已知连续函数,请问怎么求其不定积分

Question

如题所述

Answer 1

∫ (sint)^6 dx

=∫ [(sint)^2]^3 dx

=(1/8)∫ [ 1- cos2t ]^3 dx

= (1/8)∫ [ 1- 3cos2t + 3(cos2t)^2 - (cos2t)^3 ]dx

=(1/8)[ t - (3/2)sin2t] +(3/8)∫ (cos2t)^2 dx-(1/8)∫ (cos2t)^3 dx

=(1/8)[ t - (3/2)sin2t ]+(3/16)∫ (1+cos4t) dx-(1/16)∫ (cos2t)^2 dsin2t

=(1/8)[ t - (3/2)sin2t ] +(3/16)[x+(1/4)sin4t] -(1/16)∫[1- (sin2t)^2] dsin2t

=(1/8)[ t - (3/2)sin2t ] +(3/16)[x+(1/4)sin4t] -(1/16)[sin2t- (1/3)(sin2t)^3] +C

连续函数，一定存在定积分和不定积分；若在有限区间[a,b]上只有有限个间断点且函数有界，则定积分存在；若有跳跃、可去、无穷间断点，则原函数一定不存在，即不定积分一定不存在。