Find a relation between $I_n$ and $I_{n+1}$, where $I_{n}=\int_a^b(x-a)^n \sqrt{b-x}\;dx$ $n \in \Bbb{N}^*$ and $0<a<b$ and 
$$I_{n}=\int_a^b(x-a)^n \sqrt{b-x}\;dx$$
I'm tasked with finding a relation between $I_n$ and $I_{n+1}$ to prove later that 
$$I_n=\frac{2^{2n+2}(n+1)!n!}{(2n+3)!}(b-a)^{n+1}\sqrt{b-a}$$
Any ideas?
 A: Putting
$$
\left\{ \matrix{
  y = {{x - a} \over {b - a}}\quad x = a + \left( {b - a} \right)y \hfill \cr 
  dy = {1 \over {b - a}}dx \hfill \cr}  \right.
$$
we get the expression of the integral in terms of the Beta function, so
in terms of the Gamma, or Rising Factorial, etc.
$$
\eqalign{
  & \int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n} \sqrt {\left( {b - x} \right)} \,dx}  =   \cr 
  &  = \left( {b - a} \right)^{\,n + 3/2} \int_{y = 0}^{\;1} {y^{\,n} \left( {1 - y} \right)^{\,1/2} \,dy}  =   \cr 
  &  = \left( {b - a} \right)^{\,n + 3/2} {\rm B}(n + 1,3/2) =   \cr 
  &  = \left( {b - a} \right)^{\,n + 3/2} {{\Gamma \left( {n + 1} \right)\Gamma \left( {3/2} \right)} \over {\Gamma \left( {n + 5/2} \right)}} =   \cr 
  &  = {{\sqrt \pi  } \over 2}\left( {b - a} \right)^{\,n + 3/2} {1 \over {\left( {n + 1} \right)^{\,\overline {\,3/2\,} } }} =  \cdots  \cr} 
$$
Concerning the recurrence on $n$, you can put
$$
\eqalign{
  & I_{\,n + 1}  = \int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n + 1} \sqrt {\left( {b - x} \right)} \,dx}  =   \cr 
  &  =  - {2 \over 3}\int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n + 1} d\left( {\left( {b - x} \right)^{\,3/2} } \right)}  =   \cr 
  &  =  - {2 \over 3}\left( {\left. {\left( {x - a} \right)^{\,n + 1} \left( {b - x} \right)^{\,3/2} \,} \right|_{x = a}^b
  - \left( {n + 1} \right)\int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,3/2} dx} } \right) =   \cr 
  &  = {2 \over 3}\left( {\left( {n + 1} \right)\int_{x = a}^{\;b} {\left( {b - x} \right)\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,1/2} dx} } \right) =   \cr 
  &  = {{2\left( {n + 1} \right)} \over 3}\left( {\int_{x = a}^{\;b} {\left( {b - a + a - x} \right)\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,1/2} dx} } \right) =   \cr 
  &  = {{2\left( {n + 1} \right)} \over 3}\left( {\left( {b - a} \right)\int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n} \left( {b - x} \right)^{\,1/2} dx}
  - \int_{x = a}^{\;b} {\left( {x - a} \right)^{\,n + 1} \left( {b - x} \right)^{\,1/2} dx} } \right) =   \cr 
  &  = {{2\left( {n + 1} \right)} \over 3}\left( {\left( {b - a} \right)I_{\,n}  - I_{\,n + 1} } \right) \cr} 
$$
i.e.
$$
I_{\,n + 1}  = {{2\left( {n + 1} \right)} \over {3 + 2\left( {n + 1} \right)}}\left( {b - a} \right)I_{\,n} 
$$
A: Hint: Try the transformation: $t = \frac{x-a}{b-x}$.
A: $$I_{n+1}=\int_a^b(x-a)^{n+1}\sqrt{b-x}dx$$
using integration by parts, we get:
$$I_{n+1}=\left[(x-a)^{n+1}\times\frac{-2(b-x)^{3/2}}{3}\right]_a^b+\frac23(n+1)\int_a^b(x-a)^n(b-x)^{3/2}dx$$
we want to somehow fit in:
$$I_n=\int(x-a)^n\sqrt{b-x}dx$$
and so we can notice that:
$$I_{n+1}=\left[(x-a)^{n+1}\times\frac{-2(b-x)^{3/2}}{3}\right]_a^b+\frac23(n+1)\int_a^b(x-a)^n\sqrt{b-x}(b-x)dx$$
From here it is the same as G cab's method
