Question

高数不定积分，求过程

Answer 1

∫ [ (sint)^4- (sint)^6 ] dt
=∫ [(1/4)( 1- cos2t)^2 - (1/8)(1-cos2t)^3 ] dt
=(1/8)∫ [2( 1- cos2t)^2 - (1-cos2t)^3 ] dt
=(1/8)∫{ [2- 4cos2t + 2(cos2t)^2] -[ 1-3(cos2t) +3(cos2t)^2+(cos2t)^3 ] } dt
=(1/8)∫ [1- cos2t - (cos2t)^2 -(cos2t)^3 ] dt
=(1/16)∫ [1- 2cos2t - cos4t ]dt -(1/8)∫ (cos2t)^3 dt
=(1/16) [ t- sin2t - (1/4)sin4t ] -(1/16)∫ (cos2t)^2 dsin2t
=(1/16) [ t- sin2t - (1/4)sin4t ] -(1/16)∫ [1-(sin2t)^2 ] dsin2t
=(1/16) [ t- sin2t - (1/4)sin4t ] -(1/16) [sin2t - (1/3)(sin2t)^3] + C