百科知识

计算不定积分,急救!!!!!!!!!!!!!计算不定积分:∫x

2007-01-02 16:53:08草***
计算不定积分:∫【x^2/√(1-x^2)】dx 计算不定积分,急救!!!!!!!!!!!!!计算不定积分:∫【x^2/√(1-x^2)】dx:令x=sint,则∫x^2/√(1-x^2)=∫(sint)^2 ?

最佳回答

  • 令x=sint,则∫x^2/√(1-x^2)=∫(sint)^2 dt=1/2∫[1-cos2t]dt=1/2t-1/4sin2t+C=1/2arcsinx-1/2x√(1-x^2)+C
    2007-01-02 17:05:48
  • 很赞哦! (237)