Integrating factor of the differential equation cos x dy/dx + y sin x = 1 is

The correct option is sec x.

To find the integrating factor of the given differential equation, we first write it in the standard form of a linear first-order differential equation.
The given differential equation is:

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

Dividing the entire equation by $\cos x$ (assuming $\cos x\ne 0$), we get:

$$\frac{dy}{dx}+y\left(\frac{\sin x}{\cos x}\right)=\frac{1}{\cos x}$$

Simplifying the trigonometric terms, we obtain:

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

This is now in the standard first-order linear differential equation form:

$$\frac{dy}{dx}+P(x)y=Q(x)$$

By comparing the two equations, we can identify:

$$P(x)=\tan x$$

The Integrating Factor (I.F.) is given by the formula:

$$\text{I.F.}=e^{\int P(x)dx}$$

Substituting $P(x)=\tan x$ into the formula:

$$\text{I.F.}=e^{\int \tan xdx}$$

Since the integral of $\tan x$ is $\ln |\sec x|$, we have:

$$\text{I.F.}=e^{\ln |\sec x|}$$

Using the exponential property $e^{\ln u}=u$, the equation simplifies to:

$$\text{I.F.}=\sec x$$

Therefore, the integrating factor is sec x.