Question

求∫根号1-4x²dx的不定积分

Answer 1

设 2x = sint，则 2dx = cost*dt
原积分公式：
=∫√[1 - (sint)^2] * (1/2)*cost * dt
=∫1/2 * (cost)^2 * dt
=1/4 * ∫2(cost)^2 * dt
=1/4 * ∫[1 + cos(2t)] * dt
=1/4*∫dt + 1/4*∫cos(2t)*dt
=t/4 + 1/8*∫cos(2t)*d(2t)
=t/4 + 1/8 * sin(2t) + C

追问

答案是1/4arcsin2x+x/2根号1-4x²+C

追答

忘记把 x 代回去了！
t = arcsin(2x)，sin(2t) = 2sint*cost = 2*(2x)*√[1-(2x)^2] = 4x * √(1-4x^2)
所以，原积分：
= (1/4) * arcsin(2x) + (1/2)*x *√(1 - 4x^2) + C