If I=2π∫−π/4π/4dx1+esinx2−cos2x  then 27   I2 is equal to

I=2π∫−π/4π/4dx1+esinx2−cos2x               since ∫abfxdx =∫abfa+b-xdxI=2π∫-x4π4dx1+e-sinx(2-cos2x)    ⇒2I=  2π∫-π4π411+esinx2-cos2x+esinxesinx+12-cosxdx=2π∫-π4π4  dx(2-cos2x) ⇒2I=2π×2∫0π4    dx2-2cos2x-1       since 12-cos2x is even⇒I=2π∫0π4sec2xdx3sec2x-2 dx              =2π∫01dt3t2+1-2=2π∫0113t2+1dt           put tanx = t then sec2x=dt    =2π×13tan-13t101=23ππ3-0=233 ⇒ I2=427⇒27I2=4