Question

一道高数题积分题求助？

Answer 1

Answer 2

根据三角函数的周期性，本积分为零

Answer 3

Answer 4

∫(0->2π) -(1/8)(sint)^3 dt
=(1/8) ∫(0->2π) (sint)^2 dcost
=(1/8) ∫(0->2π) [1- (cost)^2] dcost
=(1/8)[ cost -(1/3)(cost)^3 ]|(0->2π)
=0
∫(0->2π) (1/4)(1-cost) dt
=(1/5)[ t -sint]|(0->2π)
=(2/5)π
∫(0->2π) (1/8)(1+cost)^2. cos(t/2) dt
=(1/8)∫(0->2π) (2[cos(x/2)]^2 )^2. cos(t/2) dt
=(1/2)∫(0->2π) [cos(x/2)]^5 dt
=-∫(0->2π) [cos(x/2)]^4 dsin(x/2)
=-∫(0->2π) {1 -[sin(x/2)]^2 }^2 dsin(x/2)
=-∫(0->2π) {1 -2[sin(x/2)]^2 +[sin(x/2)]^4 } dsin(x/2)
=-[ sin(x/2) - (2/3)[sin(x/2)]^3 +(1/5)[sin(x/2)]^5 ]|(0->2π)
=0
=>
∫(0->2π) [ -(1/8)(sint)^3 +(1/4)(1-cost) +(1/8)(1+cost)^2. cos(t/2) ] dt
=0+(2/5)π +0
=(2/5)π