(1+x^2)*sin(2x)的不定积分

∫ (1+x²)sin2x dx
=∫ sin2x dx + ∫ x²sin2x dx
=-(1/2)cos2x - (1/2)∫ x² d(cos2x)
=-(1/2)cos2x - (1/2)x²cos2x + ∫ xcos2x dx
=-(1/2)cos2x - (1/2)x²cos2x + (1/2)∫ x d(sin2x)
=-(1/2)cos2x - (1/2)x²cos2x + (1/2)xsin2x - (1/2)∫ sin2x dx
=-(1/2)cos2x - (1/2)x²cos2x + (1/2)xsin2x + (1/4)cos2x + C
=-(1/4)cos2x - (1/2)x²cos2x + (1/2)xsin2x + C

希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮.