∫√(a^2-x^2)dx=?

Answer 1

令x = asinz，dx = acosz dz
∫ √(a^2 - x^2) dx
= ∫ (acosz)(acosz) dz
= a^2/2 • ∫ (1 + cos2z) dz
= a^2/2 • [z + (sin2z)/2] + C
= (a^2/2)arcsin(x/a) + (a^2/2)sinzcosz + C
= (a^2/2)arcsin(x/a) + (a^2/2)(x/a)√(a^2 - x^2)/a + C
= (a^2/2)arcsin(x/a) + (x/2)√(a^2 - x^2) + C
= (1/2)[a^2arcsin(x/a) + x√(a^2 - x^2)] + C

Answer 2

x/2√(a^2-x^2)+a/2arcsin(x/a)+C
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