求 ∫√(a^2-x^2)dx 的不定积分,

3个回答

  • 令x=asiny,dx=acosy dy

    √(a²-x²)=√(a²-a²sin²y)=acosy

    原式=a²∫cos²y dy

    =a²/2*∫(1+cos2y) dy

    =a²/2*y+a²/2*1/2*∫cos2y d(2y)

    =a²/2*y+a²/4*sin2y

    =a²/2*y+a²/2*sinycosy

    =a²/2*arcsin(x/a)+a²/2*(x/a)*√(a²-x²)/a

    =(a²/2)arcsin(x/2)+(x/2)√(a²-x²)+C