企源知识库

计算∫₀¹x²√1-x²dx．
【正确答案】：π/16 分析 令x=sint，则dx=costdt,并且t从0→π时，x单调的从0→1，所以 ∫₀¹x²√1-x²dx=∫₀^π/2sin²tcos²tdt=(1/4)∫₀^π/2sin²2tdt =1/4∫₀^π/2[(1-cos4t)/2]dt =1/4[(1/2)t|₀^π/2-(1/8)cos4td4t] =1/4[π/4-(1/8)sin4t|₀^π/2]=π/16