Question

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。

Answer 1

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