Solve: `dy / dx + y secx = tanx`

Explanation

Given `frac{d y}{d x}+y sec x=tan x`

It is of the form `frac{d y}{d x}+y p(x)=q(x)`

Integrating factor `e^{int p(x)}`

Integrating factor `{e}^{int sec {x}}`

`=>{e}^{ln |sec x+tan x|}=|sec x+tan x|`

Solution: `y|sec x+tan x|=int tan x(sec x+tan x mid) d x`

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`=>int tan x sec x d x+int tan ^{2} x d x`

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