Recurrence relation for integral Find a recursive relation for the integral $\int(1-x^2)^\frac{n}{2}$ :
Pretty sure partial integration is the way to go since we haven't learned anything past that. My try:
The only non-tedious way to split this for a partial integration is 
$u=(1-x^2)^\frac{n}{2}$  $dx=dv$
$du=-nx(1-x^2)^{\frac{n}{2}-1}$ $x=v$
$\int(1-x^2)^\frac{n}{2} = x(1-x^2)^\frac{n}{2} + n\int x^2(1-x^2)^{\frac{n}{2}-1}$
Not quite a recurrence relation yet. I tried partial again on the second integral but that didn't work too well. Any suggestions?
 A: Write $n\int x^2(1-x^2)^{\frac{n}{2} -1}$ as $$-n\int(1-x^2)^{\frac{n}{2}}+n\int (1-x^2)^{\frac{n}{2}-1}=-nI_n+nI_{n-2}.$$
A: We shall assume $|x| \le 1$ in order to remain in the real field. 
Then we can make the substitution
$$
x = \sin \alpha \quad dx = \cos \alpha \;d\alpha 
$$
to get
$$
\eqalign{
  & I_n (x) = \int {\left( {1 - x^{\,2} } \right)^{\,n/2} dx}  = I_n (\sin \alpha ) = \int {\cos ^{\,n + 1} \alpha \;d\alpha }  =   \cr 
  &  = \int {\cos ^{\,n} \alpha \;d\left( {\sin \alpha } \right)}  =   \cr 
  &  = \cos ^{\,n} \alpha \sin \alpha  + n\int {\cos ^{\,n - 1} \alpha \,\;\sin ^{\,2} \alpha \;d\alpha }  =   \cr 
  &  = \cos ^{\,n} \alpha \sin \alpha  + n\int {\cos ^{\,n - 1} \alpha \,\;d\alpha }  - n\int {\cos ^{\,n + 1} \alpha \,\;\;d\alpha }  \cr} 
$$
which is
$$
\eqalign{
  & I_n (\sin \alpha ) = \cos ^{\,n} \alpha \sin \alpha  + nI_{n - 2} (\sin \alpha ) - nI_n (\sin \alpha )  \cr 
  & I_n (\sin \alpha ) = {1 \over {1 + n}}\cos ^{\,n} \alpha \sin \alpha  + {n \over {1 + n}}I_{n - 2} (\sin \alpha ) \cr} 
$$
and we can rewrite that as
$$
I_n (x) = {1 \over {1 + n}}x\left( {1 - x^{\,2} } \right)^{\,n/2}  + {n \over {1 + n}}I_{n - 2} (x)
$$
