Q.Integrating factor of the differential equation $\frac{dy}{dx}+y\tan x-\sec x=0$ is:
(A) $\cos x$
(B) $\sec x$
(C) $e^{\cos x}$
(D) $e^{\sec x}$

The key idea is the Integrating Factor Method for a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$. The integrating factor (I.F.) is $e^{\int P \, dx}$.

Step 1: Rewrite the given equation in standard form:

Here, $P(x) = \tan x$ and $Q(x) = \sec x$.

Step 2: Compute the integrating factor: …

The key to solving any first-order linear differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$ is the Integrating Factor Method. Why does it work? Because multiplying the entire equation by a specially chosen function $\mu(x)$ turns the left-hand side into the exact derivative of $\mu(x) y$, which we can then integrate directly. That function is $\mu(x) = e^{\int P(x) \, dx}$.

Here, the given equation is $\frac{dy}{dx} + y \tan x - \sec x = 0$. Let’s rewrite it in the standard linear form:

So $P(x) = \tan x$ and $Q(x) = \sec x$.

Now we compute the integrating factor step by step.

1. Identify $P(x)$ and set up the integral

   The integrating factor $\mu(x)$ is $e^{\int P(x) \, dx} = e^{\int \tan x \, dx}$.

2. Evaluate $\int \tan x \, dx$

   Recall that $\tan x = \frac{\sin x}{\cos x}$. Let $u = \cos x$, then $du = -\sin x \, dx$, so $\int \tan x \, dx = \int \frac{\sin x}{\cos x} dx = -\int \frac{du}{u} = -\log|u| + C = -\log|\cos x| + C$.

   A more common form is $\log|\sec x| + C$, since $-\log|\cos x| = \log|\sec x|$.

   For the integrating factor, we only need one antiderivative (the constant of integration is irrelevant — it cancels out), so we take $\int \tan x \, dx = \log|\sec x|$.

   Tip

   Remember: $\int \tan x \, dx = \log|\sec x| + C$ is a standard result. It’s faster than re-deriving every time.

3. Exponentiate to get $\mu(x)$

   $\mu(x) = e^{\log|\sec x|} = |\sec x|$ …

Method: Integrating Factor $e^{\int\tan x\,dx}$ — Watch the Exponent

The integrating factor is $e^{\int P\,dx}$, with the integral of $P$ in the exponent — not $P$ itself. Getting $\int\tan x\,dx$ right is the crux here.

Steps

Step 1: Put the equation in standard form.

$\frac{dy}{dx}+y\tan x-\sec x=0$ becomes $\frac{dy}{dx}+(\tan x)y=\sec x$, so $P=\tan x$.

Step 2: Integrate $P$ first.

$\int\tan x\,dx=\log|\sec x|.$ …

Common Mistakes

Mistake 1: Writing the integrating factor as $e^{\tan x}$ or $e^{\sec x}$.

Why it's wrong: the exponent must be $\int P\,dx$, not $P$ itself; with $P=\tan x$, the exponent is $\int\tan x\,dx=\log|\sec x|$. Correct approach: integrate $P$ before exponentiating.

Mistake 2: Not simplifying $e^{\log|\sec x|}$. …