Given differential equation is

$$\begin{aligned} \frac{d y}{d x}+y & =\sin x \\ \text { IF } & =\int e^{1 d x}=e^x \end{aligned}$$

The general solution is

$$y \cdot e^x=\int e^x \sin x d x+C\quad\text{.... (i)}$$

$$\begin{aligned} \text{Let}\quad I & =\int e^x \sin x d x \\ I & =\sin x \mathrm{e}^x-\int \cos x \mathrm{e}^x d x \\ & =\sin x \mathrm{e}^x-\cos x \mathrm{e}^x+\int(-\sin x) \mathrm{e}^x d x \\ 2 I & =\mathrm{e}^x(\sin x-\cos x) \\ I & =\frac{1}{2} e^x(\sin x-\cos x) \end{aligned}$$

From Eq. (i),

$$\begin{aligned} y \cdot e^x & =\frac{x}{2}(\sin x-\cos x)+C \\ \Rightarrow\quad y & =\frac{1}{2}(\sin x-\cos x)+C \cdot e^{-x} \end{aligned}$$

$$\begin{aligned} &\text { Given differential equation is }\\ &\begin{aligned} \cot y d x & =x d y \\ \Rightarrow\quad \frac{1}{x} d x & =\tan y d y \end{aligned} \end{aligned}$$

On integrating both sides, we get

$$\begin{array}{ll} \Rightarrow & \int \frac{1}{x} d x=\int \tan y d y \\ \Rightarrow & \log (x)=\log (\sec y)+\log C \end{array}$$

$\Rightarrow \quad \log \left(\frac{x}{\sec y}\right)=\log C$

$$\begin{aligned} \Rightarrow \quad& \frac{x}{\sec y} =C \\ \Rightarrow \quad & x =C \sec y \end{aligned}$$

Given differential equation is

$$\begin{aligned} &\text { Given differential equation is }\\ &\begin{aligned} \frac{d y}{d x}+y & =\frac{1+y}{x} \\ \frac{d y}{d x}+y & =\frac{1}{x}+\frac{y}{x} \\ \Rightarrow \quad \frac{d y}{d x}+y\left(1-\frac{1}{x}\right) & =\frac{1}{x} \end{aligned} \end{aligned}$$

$$\begin{aligned} \therefore\quad \mathrm{IF} & =e^{\int\left(1-\frac{1}{x}\right) d x} \\ & =e^{x-\log x} \\ & =e^x \cdot e^{-\log x}=\frac{e^x}{x} \end{aligned}$$