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OpenStudy (anonymous):

Given: f'(x)=cscx(cotx-sin2x) what is f(x)? note: sin2x=2sinxcosx

OpenStudy (anonymous):

need a step by step explanation

OpenStudy (anonymous):

well this question is about integration

OpenStudy (anonymous):

sure will do

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

im not sure what to do tho lol

OpenStudy (anonymous):

not to worry follow my steps you'll get to know

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

please confirm if i copied your question correctly: \[f'(x) = \csc(x)(\cot(x)-\sin(2x))\]

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

ok good

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

my net is a bit slow today i need to format my computer but i sill show you all the steps ok

OpenStudy (jamesj):

*bookmark. I am curious to see if mathsmind ever gets to the end of an explaination in a finite amount of time

OpenStudy (anonymous):

lol ok

OpenStudy (anonymous):

you have been chasing me all day long mr james bond

OpenStudy (anonymous):

ok the first step: is simplification

OpenStudy (anonymous):

ok.

OpenStudy (anonymous):

now you should know by definition what cscx =?

OpenStudy (anonymous):

ok recall

OpenStudy (anonymous):

\[\csc(x) = \frac{ 1 }{ \sin(x) }\]

OpenStudy (anonymous):

1/sin

OpenStudy (anonymous):

good .. good

OpenStudy (anonymous):

ok so now do i integrate my new equation\'?

OpenStudy (jamesj):

Let's push this along a little, shall we? \[ f'(x) = \csc(x)(\cot(x)−\sin(2x)) = \frac{\cos x}{\sin^2 x} - 2\cos x \] yes?

OpenStudy (anonymous):

\[\int\limits \csc(x)(\cot(x)-\sin(2x))dx\]

OpenStudy (anonymous):

no i will make it simpler

OpenStudy (anonymous):

\[\int\limits\limits\limits (\csc(x)\cot(x)-\csc(x)\sin(2x))dx\]

OpenStudy (anonymous):

\[\int\limits \csc(x)\cot(x) dx - \int\limits \csc(x)\sin(2x) dx\]

OpenStudy (anonymous):

now here for sin(2x) i will use the trig identity you provided in the hint which is sin(2x) =2sin(x)cos(x)

OpenStudy (anonymous):

and we will use the definition of csc(x)

OpenStudy (anonymous):

this yields to

OpenStudy (anonymous):

\[\int\limits\limits \csc(x)\cot(x) dx - \int\limits\limits \ \frac{ 1 }{ \sin(x) }2\cos(x)\sin(x) dx\]

OpenStudy (jamesj):

You see? The questioner has gone. Are you there, @bmelyk?

OpenStudy (anonymous):

look at the 2nd integral

OpenStudy (anonymous):

\[\int\limits\limits\limits \csc(x)\cot(x) dx - 2\int\limits\limits\limits \ \cos(x) dx\]

OpenStudy (anonymous):

\[\int\limits\limits\limits\limits \csc(x)\cot(x) dx= - \csc(x) + C_1\]

OpenStudy (anonymous):

\[- 2\int\limits\limits\limits\limits \ \cos(x) dx=-2\sin(x) +C_2\]

OpenStudy (anonymous):

final answer is:

OpenStudy (anonymous):

\[=-(\csc(x) + 2\sin(x)) + C\]

OpenStudy (anonymous):

i hope this helps sorry my net is frustrating today

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