Rewrite the differential equation in standard linear form.

The given equation is:

$$\sec(x) \frac{dy}{dx} = y + \sin(x)$$

Divide by $\sec(x)$ (which is equivalent to multiplying by $\cos(x)$):

$$\frac{dy}{dx} = \frac{y}{\sec(x)} + \frac{\sin(x)}{\sec(x)}$$ $$\frac{dy}{dx} = y \cos(x) + \sin(x) \cos(x)$$

Rearrange into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$:

$$\frac{dy}{dx} - \cos(x) y = \sin(x) \cos(x)$$

Here, $P(x) = -\cos(x)$ and $Q(x) = \sin(x) \cos(x)$.

The integrating factor is given by $I(x) = e^{\int P(x) dx}$.

$$I(x) = e^{\int -\cos(x) dx}$$ $$I(x) = e^{-\sin(x)}$$

$$e^{-\sin(x)} \left( \frac{dy}{dx} - \cos(x) y \right) = e^{-\sin(x)} (\sin(x) \cos(x))$$

The left side of the equation is the derivative of the product $y \cdot I(x)$:

$$\frac{d}{dx} (y e^{-\sin(x)}) = \sin(x) \cos(x) e^{-\sin(x)}$$

$$\int \frac{d}{dx} (y e^{-\sin(x)}) dx = \int \sin(x) \cos(x) e^{-\sin(x)} dx$$ $$y e^{-\sin(x)} = \int \sin(x) \cos(x) e^{-\sin(x)} dx$$

To evaluate the integral on the right side, use a substitution. Let $u = \sin(x)$, so $du = \cos(x) dx$.
The integral becomes:

$$\int u e^{-u} du$$

This integral requires integration by parts, $\int v dw = vw - \int w dv$.
Let $v = u$ and $dw = e^{-u} du$.
Then $dv = du$ and $w = -e^{-u}$.

$$\int u e^{-u} du = u(-e^{-u}) - \int (-e^{-u}) du$$ $$= -u e^{-u} + \int e^{-u} du$$ $$= -u e^{-u} - e^{-u} + C$$

Factor out $-e^{-u}$:

$$= -e^{-u}(u+1) + C$$

Substitute back $u = \sin(x)$:

$$= -e^{-\sin(x)}(\sin(x)+1) + C$$

$$y e^{-\sin(x)} = -e^{-\sin(x)}(\sin(x)+1) + C$$

Divide both sides by $e^{-\sin(x)}$:

$$y = -(\sin(x)+1) + C e^{\sin(x)}$$ $$y = C e^{\sin(x)} - \sin(x) - 1$$