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I remember a while back someone showed me how to derive the $\int \csc xdx$ by differentiating another term. The derivative would contain a $\csc x$ which could be made the subject. Both sides could then be integrated to derive the integral. I have spent this morning trying to figure it out but can't for the life of me.

[EDIT]

I didn't clarify what I meant. I am aiming for something like this: e.g.

$\frac{\mathrm{d}\ (x csc x )}{\mathrm{d} x} = cosecx - xcosecxcotx$

$\frac{\mathrm{d}\ (x csc x )}{\mathrm{d} x} + xcosecxcotx = cosecx$

Integrating both sides

$x csc x + \int (xcosecxcotx) dx= \int (cosecx)dx $

Obviously this isn't the right term to begin with but I was hunting a method that is along these lines.

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I would say:

$\int \csc x \ dx = \int \csc x \left( \frac {\csc x + \cot x}{\csc x + \cot x}\right) \ dx = \int \frac {\csc^2 x + \csc x\cot x}{\csc x + \cot x} \ dx\\ u = \csc x + \cot x\\ du = -\csc x\cot x - \csc^2 x \ dx\\ \int -\frac {1}{u} \ du\\ -\ln |\csc x + \cot x| + C$

And if you want to you could say:

$-\ln |\frac {1+\cos x}{\sin x}| + C\\ \ln |\frac {\sin x}{1+\cos x}| + C\\ \ln |\tan \frac 12 x| + C\\ $

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  • $\begingroup$ Thanks for the answer! I am aware of this proof but I was thinking of deriving it without substitution. For example $ \frac{\mathrm{d}\ (x csc x )}{\mathrm{d} x} = cosecx - xcosecxcotx $ By making the $cosecx$ a subject of the equation and then integrating both sides. I don't know if this is the term we are meant to start with but I was hoping a method similar to this was feasible. $\endgroup$ Apr 26 '18 at 1:29
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$$ \frac {d}{dx} \ln (\csc x + \cot x) = \frac {-\csc x \cot x - \csc ^2 x}{\csc x + \cot x } = -\csc x $$ Thus, $$ \int \csc xdx= -ln | \csc x + \cot x | +C$$

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