Question

求不定积分∫dx/(1+(1-(x^2))^1/2),求详细过程~

Answer 1

x = sinθ then dx = cosθdθ，cosθ = √(1-x²)
∫ dx/[1+√(1-x²)] = ∫ cosθ/(1+cosθ) dθ
= ∫ (1+cosθ-1)/(1+cosθ) dθ
= ∫ dθ - ∫ (1-cosθ)/sin²θ dθ
= θ - (-cotθ + cscθ) + C
= arcsin(x) + √(1-x²)/x - 1/x + C

Answer 2

令x=sint
dx=costdt

∫dx/(1+(1-(x^2))^1/2)=∫1/(1+cost)*costdt
=∫[1-1/(1+cost)]dt
=t-∫[1/(2*cos^2(t/2))]dt
=t-∫sec^2(t/2)d(t/2)
=t-tan(t/2)
=arcsinx-tan((arcsinx)/2)