The integrating factor of differential equation `cos x (dy)/(dx)+y sin x =1` is

Options

A. cos x
B. tan x
C. sec x
D. sin x

Explanation

To find the integrating factor of the differential equation \( \cos x \frac{dy}{dx} + y \sin x = 1 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the given equation:

\[ \cos x \frac{dy}{dx} + y \sin x = 1 \]

We can divide the entire equation by \( \cos x \) to simplify it:

\[ \frac{dy}{dx} + y \frac{\sin x}{\cos x} = \frac{1}{\cos x} \]

This simplifies to:

\[ \frac{dy}{dx} + y \tan x = \sec x \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
Now, we can identify \( p(x) \) and \( q(x) \) from the standard form \( \frac{dy}{dx} + p(x)y = q(x) \):

- \( p(x) = \tan x \)
- \( q(x) = \sec x \)

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by the formula:

\[ \mu(x) = e^{\int p(x) \, dx} \]

Substituting \( p(x) = \tan x \):

\[ \mu(x) = e^{\int \tan x \, dx} \]

### Step 4: Integrate \( \tan x \)
We know that:

\[ \int \tan x \, dx = -\log(\cos x) + C \]

Thus, we have:

\[ \mu(x) = e^{-\log(\cos x)} = \frac{1}{\cos x} = \sec x \]

### Step 5: Conclusion
The integrating factor for the differential equation \( \cos x \frac{dy}{dx} + y \sin x = 1 \) is:

\[ \mu(x) = \sec x \]