Question

计算不定积分∫根号下（4-x²）dx 注：4-x平方；dx在根号外 各位大神速度啊、

Answer 1

解：令x=2cost，
∴dx=-2sintdt，sint=√[1-(x/2)^2]=(1/2)√4-x^2，t=arccos(x/2)
∫√（4-x^2)dx=∫2sint*(-2sint)dt

=2∫(cos2t-1)dt
=sin2t-2t+C
=(1/2)x√4-x^2-2arccos(x/2)+C

Answer 2

令x=2sint
则dx=2costdt
原式=∫2cost*2costdt
=2∫(1+cos2t)dt
=2[t+0.5sin2t]+C
=2t+sin2t+C
=2arcsin(x/2)+2*x/2*√(1-x^2/4)+C
=2arcsin(x/2)+x√(1-x^2/4)+C

Answer 3

令x = 2sinφ，dx = 2cosφ dφ
∫ √(4 - x²) dx
= ∫ √(4 - 4sin²φ) • (2cosφ dφ)
= ∫ √(4cos²φ) • (2cosφ dφ)
= 4∫ cos²φ dφ
= 2∫ (1 + cos2φ) dφ
= 2[φ + (1/2)sin2φ] + C
= 2φ + 2sinφcosφ + C
= 2arcsin(x/2) + 2(x/2)√(4 - x²)/2 + C
= 2arcsin(x/2) + (x/2)√(4 - x²) + C