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So simple it has no answer

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

do you know the integration of sec(t) ?

Why is that necessary?

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

Im not quite sure where sec come into this.

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

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

The derivative of csc i believe is -cscxcotx

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

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

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

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

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

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

it is designed to make you apply the fundamental thrm yes

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

Hmm?

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

Or b

the negation flips the limits, correct?

never mind

IE, b or d

a or d is my thought

A or D was what i started stuck between XD

im sorry b or d

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

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

Essentially there youre using the intergral to find the antiderivative?

im using the integral to determine that solution yes :)

let a=4, and C=-9

Thanks :) It really was simple XD

I figured

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

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

good luck :)

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