Solution :

Given differential equation is

$$ \begin {aligned} \frac{dy}{dx} +\frac{x+ycosx}{1+sinx} =0 \end{aligned}$$

$$\begin {aligned} \frac{d y}{d x} =-\frac{x+y \cos x}{1+\sin x} \\ \end{aligned}$$ $\begin {aligned}\Rightarrow \frac{d y}{d x}+\frac{\cos x}{1+\sin x} y=-\frac{x}{1+\sin x} \\....(1) \end{aligned}$

The given linear differential equation is in the form of

$$\frac{dy}{dx} +Py=Q $$

Comparing the given linear differential equation, we get

$ \begin{aligned} P &=\frac{\cos x}{1+\sin x}, Q=-\frac{x}{1+\sin x} \\IF &=e^{\int} \frac{\cos x}{1+\sin x} d x \\ &=e^{\log (1+\sin x)} \\ &=1+\sin x \end{aligned} $

Multiplying equation (1) both the sides by I.F. $=1+$ $\sin x$, we get

$$ (1+\sin x) \frac{d y}{d x}+y \cos x=-x $$

Integrating with respect to $x$, we get $$ \begin{aligned} y(1+\sin x)=\int-x d x+C \ \Rightarrow y=\frac{2 C-x^{2}}{2(1+\sin x)} \end{aligned} $$