The correct option is
B π4∫a0dxx+√a2−x2But x=asinθ x=0 θ=0
1=acosθdθdx x=a θ=π2
dx=acosθdθ
I=∫π20acosθasinθ+√a2−a2sin2θ=∫π20acosθasinθ+√a2(1−sin2θ)dθ
I=∫π20acosθasinθ+acosθdθ
I=∫π20cosθsinθ+cosθdθ ⟶(1)
∫a0f(x)dx=∫a0f(0−x)dx
∴ I=∫π20cos(π2−θ)sin(π2−θ)+cos(π2−θ)dθ
I=∫π20sinθcosθ+sinθdθ ⟶(2)
(1)+(2)
2I=∫π20(cosθsinθ+cosθ+sinθcosθ+sinθ)dθ
2I=∫π201⋅dθ
2I=|θ∫π20
2I=π2
I=π4