大一高数 定积分的题

1个回答

  • 原式=(-1/3)*∫ arcsinx d[(1-x^2)^(3/2)]

    =(-1/3)*[arcsinx*(1-x^2)^(3/2)-∫ (1-x^2)^(3/2) d(arcsinx)]

    =(-1/3)*[arcsinx*(1-x^2)^(3/2)-∫ (1-x^2)^(3/2)*(1-x^2)^(-1/2) dx]

    =(-1/3)*[arcsinx*(1-x^2)^(3/2)-∫ (1-x^2) dx]

    =(-1/3)*[arcsinx*(1-x^2)^(3/2)-x+x^3/3]+C

    =x/3-(1/3)*arcsinx*(1-x^2)^(3/2)-(1/9)*x^3+C