求不定积分x平方sinx平方dx…谢谢

发型不亮2022-10-04 11:39:541条回答

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qqq小子共回答了27个问题 | 采纳率92.6%: ∫ 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; 1年前