求不定积分∫x.sinx^2.cosx^2dx

先利用倍角公式，然后利用分部积分法及第一换元积分法：
∫x(sinx)^2 (cosx)^2dx
=1/4 ∫x(sin2x)^2dx
=1/8 ∫x(1-cos4x)dx
=1/8 (∫xdx-∫xcos4xdx)
=1/16 x^2 - 1/64 xsin4x + 1/64 ∫sin4xdx
= 1/16 x^2 - 1/64 xsin4x - 1/256 cos4x + C

∫x.sinx^2.cosx^2dx
=∫x.sinx^2.cosx^2dx
=∫x/4*(sin2x)^2dx
=1/8∫x(1-(cos2x)^2)dx
=1/8[∫xdx-∫x(cos2x)^2dx]
=1/8[∫xdx-∫x(1/2+1/2cos4x)dx]
=1/8∫xdx-1/16∫xdx-1/16∫xco4xdx
=1/16∫xdx-1/64∫xd(sin4x)
=1/16∫xdx-1/64(xsin4x-∫sin4xdx)
=1/16∫xdx-1/64xsin4x-1/256∫d(cos4x)
=1/32x^2-1/64xsin4x-1/256cos4x+c

∫x.sinx^2.cosx^2dx
=(1/2)∫xsin2x^2dx
令u=2x^2
du=4x
原式=(1/8)∫sinudu
=-(1/8)cosu+C
=-(1/8)cos2x^2+C

先把x移到微元中
变成dx^2
结果-1/8cos2x^2+c