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.
人气:166 ℃ 时间:2019-10-10 01:16:45
解答
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...
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