Question

x的4次方/根号下(1-x的2次方)的不定积分是多少

要过程

Answer 1

令 x = sint，
I = ∫x^4dx/√(1-x^2) = ∫(sint)^4 costdx/cost = ∫(sint)^4 dx
= (1/4)∫(1-cos2t)^2 dt = (1/4)∫[1-2cos2t+(cos2t)^2] dt
= (1/4)∫[1-2cos2t+(1/2)(1+cos4t)] dt
= (1/4)∫[3/2-2cos2t+(1/2)cos4t] dt
= (1/4)[3t/2 - sin2t + (1/8)sin4t] + C
= (1/4)[3t/2 - 2sintcost + (1/4)sin2tcos2t]
= (1/4){3t/2 - 2sintcost + (1/2)sintcost[1-2(sint)^2]} + C
= (1/4){(3/2)arcsinx - 2x√(1-x^2) + (1/2)x(1-2x^2)√(1-x^2)} + C

Answer 2

望能给与帮助