Question

高等数学定积分求0到1的定积分√[(1-x^2)^3]

Answer 1

设x=sint,dx=costdt,原式=∫[0,π/2](cost)^3*costdt=(1/4)∫[0,π/2](1+cos2t)^2dt

={t/4+(sin2t)/4+t/8+(sin4t)/32})[0,π/2]

=(3t/8)[0,π/2]+(0-0)/4+(0-0)/32

=3π/16.