求定积分,【从-π/2到π/2】[（1+x）cosx]/(1+sinx^2) dx

题目

求定积分,【从-π/2到π/2】[（1+x）cosx]/(1+sinx^2) dx
答案是π/2,原题中还有一部分是ln[x+(1+x^2)^1/2]因为是奇函数等于0,就不用再算了.

答案

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...