Question

这个式子怎么求，高等数学

Answer 1

没有那个-1/2

Answer 2

原式 = ∫<-π/2, π/2> sint(cost)^2dt + ∫<-π/2, π/2> (cost)^2dt
= 0 + 2 ∫<0, π/2> (cost)^2dt = ∫<0, π/2> (1+cos2t)dt
= [t +(1/2)sin2t]<0, π/2> = π/2

Answer 3

原式=∫(-π/2,π/2)sintcos^2tdt+∫(-π/2,π/2)cos^2tdt
因为sintcos^2t是奇函数，cos^2t是偶函数
所以原式=0+2∫(0,π/2)cos^2tdt
=∫(0,π/2)(1+cos2t)dt
=[t+(1/2)*sin2t]|(0,π/2)
=π/2