数学积分题

设x=asint,dx=acostdt,√（a^2-x^2=acost
则t=arcsin(x/a)
所以原式=∫x^2/√（a^2-x^2）dx
=∫a^2(sint)^2*acostdt/acost
=a^2∫(sint)^2dt
=a^2∫[1-cos2t]dt/2
=(a^2/2)[∫dt-∫cos2tdt]
=(a^2/2)[t-∫cos2td(sin2t/2cos2t]
=(ta^2/2)-(a^2*sin2t/4)
=(a^2/2)*arcsin(x/a)-(a^2/4)*[sin2(arcsinx/a)]

不懂

这个题~选C