Question

___________________??

Answer 1

$I=2π∫π/4−π4dx(1+esinx)(2−cos2x) ..(i)$

by a+b-x property

$I=2π∫π/4−π4dx(1+e−sinx)(2−cos2x)=2π∫π/4−π4esinxdx(1+esinx)(2−cos2x)dx ...(ii)$

adding (1) and (2)

$2I=2π∫π/4−π4(1+esinx)(1+esinx)(2−cos2x)dx⇒I=1π∫π/4−π412(2cos2x−1)dx=1πsec2xπ/4−π43sec2x−2dx$

$put tanx=t.sec2xdx=dt=2π∫10dt3t2+1=23π1(13√)(tan−1(t1/3–√))10=23π(tan−1(3–√)−tan−1(0))=23–√π(π3)=233–√$

$Now 27I2=27×427=4$