Step 1: 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 $Q(x) = \sin(x) \cos(x)$. Step 2: Calculate the integrating factor $I(x)$. The integrating factor is given by $I(x) = e^{\int P(x) dx}$. $$I(x) = e^{\int -\cos(x) dx}$$ $$I(x) = e^{-\sin(x)}$$ Step 3: Multiply the standard form equation by the integrating factor. $$e^{-\sin(x)} \left( \frac{dy}{dx} - \cos(x) y \right) = e^{-\sin(x)} (\sin(x) \cos(x))$$ The left side of the equation is the derivative of the product $y \cdot I(x)$: $$\frac{d}{dx} (y e^{-\sin(x)}) = \sin(x) \cos(x) e^{-\sin(x)}$$ Step 4: Integrate both sides with respect to $x$. $$\int \frac{d}{dx} (y e^{-\sin(x)}) dx = \int \sin(x) \cos(x) e^{-\sin(x)} dx$$ $$y e^{-\sin(x)} = \int \sin(x) \cos(x) e^{-\sin(x)} dx$$ 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 $u = \sin(x)$: $$= -e^{-\sin(x)}(\sin(x)+1) + C$$ Step 5: Substitute the result of the integral back into the equation and solve for $y$. $$y e^{-\sin(x)} = -e^{-\sin(x)}(\sin(x)+1) + C$$ Divide both sides by $e^{-\sin(x)}$: $$y = -(\sin(x)+1) + C e^{\sin(x)}$$ $$y = C e^{\sin(x)} - \sin(x) - 1$$ The final answer is $\boxed{y = C e^{\sin(x)} - \sin(x) - 1}$.