Question

求(cos2x)(sinx)^2的不定积分

Answer 1

本题的关键是先进行三角恒等变换。
(cos2x)(sinx)^2
=(cos2x)[1-cos(2x)]/2
=(1/2)cos(2x) -(1/2)[cos(2x)]^2
=(1/2)cos(2x) -(1/2)[1+cos(4x)]/2
=(1/2)cos(2x) -(1/4)cos(4x) -1/4
∫(cos2x)(sinx)^2 dx
=∫[(1/2)cos(2x) -(1/4)cos(4x) -1/4]dx
=(1/4)sin(2x) - (1/16)sin(4x) -x/4 +C

Answer 2

cos2x=1-2*(sinX)^2

(cos2x)(sinx)^2
=[1-2*(sinX)^2](sinx)^2

=-2(sinx)^4+(sinx)^2
令sinx=u
则 x=arcsinu
则dx=darcsinu=1/√（1-u^2）du

∫【-2(sinx)^4+(sinx)^2】dx
=∫【-2u^4+u^2)】【1/√（1-u^2）】du