If y=y(x) is the solution of the equation


If $y=y(x)$ is the solution of the equation $e^{\sin y} \cos y$ $\frac{d y}{d x}+e^{\sin y} \cos x=\cos x, y(0)=0$; then

$1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)$ is equal to


$e^{\sin y} \cos y \frac{d y}{d x}+e^{\sin y} \cos x=\cos x$

Put $e^{\sin y}=t$

$e^{\sin y} \times \cos y \frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow \frac{d t}{d x}+t \cos x=\cos x$

I. $F .=e^{\int \cos x d x}=e^{\sin x}$

Solution of diff equation:

$\mathrm{t} \cdot \mathrm{e}^{\sin \mathrm{x}}=\int \mathrm{e}^{\sin \mathrm{x}} \cdot \cos \mathrm{x} \mathrm{dx}$

$e^{\sin y} \cdot e^{\sin x}=e^{\sin x}+c$

at $x=0, y=0$]

$1=1+c \quad \Rightarrow c=0$

$e^{\sin x+\sin y}$ $e^{\sin x}$

$\sin x+\sin y=\sin x$


$\Rightarrow y\left(\frac{\pi}{6}\right)=0, \quad y\left(\frac{\pi}{3}\right)=0, \quad y\left(\frac{\pi}{4}\right)=0$

$\Rightarrow 1+0+0+0=1$

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