Question

请问这个积分怎么做啊？谢谢！

Answer 1

详细过程如图rt……希望能帮到你解决问题

Answer 2

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 3

Answer 4

Answer 5

设 x = a*sinα。那么，dx = a * cosα dα, α = arcsin(x/a)
则原积分就可以变换为：
∫√(a² - a² * sin²α) * a * cosα *dα
=∫√(a² * cos²α) * a * cosα * dα
=∫a²*cos²α * dα
=a²/2 * ∫(2cos²α) * dα
=a²/2 * ∫[1 + cos(2α)] * dα
=α²/2 * [∫dα + ∫cos(2α) * dα]
=a²/2 * [α + 1/2 * ∫cos(2α) * d(2α)]
=a²/2 * α + a²/4 * sin(2α) + C
=a²/2 * α + a²/4 * 2sinα * cosα + C
=a²/2 * arcsin(x/a) + [(asinα) * (acosα)]/2 + C
=a²/2 * arcsin(x/a) + x * √(a²-x²) /2 + C