I=2ππ4∫−π4dx(1+esinx)(2−cos2x)⋯(i)
Using the property:
a∫bf(x) dx=a∫bf(a+b−x) dx
I=2ππ4∫−π4dx(1+e−sinx)(2−cos2x)
I=2ππ4∫−π4esinx dx(1+esinx)(2−cos2x)⋯(ii)
Adding equation (i) and (ii), we get-
2I=2ππ4∫−π4dx(2−cos2x)
∵a∫−af(x)dx=2a∫0f(x)dx⇒Even function
2I=4ππ4∫0dx(1+2sin2x)
I=2ππ4∫0sec2x dx(sec2x+2tan2x)
I=2ππ4∫0sec2x dx(1+tan2x+2tan2x)
Put tanx=u⇒sec2x dx=du
I=2π1∫0du(1+3u2)=[2√3πtan−1(√3u)]10
I=2√3π×(π3−0)=23√3
∴27I2=27×(23√3)2=4