#d/(dx)(sqrt[x] sin^2x/x^2)=(2 cosx sinx)/x^(3/2) - (3 sin^2x)/(2 x^(5/2))#
so
#int_0^a (sin^2x)/x^(5/2)dx =2/3(int_0^a(2 cosx sinx)/x^(3/2)dx-(sqrt[x] sin^2x/x^2)_0^a)#
but
#d/(dx)(sqrt x cosx sinx/x) =cos^2x/sqrt[x] - (cosx sinx)/(2 x^(3/2)) - sin^2x/sqrt[x] #
so
#int_0^a (cosx sinx)/(2 x^(3/2)) dx = int_0^acos^2x/sqrt[x]dx-int_0^asin^2x/sqrt[x] dx-(sqrt x cosx sinx/x)_0^a #
so the problem is with the integrals
#int_0^a cos^2x/sqrt[x]dx #
and
#int_0^a sin^2x/sqrt[x]#
but
#int_0^a abs(cos^2x/sqrt[x])dx le int_0^aabs(dx/sqrt(x))=2sqrt(a)# and
#int_0^a abs(sin^2x/sqrt[x])dx le int_0^aabs(dx/sqrt(x))=2sqrt(a)#
so the integral is convergent.
Note: It is assumed that #lim_(x->0)sinx/x=1#