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∫(t-sint)(1-cost)^2dt
=∫t[1-2cost+cos^2(t) ]dt-∫sint(1-cost)^2 dt
=∫T[1-2cost+(1+cos2t)/2]dt-∫(1-cost)^2 d(1-cost)
=3t/2-2∫tdsint+1/4∫tdsin2t-(1-cost)^3/3
=3t/2-(1-cosT)^3/3-2tsint+2∫sintdt+tsin2t/4-1/4∫sin2tdt
=3t/2-(1-cost)^3/3-2tsint-2cost+tsin2t/4+1/8cos2t+C
∫(1-cost)^3dt
=∫[1-3cost+3(cost)^2 -(cost)^3]dt
=t - 3sint +3∫(cost)^2dt - ∫(cost)^3dt
=t - 3sint +(3/2)∫(1+cos(2t) )dt - ∫[1-(sint)^2]d(sint)
=t - 3sint +(3/2)[t +sin(2t)/2] - [sint - (sint)^3/3] + C
带入积分区间
=3Π+2Π+3Π
=8Π
=∫t[1-2cost+cos^2(t) ]dt-∫sint(1-cost)^2 dt
=∫T[1-2cost+(1+cos2t)/2]dt-∫(1-cost)^2 d(1-cost)
=3t/2-2∫tdsint+1/4∫tdsin2t-(1-cost)^3/3
=3t/2-(1-cosT)^3/3-2tsint+2∫sintdt+tsin2t/4-1/4∫sin2tdt
=3t/2-(1-cost)^3/3-2tsint-2cost+tsin2t/4+1/8cos2t+C
∫(1-cost)^3dt
=∫[1-3cost+3(cost)^2 -(cost)^3]dt
=t - 3sint +3∫(cost)^2dt - ∫(cost)^3dt
=t - 3sint +(3/2)∫(1+cos(2t) )dt - ∫[1-(sint)^2]d(sint)
=t - 3sint +(3/2)[t +sin(2t)/2] - [sint - (sint)^3/3] + C
带入积分区间
=3Π+2Π+3Π
=8Π
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请问这个(1-cost)^3怎么变到下一步的?
用了什么公式
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努力学习就有好的成绩
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