As many of you may know the integral of the form $\int{\frac{P_n(x)}{\sqrt{ax^2+bx+c}} dx}$ can be rewritten as
$$\int{\frac{P_n(x)}{\sqrt{ax^2+bx+c}} dx} = Q_{n-1}(x)\sqrt{ax^2+bx+c}+\lambda\int{\frac{dx}{\sqrt {ax^2+bx+c}}}$$
In this case $Q_{n-1}(x)$ is a polynomial with degree $n-1$. Specifically for my problem:
$$\int{x^4\sqrt{a^2-x^2}dx} = \int{\frac{x^4(a^2-x^2)}{\sqrt{a^2-x^2}}} = $$
$$= (Ax^5+Bx^4+Cx^3+Dx^2+Ex+F)\sqrt{a^2-x^2}+ \lambda\int{\frac{dx}{\sqrt{a^2-x^2}}}$$
By differentiating this problem becomes more of system of equations problem than an integral problem. What I stumbled upon is this same problem done almost identically with only one difference:
$$\int{x^4\sqrt{a^2-x^2}dx}= (Ax^5+Bx^3+Cx)\sqrt{a^2-x^2}+ \lambda\int{\frac{dx}{\sqrt{a^2-x^2}}}$$
The author of the book came to conclusion that the extra coefficients are unnecessary. There are no notes about why this was done. I see the same thing on other similar integrals that were simplified to half of the actual number of coeficients you would use. Every even coeficient was terminated. This can help tremendously for some higher degree polynomials. I'm trying to understand why. I checked both solutions and the answer is exactly the same. Even if you try to calculate the rest of the missing coeficients you will get that they are indeed $0$. However this was all done before actually doing any calculation. What's the inspiration behind this? Another similar problem where this was done:
$$\int{\frac{x^{10}dx}{\sqrt{1+x^2}}}= (Ax^9+Bx^7+Cx^5+Dx^3+Ex)\sqrt{1+x^2}+\lambda\int{\frac{dx}{\sqrt{1+x^2}}}$$