求下列定积分
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J = 4∫_(-π/2)^(π/2) cos⁴x dx
=4 ∫_(-π/2)^(π/2) [(1 + cos2x)/2]² dx
= ∫_(-π/2)^(π/2) (1 + 2cos2x + cos²2x) dx
= ∫_(-π/2)^(π/2) (1 + 2cos2x) dx +(1/2)∫_(-π/2)^(π/2) (1 + cos4x) dx
= (x + sin2x)_(-π/2)^(π/2) + (1/2)(x + 1/4 * sin4x)_(-π/2)^(π/2)
= π + π/2
= 3π/2
=4 ∫_(-π/2)^(π/2) [(1 + cos2x)/2]² dx
= ∫_(-π/2)^(π/2) (1 + 2cos2x + cos²2x) dx
= ∫_(-π/2)^(π/2) (1 + 2cos2x) dx +(1/2)∫_(-π/2)^(π/2) (1 + cos4x) dx
= (x + sin2x)_(-π/2)^(π/2) + (1/2)(x + 1/4 * sin4x)_(-π/2)^(π/2)
= π + π/2
= 3π/2
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