∫cos(1-cos2x)dx
=∫2sin^xcosxdx
=∫2sin^xdsinx
=2/3 sin³x+C你倒过来求导看对吗看错题了，对不起。
∫cos(1-cos2x)dx

=∫cos(2sin^x)dx

=cos(2sin^x)x-∫x dcos(2sin^x)
=cos(2sin^x)x+∫x sin(2sin^x)*4sinxcosx dx
=cos(2sin^x)x+∫x sin(1-cos2x)*2sin2x dx
=cos(2sin^x)x-∫x sin(1-cos2x)*dcos2x
=cos(2sin^x)x+∫x dcos(1-cos2x)
=cos(2sin^x)x+x cos(1-cos2x)-∫cos(1-cos2)dx
2∫cos(1-cos2)dx=cos(2sin^x)x+x cos(1-cos2x)
∫cos(1-cos2x)dx=1/2[cos(2sin^x)x+x cos(1-cos2x)]+C