If y=yx   is the solution of the equation esinycosydydx+esinycosx=cosx,y0=0; then 1+yπ6+32yπ3+12yπ4 is equal to __

The given differential equation is esinycosydydx+esinycosx=cosx,y0=0Suppose that   esiny=t⇒esinycosydydx=dtdxSubstitute the above values in the given differential equationdtdx+tcosx=cosxThis is linear differential equationHence the integrating factor is e∫cosxdx=esinx                        Solution is tesinx=∫esinx.cosxdx=esinx+c Hence, the solution for the given differential equation is esiny=1+ce−sinx,y0=0⇒c=0                                                                                 ⇒    esiny =1   ⇒ y=0Therefore, the expression 1+yπ6+32yπ3+12y(π4)=1+0=1