求不定积分x平方sinx平方dx…谢谢

1个回答

  • ∫ x^2sin^2x dx

    = ∫ x^2*(1 - cos2x)/2 dx

    = (1/2)∫ x^2 dx - (1/2)∫ x^2cos2x dx

    = x^3/6 - (1/4)∫ x^2 d(sin2x)

    = x^3/6 - (1/4)x^2sin2x + (1/4)∫ 2xsin2x dx

    = x^3/6 - (1/4)x^2sin2x - (1/4)∫ x d(cos2x)

    = x^3/6 - (1/4)x^2sin2x - (1/4)xcos2x + (1/4)∫ cos2x dx

    = x^3/6 - (1/4)x^2sin2x - (1/4)xcos2x + (1/8)sin2x + C