The integrating factor of the differential equation \(\displaystyle\frac{dy}{dx} + 2y\sec x = x\), \(0 \lt x \lt \displaystyle\frac{\pi}{2}\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((\sec x + \tan x)^2\)
Here \(P = 2\sec x\). Using \(\int \sec x\,dx = \log|\sec x + \tan x|\), and \(\sec x + \tan x \gt 0\) on \(0 \lt x \lt \frac{\pi}{2}\):
\( \displaystyle \int 2\sec x\,dx = 2\log(\sec x + \tan x) = \log(\sec x + \tan x)^2 \)
The integrating factor is
\( e^{\log(\sec x + \tan x)^2} = (\sec x + \tan x)^2 \)
Why the other options are incorrect
Option (1) \(\sec^2 x + \tan^2 x\) — Incorrect. This squares each term separately; \((\sec x + \tan x)^2\) also has the term \(2\sec x\tan x\).
Option (2) \(2(\sec x + \tan x)\) — Incorrect. The factor 2 multiplies the logarithm, so it becomes a power: \(e^{2\log u} = u^2\), not \(2u\).
Option (4) \((\sec x\tan x)^2\) — Incorrect. The integral of \(\sec x\) is \(\log|\sec x + \tan x|\), a sum, not \(\log|\sec x\tan x|\).