For this problem to make sense #4-9x^2>=0#, so #-2/3<=x<=2/3#. Therefore we can choose a #0<=u<=pi# such that #x=2/3cosu#. Using this, we can subsitute the variable x in the integral using #dx=-2/3sinudu#: #int x^2/sqrt(4-9x^2)dx=-4/27intcos^2u/(sqrt(1-cos^2u))sinudu=-4/27intcos^2udu# here we use that #1-cos^2u=sin^2u# and that for #0<=u<=pi# #sinu>=0#.
Now we use integration by parts to find #intcos^2udu=intcosudsinu=sinucosu-intsinudcosu=sinucosu+intsin^2u=sinucosu+intdu-intcos^2udu=sinucosu+u+c-intcos^2udu#. Therefore #intcos^2udu=1/2(sinucosu+u+c)#.
So we have found #int x^2/sqrt(4-9x^2)dx=-2/27(sinucosu+u+c)#, now we substitute #x# back for #u#, using #u=cos^(-1)((3x)/2)#, so #int x^2/sqrt(4-9x^2)dx=-1/9xsin(cos^(-1)((3x)/2))-2/27cos^(-1)((3x)/2)+c#.
We can further simplify this by using the definition of sines and cosines in terms of triangles. For a right triangle with an angle #u# at one of the non-right corners, #sinu="opposite side"/"longest side"#, while #cosu="adjacent side"/"longest side"#, since we know #cosu=(3x)/2#, we can pick the adjacent side to be #3x# and the longest side to be #2#. Using Pythagoras' theorem, we find the opposite side to be #sqrt(4-9x^2)#, so #sin(cos^(-1)((3x)/2))=sinu=1/2sqrt(4-9x^2)#. Therefore #int x^2/sqrt(4-9x^2)dx=-1/18xsqrt(4-9x^2)-2/27cos^(-1)((3x)/2)+c#.
