求不定积分∫x3(1-x2)1/2dx．

求不定积分∫x³(1-x²)^1/2dx．
【正确答案】：∫x³(1-x²)^1/2dx=∫x³√(1-x³)dx，令x=sint， 原式=∫sin³tcos²tdt =∫(cos²t-1)cos²td(cost) =(1/5)cos⁵t-(1/3)cos³t+C =(1/5)(1-x)^5/2-(1/3)(1-x²)^3/2+C．