In an exercise I'm asked the following question:
Find a formula for the primitive $\int(1-x^2)^n dx$ with $n \in \mathbb N$ using recursion.
So, I already asked a question on this topic but that question is about the integral $\int_0^1(1-x^2)^n dx$.
So what I did was:
Let $I_n:=\int(1-x^2)^n dx$, then: $$I_n=\int(1-x^2)(1-x^2)^{n-1} dx = I_{n - 1} - \int x^2(1-x^2)^{n - 1}dx$$
How can I now complete this in order to ger a formula for $I_n$ using recursion? Thanks
Hint $u=(1-x^2)^n, dv= dx $ gives $$ I_n:=\int(1-x^2)^n dx=x(1-x^2)^n +2n \int x^2(1-x^2)^{n-1} dx $$
Substitute $\sin\theta=x$ so that $\cos \theta\, d\theta=dx$ and the integral becomes
$\begin{align}I_n&=\int \cos^{2n+1}\theta\, d\theta\\&=\int \cos^{2n}\theta(\sin\theta)'\,d\theta\\&=\cos^{2n}\theta \sin \theta-\int 2n \cos ^{2n-1}\theta(\cos^2\theta-1)\,d\theta\\&=\cos^{2n}\theta \sin \theta-2n I_n+2n I_{n-1}\\&=(\sqrt {1-x^2})^nx-2n I_n+2n I_{n-1}\implies (2n+1)I_n=(\sqrt {1-x^2})^nx+2n I_{n-1}\end{align}$.
Hint: evaluate the second integral (including the $-$ sign in front of it with integrate by parts viz. $u=\tfrac{x}{2n},\,v=(1-x^2)^n$ to obtain a condition of the form $I_n=I_{n-1}+f(x)+c_nI_n$ with $c_n\ne1$.
