分别求x(1+x^2)和(sinx*cosx)/(sinx+cosx)两个式子的不定积分.

1个回答

  • 第一题没搞错题吧?

    ∫ x(1+x²)dx

    =∫ (x+x³)dx

    =(1/2)x²+(1/4)x^4+C

    ∫ (sinx*cosx)/(sinx+cosx)dx

    =(1/2)∫ (2sinx*cosx)/(sinx+cosx)dx

    =(1/2)∫ (2sinx*cosx+1-1)/(sinx+cosx)dx

    =(1/2)∫ (2sinx*cosx+1)/(sinx+cosx)dx-(1/2)∫ 1/(sinx+cosx)dx

    =(1/2)∫ (2sinx*cosx+sin²x+cos²x)/(sinx+cosx)dx-(1/2)∫ 1/(sinx+cosx)dx

    =(1/2)∫ (sinx+cosx)²/(sinx+cosx)dx-(1/(2√2))∫ 1/((1/√2)sinx+(1/√2)cosx)dx

    =(1/2)∫ (sinx+cosx)dx-(1/(2√2))∫ 1/sin(x+π/4)dx

    =(1/2)(sinx-cosx)-(1/(2√2))ln|csc(x+π/4)-cot(x+π/4)|+C