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∫(π/2,-π/2)(x³+1)cos²xdx=
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本题涉及到奇函数在对称的积分区域的定积分值为0,这一规律的应用:
∫(π/2,-π/2)(x³+1)cos²xdx
=∫(π/2,-π/2)x³cos²xdx+∫(π/2,-π/2)cos²xdx
=0+∫(π/2,-π/2)cos²xdx
=2∫(π/2,0)cos²xdx
=2∫(π/2,0)(1+cos2x)/2dx
=∫(π/2,0)(1+cos2x)dx
=∫(π/2,0)dx+∫(π/2,0)cos2xdx
=π/2+(1/2)∫(π/2,0)cos2xd2x
=π/2+(1/2)sin2x(π/2,0)
=π/2.
∫(π/2,-π/2)(x³+1)cos²xdx
=∫(π/2,-π/2)x³cos²xdx+∫(π/2,-π/2)cos²xdx
=0+∫(π/2,-π/2)cos²xdx
=2∫(π/2,0)cos²xdx
=2∫(π/2,0)(1+cos2x)/2dx
=∫(π/2,0)(1+cos2x)dx
=∫(π/2,0)dx+∫(π/2,0)cos2xdx
=π/2+(1/2)∫(π/2,0)cos2xd2x
=π/2+(1/2)sin2x(π/2,0)
=π/2.
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