Question

x^2arcsinxdx的定积分怎么求 其中x^2 是 x的平方

Answer 1

∫ x^2 arcsinx dx 令 arcsinx = t, x=sint dx=cost dt
= ∫ t d (sint)^3 /3
= t * (sint)^3 /3 - (1/3) ∫ (sint)^3 dt u = cost, du = -sint dt
= t * (sint)^3 /3 - (1/3) ∫ (u^2-1) du
= t * (sint)^3 /3 - (1/3) ( u^3 /3 - u) + C
= x^3 arcsinx /3 - (1/9) (1-x^2)^(3/2) + (1/3) (1-x^2)^(1/2) + C
= x^3 arcsinx /3 +(1/9) (1-x^2)^(1/2) (2+x^2) + C

Answer 2

∫x^2arcsinxdx
=∫arcsinxdx³/3
=arcsinxx³/3-1/3∫x³/√(1-x²)dx
x=sint dx=costdt
√1-x²=cost
∫x³/√(1-x²)dx
=∫sin³t/cost costdt
=∫sin³tdt
=∫sin²tsintdt
=-∫(1-cos²t)dcost
=-(cost-1/3cos³t)
=1/3cos³t-cost
∴∫x^2arcsinxdx
=arcsinxx³/3-1/3∫x³/√(1-x²)dx
=arcsinxx³/3-1/3(1/3cos³x-cosx)
=arcsinxx³/3-1/9cos³x+1/3cosx+C