Question

一道求不定积分的题

如图求解

Answer 1

方法如下

Answer 2

以上请采纳。

Answer 3

∫(2sint-sin2t)(2sin2t-2sint) dt
=2∫(2sint-sin2t)(sin2t-sint) dt
=2∫ [2sint.sin2t - 2(sint)^2 - (sin2t)^2 + sin2t.sint ]dt
=2∫ [3sint.sin2t - 2(sint)^2 - (sin2t)^2 ]dt
=2∫ [ (3/2)(cost - cos3t) - (1- cos2t) - (1/4)(1-cos4t) ]dt
=(1/2)∫ [ 6(cost - cos3t) - 4(1- cos2t) - (1-cos4t) ] dt
=(1/2)∫ [ cos4t -6cos3t +4cos2t +6cost -5 ] dt
=(1/2 [ (1/4)sin4t -2sin3t +2sin2t +6sint -5t ] + C

Answer 4

I = ∫(2sint-sin2t)(2sin2t-2sint)dt = ∫[6sintsin2t-2(sin2t)^2-4(sint)^2]dt
= ∫[3(cost-cos2t)-(1-cos4t)-2(1-cos2t)]dt
= ∫(3cost - cos2t + cos4t - 3)dt
= 3sint - (1/2)sin2t + (1/4)sin4t - 3t + C