Question

高数 定积分问题，求大神详解

Answer 1

∫[0：1][1/(1+x²)]dx
=arctanx|[0：1]
=arctan1 -arctan0
=π/4 -0
=π/4

令x=sint
x：0→1，则t：0→π/2
∫[0：1]√(1-x²)dx
=∫[0：π/2]√(1-sin²t)d(sint)
=∫[0：π/2]cos²tdt
=¼∫[0：π/2](1+cos2t)d(2t)
=¼(2t+sin2t)|[0：π/2]
=¼[(π+sinπ)-(0+sin0)]
=π/4

原式=(π/4)/(1-π/4)=π/(4-π)