Question

求积分(1-x^2)^(1/2)arcsinxdx

求详细。

Answer 1

令x=sinu，-π/2<=u<=π/2，则(1-x^2)^(1/2)=cosu，arcsinx=arcsin(sinu)=u
dx=cosudu
原式=∫ u(cosu)^2du=(1/2)∫ u(1+cos2u)du
=(1/2)∫ (u+ucos2u)du=(1/4)u^2+(1/4)∫ ud(sin2u)
=(1/4)u^2+(1/4)u*sin2u-(1/4)∫ sin2udu
=(1/4)u^2+(1/2)u*sinucosu+(1/8)cos2u+C
=(1/4)u^2+(1/2)u*sinucosu+(1/4)(1-2(sinu)^2)+C
=(1/4)(arcsinx)^2+(1/2)x√(1-x^2)*arcsinx+(1/4)(1-2x^2)+C
=(1/4)(arcsinx)^2+(1/2)x√(1-x^2)*arcsinx-(1/2)x^2+C1