Given (x + tan y)dy = sin 2y dx

This is a first order linear differential equation of the form

Here, P = –cosec 2y and

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

[∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let tan y = t

⇒ sec²y dy = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

[∵ t = tan y]

Thus, the solution of the given differential equation is