求不定積分∫x.sinx^2.cosx^2dx

求不定積分∫x.sinx^2.cosx^2dx


∫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