If the value of the integral ∫−2π2π(1+πxx2cosx+1+esinx21231+sin2x)dx=4π(π+a)−2, then the value of a is
Solutions · 1
Official solution
I=∫−π/2π/2(1+πxx2cosx+1+esinx20231+sin2x)dxI=∫−π/2π/2(1+π−xx2cosx+1+esin(−x)20231+sin2x)dx
On Adding, we get
2I=∫−π/2π/2(x2cosx+1+sin2x)dx
On solving
I=4π2+43π−2a=3