Yefim S. answered 04/17/21
Math Tutor with Experience
Limits of integration: 2 + sinθ = 2 + cos2θ; sinθ = 1 - 2sin2θ; 2sin2θ + sinθ - 1 = 0; (2sinθ - 1)(sinθ + 1) = 0;
sinθ = 1/2; θ = π/6 or θ = 5π/6; sinθ = - 1; θ = 3π/2.
Area A = 1/2∫π/65π/6[(2 + sinθ)2 - (2 + cos2θ)2]dθ = 1/2∫π/65π/6(4sinθ + sin2θ - 4cos2θ - cos22θ)dθ =
1/2∫π/65π/6(4sinθ - 1/2cos2θ - 4cos2θ - 1/2cos4θ)dθ= 1/2(- 4cosθ - 1/4sin2θ - 2sin2θ - 1/8sin4θ)π/65π/6 =
1/2[(2√3 + 9√3/8 + √3/16) - (- 2√3 - 9√3/8 - √3/16) = 51√3/16 = 5.52