

never mind

- anonymous

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

- anonymous

IE, b or d

- amistre64

-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)

- amistre64

a or d is my thought

- anonymous

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

- anonymous

A or D was what i started stuck between XD

- anonymous

im sorry b or d

- amistre64

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

- anonymous

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

- anonymous

Essentially there youre using the intergral to find the antiderivative?

- amistre64

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
\[-9+ln(csc(4)+cot(4))=ln(csc(a)+cot(a))+C\]

- amistre64

im using the integral to determine that solution yes :)

- amistre64

let a=4, and C=-9

- anonymous

Thanks :) It really was simple XD

- anonymous

I figured

- amistre64

notice that by comparing like parts:
ln(csc(4)+cot(4)) = ln(csc(a)+cot(a)) , let a=4
-9 = C ...

- amistre64

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

- amistre64

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

- anonymous

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.

- amistre64

good luck :)

- anonymous

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

- anonymous

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

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