Question

如图,高数积分问题...求解

Answer 1

解法一：∵ ∫<-π/2,π/2>[2sint*(cost)^2]dt
=-2∫<-π/2,π/2>(cost)^2d(cost)
=-2[(cost)^3/3]│<-π/2,π/2>
=-(2/3)[(cos(π/2))^3-(cos(-π/2))^3]
=-(2/3)(0-0)
=0
∴ 4∫<-π/2,π/2>[3(cost)^2+2sint*(cost)^2]dt
=4∫<-π/2,π/2>[3(cost)^2]dt+∫<-π/2,π/2>[2sint*(cost)^2]dt
=4∫<-π/2,π/2>[3(cost)^2]dt
=12∫<-π/2,π/2>(cost)^2dt。
解法二：∵2sint*(cost)^2是奇函数
∴∫<-π/2,π/2>[2sint*(cost)^2]dt=0 (应用对称区间定积分公式)
故 4∫<-π/2,π/2>[3(cost)^2+2sint*(cost)^2]dt
=4∫<-π/2,π/2>[3(cost)^2]dt+∫<-π/2,π/2>[2sint*(cost)^2]dt
=4∫<-π/2,π/2>[3(cost)^2]dt
=12∫<-π/2,π/2>(cost)^2dt。

Answer 2

后面的sint(cost)^2是奇函数，积分为0