求指导,求∫[π/2-π／2]sin^4xdx的定积分∫[π/2-π／2]应为∫[-π/2π／2]

Question

求指导,求∫[π/2 -π／2]sin^4xdx的定积分
∫[π/2 -π／2]应为∫[-π/2 π／2]

Answer 1

关键是三角恒等变换.
sin⁴x={[1-cos(2x)]/2}²
=1-cos(2x)+cos²(2x) /4
=1-cos(2x)+[1+cos(4x)]/8
=9/8-cos(2x) +cos(4x)/8

∫[π/2 -π／2]sin⁴xdx
=∫[π/2 -π／2][9/8-cos(2x) +cos(4x)/8]dx
=9x/8 -sin(2x)/2 +sin(4x)/32|(π/2,-π/2) 积分上下限很不清楚,按你写的,则-π/2是积分上限,就不再写了.你直接按上下限代进去就可以了.