$\begin{aligned} & \sin ^{-1}\left(\frac{d y}{d x}\right)=x+y \\ & \therefore \frac{d y}{d x}=\sin (x+y) \\ & \text { Put } x+y=t \quad \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \\ & \therefore \frac{d t}{d x}-1=\sin t \Rightarrow \frac{d t}{d x}=1+\sin t \\ & \therefore \int \frac{d t}{1+\sin t}=\int d x \\ & \therefore \int \frac{(1-\sin t) d t}{1-\sin } t=\int d x \Rightarrow \int \frac{1-\sin t}{\cos ^2 t} d t=\int d x \\ & \Rightarrow \int\left(\sec ^2 t-\operatorname{sect} \tan t\right) d t=\int d x \\ & \therefore \tan t-\sec t=x+c \\ & \therefore \tan (x+y)-\sec (x+y)=x+c\end{aligned}$