Question

积分求解！

如何求的？想要详细过程T_T

Answer 1

应该是把式子拆开来做吧，上下限对称，看每个分式子是奇函数还是偶函数，奇函数就可以消了

Answer 2

I = 8a^3∫<-π/2, π/2> (cosx)^2(1+2cosxsinx)[cosx)^2-(sinx)^2]dx
= 4a^3∫<-π/2, π/2> (1+cos2x)(1+sin2x)cos2xdx
= 4a^3[∫<-π/2, π/2> (1+cos2x)cos2xdx +∫<-π/2, π/2> (1+cos2x)sin2xcos2xdx]
(对称区间上积分 “偶倍奇零”）
I = 8a^3∫<0, π/2> (1+cos2x)cos2xdx
= 8a^3∫<0, π/2> cos2xdx + 4a^3∫<0, π/2> (1+cos4x)dx
= 4a^3[sin2x]<0, π/2> + 4a^3[x+(1/4)sin4x]<0, π/2> = 2πa^3