Question

计算定积分，最好手写谢谢

Answer 1

解：令x=2sint x：0→2，t：0→π/2 ∫[0：2]x3·√(4-x2)dx =∫[0：π/2](2sint)3·√(4-4sin2t)d(2sint) =∫[0：π/2](2sint)3·√(4cos2t)d(2sint) =∫[0：π/2](2sint)3·2cost·2costdt =8∫[0：π/2]sin3t·cos2tdt =8∫[0：π/2]sint·(1-cos2t)·cos2tdt =8∫[0：π/2](cos2t-cos?t)·sintdt =8∫[0：π/2](cos?t-cos2t)d(cost) =8[(1/5)cos?t -?cos3t]|[0：π/2] =8[(1/5)cos?(π/2) -?cos3(π/2)]-8[(1/5)cos?0 -?cos30] =8(0-0)-8(1/5 -?) =16/15