What is a solution to the differential equation dy/dx=2csc^2x/cotx?

What is a solution to the differential equation $\frac{dy}{dx}=\frac{2{\mathrm{csc}}^{2}x}{\mathrm{cot}x}$?
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Ashlee Ramos
The equation simplifies:
$\frac{dy}{dx}=2\mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)$
Separate variables:
$dy=2\mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)dx$
Integrate:
$\int dy=2\int \mathrm{csc}\left(x\right)\mathrm{sec}\left(x\right)dx$
$y=2\mathrm{ln}|\mathrm{sin}\left(x\right)|-2\mathrm{ln}|\mathrm{cos}\left(x\right)|+C$
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equipokypip1
More details
First we should separate the variables, which means that we can treat $\frac{dy}{dx}$ like division. We can move the dx to the right hand side of the equation to be with all the other terms including x.
$dy=\frac{2{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$
Now integrate both sides:
$\int dy=2\int \frac{{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$
On the right hand side, let u=cotx. This implies that $du=-{\mathrm{csc}}^{2}xdx$.
$y=-2\int \frac{-{\mathrm{csc}}^{2}x}{\mathrm{cot}x}dx$
$y=-2\int \frac{du}{u}$
$y=-2\mathrm{ln}|u|+C$
$y=-2\mathrm{ln}|\mathrm{cot}x|+C$
One possible simplification we could make if we wanted would be to bring the −1 outside the logarithm into the logarithm as a −1 power:
$y=2\mathrm{ln}|\mathrm{tan}x|+C$