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S3 1995 Q2

I am stuck trying to prove $$(2n+4)I_n =(2n+1)aI_n-1$$

I think it should be done by parts, as change of variable would mess the limits, but I really wonder how the $a$ ended there.

A general principle in mathematics is: don't stop. If you had gone with your gut and gone by parts you'd have seen the factor of $a$ appearing. In particular if you let $u = x^{n+1/2}$ (the natural choice) and $\mathrm{d}v = (a-x)^{1/2} \implies v = -\frac{2}{3}(a-x)^{3/2}$ so $$\begin{align}I_n = \frac{2n+1}{3}\int_0^a \underbrace{(a-x)^{3/2}}_{(a-x)(a-x)^{1/2}} x^{n-1/2} \, \mathrm{d}x &=\frac{2n+1}{3}\int_0^a (a-x)(a-x)^{1/2}x^{n-1/2} \, \mathrm{d}x \\&= \frac{2n+1}{3}\int_0^a \color{blue}{a}(a-x)^{1/2}x^{n-1/2} \, \mathrm{d}x - \frac{2n+1}{3}\int_0^1 (a-x)^{1/2} x^{n+1/2} \, \mathrm{d}x\end{align}$$

by distributing $(a-x)(a-x)^{1/2}x^{n-1/2}$ into $\color{purple}{a}(a-x)^{-1/2}x^{n-1/2} - x(a-x)^{1/2}x^{n-1/2}$ and using the fact that $x\cdot x^{n-1/2} = x^{n+1/2}$

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