As in the question I completed the first part by a substituion of $x=a\sin(t)$ after which the answer was easy to obtain. With regards to the second part I integrated the expression for $I _{n-1}$ by parts to give
$(2n+1)I_{n-1}=\int_{0}^{a}x^{n+\frac{1}{2}} (a-x)^{-\frac{1}{2}}dx
$
so how do I show the equivalence of the integral on the right to :
$\frac{2n+4}{a}I_{n}
$
I have spent a long time and cannot find how to do it so would appreciate some guidance through this part of the question. Thanks in advance
Best Answer
Integrate $I_n$ by parts \begin{eqnarray*} \int (a-x)^{1/2}dx =-\frac{2}{3}(a-x)^{3/2} \\ \frac{d}{dx} x^{n+1/2} =\left(n+\frac{1}{2} \right)x^{n-1/2}. \end{eqnarray*} The first term evaluates to zero \begin{eqnarray*} I_n= \frac{2}{3}\left(n+\frac{1}{2}\right) \int_0^a (a-x) x^{n-1/2} (a-x)^{1/2} dx. \end{eqnarray*} Now multiply the out the bracket, rearrange and you will have the result.