$\cos^2 x \frac{dy}{dx}$ + y = tan x ⇒ $\frac{dy}{dx}$ + $\sec^2$ x.y = $\sec^2$ x tan x This equation is in the form of $\frac{dy}{dx}$ + py = Q here p = $\sec^2$ x and Q = $\sec^2$ x tan x Integrating Factor , I.F = $e^{\int p\,dx}$ = $e^{\int \sec^2 x\,dx}$ = $e^{\tan x}$ The general solution can be given by y (I.F.) = ∫ (Q × I.F.) dx + C ... (1) Let tanx = t ⇒ $\frac{d}{dx}$ (tan x) = $\frac{dt}{dx}$ ⇒ $\sec^2$ x = $\frac{dt}{dx}$ ⇒ $\sec^2$ x dx = dt Therefore, equation 1 becomes : $y \cdot e^{\tan x}$ = ∫ $(e^t \cdot t)$ dt ⇒ y . $e^{\tan x}$ = ∫ $(e^t \cdot t)$ dt + C ⇒ y . $e^{\tan x}$ = t . ∫ $e^t$ dt - ∫ $\left(\frac{d}{dt}(t) \cdot \int e^t\,dt\right)dt$ + C ⇒ y . $e^{\tan x}$ = $t \cdot e^t - \int e^t\,dt$ + C ⇒ y . $e^{\tan x}$ = $t \cdot e^t - e^t$ + C ⇒ y . $e^{\tan x}$ = (t - 1) $e^t$ + C ⇒ y . $e^{\tan x}$ = (tan x - 1) $e^{\tan x}$ + C ⇒ y = (tan x - 1) + $C e^{-\tan x}$ , where C is an arbitary constant