Calculust-2

The differential equation becomes $\begin{array}{rcl} \frac{dy}{dx} & = & \frac{\tan y}{x+1} + (x+1)e^x \sec y \\ \frac{\cos y}{x+1} \frac{dy}{dx} - \frac{\sin y}{(x+1)^2} & = & e^x \\ \frac{d}{dx}\left(\frac{\sin y}{x+1}\right) & = & e^x \\ \frac{\sin y}{x+1} & = & e^x + c \end{array}$ Assuming that we are meant to put $c=0$ , we deduce that $\sin y \,=\, (x+1)e^x$ , and hence that $\sin a = 1$ , and hence $a = \tfrac12\pi = \boxed{1.57...}$ .

It would have been better not to talk about the constant of integration, since it is not necessarily unique. I could, in a fit of madness, have integrated the differential equation as $\frac{\sin y}{x+1} \,=\, e^x + c + 1$ . I would not then want to choose $c=0$ . It would be better to specify the constant on integration in a clearer way. For example, you could require that the curve passes through $(-1,0)$ .

This can be easily done if we substitute $siny=t$ . And then it converts to a simple differential equation which can be solved easily.