Question

第三题计算定积分怎么做

Answer 1

如图所示

追答

Answer 2

3.⑴∫-1→1 (x+√(1-x²))²dx
令x=sinα α∈[-π/2，π/2]
则原式=∫-π/2→π/2 (sinα+cosα)²dsinα
=∫-π/2→π/2 (cosα+2sinαcos²α)dα
=∫-π/2→π/2 cosαdα-2∫-π/2→π/2 cos²αdcosα
=(sinα-2/3 cos³α)|-π/2→π/2
=1-(-1)=2