If a function y=f(x) satisfies the differential equation f(x)⋅sin⁡2x−cos⁡x+1+sin2⁡xf′(x)=0 with initial condition y(0)=0, then the value  of fπ6 is equal to

The given equation is f(x)⋅sin⁡2x−cos⁡x+1+sin2⁡xf′(x)=0⇒ysin⁡2x−cos⁡x+1+sin2⁡xdydx=0⇒dydx+sin⁡2x1+sin2⁡xy=cos⁡x1+sin2⁡x is in the form of dydx+Py=Q Where P=sin⁡2x1+sin2⁡x and Q=cos⁡x1+sin2⁡x Integrating Factor (IF)=e∫sin⁡2x1+sin2⁡xdx=eln⁡1+sin2⁡x ∵∫f′(x)f(x)dx=ln⁡|f(x)|+C=1+sin2⁡x General solution is y⋅(IF)=∫(IF)Qdx⇒y1+sin2⁡x=∫1+sin2⁡xcos⁡x1+sin2⁡xdx⇒y1+sin2⁡x=∫cos⁡xdx=sin⁡x+C When x=0,y=0⇒C=0 When x=π6, then y1+14=12⇒y54=12⇒y=25∴fπ6=25Therefore, the correct answer is (4).