IF I=2π∫−π/4π/4dx1+esinx2−cos2x then the value of 27I2 is

I=2π∫−π/4π/4dx1+esinx2−cos2xI=2π∫−π/4π/4esinx1+esinx2−cos2x dx     since ∫abfxdx=∫abfa+b-xdxAdding the above both integrals2I=2π∫−π/4π/4dx2−cos2xI=1π∫−π/4π/4sec2xdx1+3tan2x=23327I2=4