(1-cost)5/2dt的积分怎么求
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∫√(1-cost)^5 dt
=∫√[2sin(t/2)^2]^5 dt
=16√2∫|sin(t/2)|^5 dt
sin(t/2)>0
= -32√2∫[(sint/2)^2]^2dcos(t/2)
= -32√2∫1-2(cost/2)^2+(cost/2)^4 dcos(t/2)
=-32√2cos(t/2) +(64√2/3)(cost/2)^3-(32√2/5)(cost/2)^5+C
sint/2<0
=32√2cos(t/2)-(64√2/3)(cost/2)^3+(32√2/5)(cost/2)^5+C
=∫√[2sin(t/2)^2]^5 dt
=16√2∫|sin(t/2)|^5 dt
sin(t/2)>0
= -32√2∫[(sint/2)^2]^2dcos(t/2)
= -32√2∫1-2(cost/2)^2+(cost/2)^4 dcos(t/2)
=-32√2cos(t/2) +(64√2/3)(cost/2)^3-(32√2/5)(cost/2)^5+C
sint/2<0
=32√2cos(t/2)-(64√2/3)(cost/2)^3+(32√2/5)(cost/2)^5+C
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