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

${e^{\sin y}}\cos y{{dy} \over {dx}} + {e^{\sin y}}\cos x = \cos x$, y(0) = 0; then

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

e^{sin y} cos y${{dy} \over {dx}}$ + e^{sin y} cos x = cos x<br><br>Put e^{sin y} = t<br><br>e^{sin y} $\times$ cos y${{dy} \over {dx}}$ = ${{dt} \over {dx}}$<br><br>$\Rightarrow$ ${{dt} \over {dx}}$ + t cos x = cos x<br><br>I. F. = ${e^{\int {\cos x\,dx} }} = {e^{\sin x}}$<br><br>Solution of differential equation :<br><br>$t.{e^{\sin x}} = \int {{e^{\sin x}}.\cos x\,dx}$<br><br>${e^{\sin y}}.{e^{\sin x}} = {e^{\sin x}} + c$<br><br>at x = 0, y = 0<br><br>1 = 1 + c $\Rightarrow$ c = 0<br><br>$\therefore$ e^{sin x + sin y} = e^{sin x}<br><br>$\Rightarrow$ sin x + sin y = sin x<br><br>$\Rightarrow$ sin y = 0 $\Rightarrow$ y = 0<br><br>$$ \Rightarrow y\left( {{\pi \over 6}} \right) = 0,y\left( {{\pi \over 3}} \right) = 0,y\left( {{\pi \over 4}} \right) = 0$$ <br><br>$\therefore$ $$1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$$ <br><br>= 1 + 0 + 0 + 0 = 1