firstly note that #d/dx cos x^(3/2) = -3/2 x^(1/2) sin x^(3/2#
So
#I = int \ x^2 \ sinx^(3/2) \ dx#
can be written as
#I = -2/3 int \ x^(3/2) \d/dx ( cos x^(3/2) )\ dx#
which by IBP gives
#-3/2 I = x^(3/2) cos x^(3/2)- int \d/dx ( x^(3/2) ) cos x^(3/2) \ dx#
#-3/2 I = x^(3/2) cos x^(3/2)- int \3/2 x^(1/2) cos x^(3/2) \ dx#
noting that #d/dx sin(x^(3/2)) = 3/2x^(1/2) cos x^(3/2#
we have #-3/2 I = x^(3/2) cos x^(3/2)- int \ d/dx sin x^(3/2) \ dx#
#-3/2 I = x^(3/2) cos x^(3/2)- sin x^(3/2) + C#
# I = -2/3 ( x^(3/2) cos x^(3/2)- sin x^(3/2)) + C#
#= 2/3 ( sin x^(3/2) - x^(3/2) cos x^(3/2) ) + C#
Here a sub may have been better
#I(x) = int \ x^2 \ sinx^(3/2) \ dx#
# u = x^(3/2), x = u^(2/3)#
So #du = 3/2 x^(1/2) dx = 3/2 u^(1/3) dx#
So the integration becomes
#I(u) = int \u^(4/3) \ sin u * \3/2 u^(- 1/3)\ du#
#I(u) =3/2 int \u \ sin u \ \ du#
But you still need IBP
#2/3 I(u) = int \ u d/(du) (- cos u) \ du#
#2/3 I(u) = - u cos u - int \d/(du) u (- cos u) \ du#
#2/3 I(u) = - u cos u + int \ cos u \ du#
#I(u) =3/2( - u cos u + sin u) + C#
so
#I(x) = 3/2( sin x^(3/2) - x^(3/2) cos x^(3/2) ) + C#
Not really.
