xcosx/(1+sinx^2)这项也是奇函数,所以是0
只剩下cosx/(1+sinx^2)了
积分(-π/2到π/2) [ cosx/(1+sinx^2) ]dx
=积分(-π/2到π/2) [ 1/(1+sinx^2) ]dsinx
=arctan(sinx) | (-π/2到π/2)
=2arctan1
=π/2
求定积分,【从-π/2到π/2】[（1+x）cosx]/(1+sinx^2) dx
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