Here,
#I=intx^3 sqrt(4-9x^2)dx#
let,
#3x=2sinu=>x=2/3sinu=>dx=2/3cosudu and #
#9x^2=4sin^2u andsinu=(3x)/2=>cos^2u=1-(9x^2)/4#
#=>cos^2u=(4-9x^2)/4=>cosu=1/2sqrt(4-9x^2)#
So,
#I=int(2/3sinu)^3sqrt(4-4sin^2u)xx2/3cosudu#
#=int8/27sin^3u(2cosu)2/3cosudu#
#=32/81intsin^3ucos^2udu#
#=32/81intsin^2ucos^2usinudu#
#=-32/81int[(1-cos^2u)cos^2u(-sinu)du#
#=-32/81[intcos^2u(-sinu)du-intcos^4u(-sinu)du]#
#=-32/81[int(cosu)^2d/(dx)(cosu)du-int(cosu)^4d/(dx)(cosu)du]#
#=-32/81[(cosu)^3/3-(cosu)^5/5]+c#
#=-32/(81xx15)[5(cosu)^3-3(cosu)^5]+c#
Subst. #cosu=1/2sqrt(4-9x^2#
#I=-32/1215[5xx(sqrt(4-9x^2))^3/(8)-3xx(sqrt(4-
9x^2))^5/(32)]+c#
#I=-1/1215[20(sqrt(4-9x^2))^3-3(sqrt(4-9x^2))^5]+c#
#I=1/1215[3sqrt(4-9x^2))^5-20(sqrt(4-9x^2))^3]+c#