高数求定积分

Question

如题所述

Answer 1

原式=∫(-π/2,π/2) dx/(1+cos^2x) + ∫(-π/2,π/2) xcosxdx/(1+cos^2x)
显然y=1/(1+cos^2x)是偶函数 y=xcosx/(1+cos^2x)是奇函数
且积分区间根据原点对称
所以原式=2∫(0,π/2) dx/(1+cos^2x) + 0
=2∫(0,π/2) dx/(sin^2x+2cos^2x)
=2∫(0,π/2) dx/[sin^2x(1+2cot^2x)]
=-∫(0,π/2) d(cotx)/(1/2+cot^2x)
=-√2*arctan(√2cotx) |(0,π/2)
=0-(-π/√2)
=π/√2