求过程与结果

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求过程与结果

如图,用半角公式

cos^4t
=(cos^2t)^2
=[(1+cos2t)/2]^2
=1/4(1+cos2t)^2
=1/4[1+2cos2t+(cos2t)^2]
=1/4[1+2cos2t+(1+cos4t)/2]
然后积分就可以了

求定积分【0,arcsin(√2/2)】(16/3)∫cos⁴tdt
原式=【0,π/4】(16/3)∫cos⁴tdt=【0,π/4】(4/3)∫(1+cos2t)²dt
=【0,π/4】(4/3)∫(1+2cos2t+cos²2t)dt=【0,π/4】(4/3)∫[1+2cos2t+(1/2)(1+cos4t)]dt
=【0,π...

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求定积分【0,arcsin(√2/2)】(16/3)∫cos⁴tdt
原式=【0,π/4】(16/3)∫cos⁴tdt=【0,π/4】(4/3)∫(1+cos2t)²dt
=【0,π/4】(4/3)∫(1+2cos2t+cos²2t)dt=【0,π/4】(4/3)∫[1+2cos2t+(1/2)(1+cos4t)]dt
=【0,π/4】(4/3)[∫dt+∫cos2td(2t)+(1/2)∫dt+(1/8)∫cos4td(4t)]
=【0,π/4】(4/3)[t+sin2t+(1/2)t+(1/8)sin4t]
=【0,π/4】(4/3)[(3/2)t+sin2t+(1/8)sin4t]
=(4/3)(3π/8+1)=(π/2)+(4/3).

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