根号下a^2-x^2不定积分中的步骤详解

I = ∫√(a^2-x^2)dx
= x√(a^2-x^2) - ∫[x(-x)/√(a^2-x^2)]dx
= x√(a^2-x^2) - ∫[(a^2-x^2-a^2)/√(a^2-x^2)]dx
= x√(a^2-x^2) - I + ∫[a^2/√(a^2-x^2)]dx
2I = x√(a^2-x^2) + a^2∫d(x/a)/√[1-(x/a)^2]
I = (x/2)√(a^2-x^2) + (a^2/2)arcsin(x/a) + C

∫sqrt(a^2+x^2)dx=xsqrt(a^2+x^2)-∫x^2dx/sqrt(a^2+x^2)
=xsqrt(a^2+x^2)-∫sqrt(a^2+x^2)dx+a^2∫dx/sqrt(a^2+x^2)
∫sqrt(a^2+x^2)dx=(1/2)[xsqrt(a^2+x^2)+a^2∫dx/sqrt(a^2+x^2)]
=(1/2)[xsqrt(a^2+x^2)+a^2ln(x+sqrt(a^2+x^2))]

详情如图所示，

有任何疑惑，欢迎追问

供参考，请笑纳。

cos²t=(1 + cos2t)/2
∫a²cos²tdt=∫(a²/2)(1 + cos2t)dt
=(a²/2)∫(1 + cos2t)dt
=(a²/2)[∫1 dt + ∫cos2t dt]
=(a²/2)[∫1 dt + ∫(1/2)cos2t d(2t)]
=(a²/2)[∫1 dt + (1/2)∫cos2t d(2t)]
=(a²/2)[t + (1/2)sin2t]
=(a²/2)t + (a²/4)sin2t + C