Question

一道大一高数定积分题

Answer 1

原式=∫[-π/2,π/2] costsin^4tdt+∫[-π/2,π/2] x^2sin^7tdt
(注意后一项，被积函数是奇函数，积分区间关于原点对称，所以值为0)
=∫[-π/2,π/2] costsin^4tdt
=∫[-π/2,π/2] sin^4tdsint
=1/5sin^5t[-π/2,π/2]
=2/5

Answer 2

原式=∫(-π/2,π/2) costsin^4t dt + ∫(-π/2,π/2) x^2*sin^7t dt
因为x^2*sin^7t是奇函数，且积分区间关于原点对称，所以 ∫(-π/2,π/2) x^2*sin^7t dt=0
原式=∫(-π/2,π/2) sin^4t d(sint) + 0
=1/5*sin^5t|(-π/2,π/2)
=1/5+1/5
=2/5

Answer 3

Answer 4

前面半部分为sint^4 dsint,后部分为奇函数，结果为2/5