Recurrence relation for integral

Question

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?

Answer 1

Write $n\int x^2(1-x^2)^{\frac{n}{2} -1}$ as $$-n\int(1-x^2)^{\frac{n}{2}}+n\int (1-x^2)^{\frac{n}{2}-1}=-nI_n+nI_{n-2}.$$

Answer 2

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 $$ I_n (x) = {1 \over {1 + n}}x\left( {1 - x^{\,2} } \right)^{\,n/2} + {n \over {1 + n}}I_{n - 2} (x) $$