Question

∫(0,1)[4√1-x^2 - 2x√1-x^2]dx

Answer 1

∫(0~1) [4√(1 - x²) - 2x√(1 - x²)] dx

= 4∫(0~1) √(1 - x²) dx - 2∫(0~1) x√(1 - x²) dx

对於前一项积分,令x = sinz,dx = cosz dz

= 4∫(0~π/2) cos²z dz - 2∫(0~1) √(1 - x²) d(x²/2)

= 2∫(0~π/2) (1 + cos2z) dz + ∫(0~1) √(1 - x²) d(1 - x²)

= 2[z + 1/2 * sin2z] |(0~π/2) + (2/3)(1 - x²)^(3/2) |(0~1)

= 2(π/2) + (2/3)[0 - 1]

= π - 2/3