Given

This is a first order linear differential equation of the form

Here, P = sec²x and Q = tan x sec²x

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

We have

∴ I.F = e^{tan x}

Hence, the solution of the differential equation is,

Let tan x = t

⇒ sec²xdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ ye^t = te^t – e^t + c

⇒ ye^t × e^–t = (te^t – e^t + c)e^–t

⇒ y = t – 1 + ce^–t

∴ y = tan x – 1 + ce^{–tan x} [∵ t = tan x]

Thus, the solution of the given differential equation is y = tan x – 1 + ce^{–tan x}