请问这不定积分怎么做,解到后面有个sin^2dt.
1个回答

设 x = asint,dx = a cost dt,t = arcsin(x/a)

则:

∫x^2/√(a^2-x^2) dx

=∫a^2*(sint)^2 * acost dt/(acost)

=a^2*∫(sint)^2 dt

=a^2/2*∫(1-cos2t)dt 注:cos2t = 1 - 2(sint)^2

=a^2/2*[t - ∫cos2tdt]

=a^2*t/2 - a^2/4*∫cos2t*d(2t)

=a^2*t/2 -a^2/4*sin2t + C

=a^2*arcsin(x/a)/2 -a^2/2 * sint*cost + C

=a^2*arcsin(x/a)/2 - x*√(a^2-x^2)/2 + C