# If $y=y(x)$ is the solution of the equation  then $\mathrm{e}^{\sin y} \cos \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{e}^{\sin y} \cos \mathrm{x}=\cos \mathrm{x}, \mathrm{y}(0)=0;$$1+y\left ( \frac{\pi}{6} \right )+\frac{\sqrt{3}}{2}y\left ( \frac{\pi}{3} \right )+\frac{1}{\sqrt{2}}y\left ( \frac{\pi}{4} \right )$ is equal to ________. Option: 1 1 Option: 2 2 Option: 3 4 Option: 4 6

$\\\text { Put } \mathrm{e}^{\sin y}=\mathrm{t} \\ \Rightarrow \mathrm{e}^{\sin y} \cos \mathrm{y} \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{dt}}{\mathrm{d} \mathrm{x}} \\$

Now, we have given equation is

$\mathrm{e}^{\sin y} \cos \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{e}^{\sin y} \cos \mathrm{x}=\cos \mathrm{x}$

$\Rightarrow \mathrm{D} . \mathrm{E} \text { is } \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \cos \mathrm{x}=\cos \mathrm{x}$

$\\\text { I.F. }=\mathrm{e}^{\int \cos \mathrm{x} \mathrm{dx}}=\mathrm{e}^{\sin \mathrm{x}} \\ \Rightarrow \text { solution is } \mathrm{t\cdot e}^{\sin \mathrm{x}}=\int \cos \mathrm{xe}^{\sin \mathrm{x}}dx \\ \Rightarrow \mathrm{e}^{\sin \mathrm{y}} \mathrm{e}^{\sin \mathrm{x}}=\mathrm{e}^{\sin \mathrm{x}}+\mathrm{c} \\ \because \mathrm{x}=0, \mathrm{y}=0 \Rightarrow \mathrm{c}=0 \\ \Rightarrow \mathrm{e}^{\sin y}=1 \\ \Rightarrow \mathrm{y}=0$

$\Rightarrow 1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)=1$

Note that:

$\\I=\int \cos x e^{\sin x}dx\\\text{put }\sin x=u\Rightarrow \cos xdx=du\\I=\int e^udu=e^u+C\\I=e^{\sin x} +C$

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