For this problem to make sense 4-9x^2>=04−9x2≥0, so -2/3<=x<=2/3−23≤x≤23. Therefore we can choose a 0<=u<=pi0≤u≤π such that x=2/3cosux=23cosu. Using this, we can subsitute the variable x in the integral using dx=-2/3sinududx=−23sinudu: int x^2/sqrt(4-9x^2)dx=-4/27intcos^2u/(sqrt(1-cos^2u))sinudu=-4/27intcos^2udu∫x2√4−9x2dx=−427∫cos2u√1−cos2usinudu=−427∫cos2udu here we use that 1-cos^2u=sin^2u1−cos2u=sin2u and that for 0<=u<=pi0≤u≤π sinu>=0sinu≥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.
