Question

高数题目求解

Answer 1

Answer 2

∫(0->π/3) ( 1+cosx)^2 dx

=∫(0->π/3) [ 1+2cosx + (cosx)^2 ] dx

=(1/2) ∫(0->π/3)  ( 3+4cosx + cos2x ) dx

=(1/2)[ 3x+4sinx +(1/2)sin2x]|(0->π/3)

=(1/2) [ π + 2√3 + (1/4)√3 ]

=(1/2) [ π + (9/4)√3 ]

=(1/2)π + (9/8)√3

∫(π/3->π/2) (3cosx)^2 dx

=(9/2)∫(π/3->π/2) (1+cos2x) dx

=(9/2)[ x +(1/2)sin2x] |(π/3->π/2)

=(9/2) [ ( π/2+0) -(π/3 +(1/4)√3) ]

=(9/2) [  π/6 -(1/4)√3 ]

=(3/4)π - (9/8)√3

2[∫(0->π/3) ( 1+cosx)^2 dx + ∫(π/3->π/2) (3cosx)^2 dx ]

=2[ (1/2)π + (9/8)√3 +(3/4)π - (9/8)√3 ]

= (5/2)π