Linear First Order ODE/(1 + x^2) dy + 2 x y dx = cotangent x dx

Theorem
The linear first order ODE:
 * $(1): \quad \left({1 + x^2}\right) \mathrm d y + 2 x y \, \mathrm d x = \cot x \, \mathrm d x$

has the solution:
 * $y = \dfrac {\ln \left({\sin x}\right)} {1 + x^2} + \dfrac {C} {1 + x^2}$

Proof
$(1)$ can be written as:
 * $(2): \quad \left({1 + x^2}\right) \dfrac {\mathrm d y} {\mathrm d x} + 2 x y = \cot x$

We notice straight away that:
 * $\dfrac {\mathrm d} {\mathrm d x} \left({1 + x^2}\right) = 2 x$

and so:
 * $\dfrac {\mathrm d} {\mathrm d x} \left({1 + x^2}\right) y = \cot x$

Thus the general solution becomes:

or:
 * $y = \dfrac {\ln \left({\sin x}\right)} {1 + x^2} + \dfrac {C} {1 + x^2}$