Problem source

If 
\[
\mathrm{I}_{n}=\int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x,
\]
show that $\mathrm{I}_{0}=\pi a^{2}/8.$

Show that $(2n+4)\mathrm{I}_{n}=(2n+1)a\mathrm{I}_{n-1}$ and hence
evaluate $\mathrm{I}_{n}$.

Solution source

\begin{align*}
&& I_n &= \int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x\\
&& I_0 &= \int_0^a x^{\frac12}(a-x)^{\frac12} \d x \\
x = a \sin^2 \theta, \d x = 2a \sin \theta \cos \theta \d \theta &&&= \int_{\theta =0}^{\theta = \pi/2} \sqrt{a}\sin \theta\sqrt{a} \cos \theta 2a \sin \theta \cos \theta \d \theta \\
&&&= \frac{a^2}{2} \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\
&&&= \frac{a^2}{4} \int_0^{\pi/2}(1- \underbrace{\cos 4\theta}_{\text{runs round the whole unit circle}}) \d \theta \\
&&&= \frac{\pi a^2}{8} \\
\\
&& I_n &= \int_0^a x^{n+\frac12}(a-x)^{\frac12} \d x \\
&&&=\underbrace{\left [-\frac23x^{n+\frac12}(a-x)^\frac32 \right]_0^a}_{=0} + \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)^\frac32 \d x \\
&&&=  \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)(a-x)^\frac12 \d x \\
&&&=  \frac23 \left(n+\frac12\right)aI_{n-1}-\frac23 \left(n+\frac12\right)I_{n} \\
\Rightarrow && \left(n+\frac12+\frac32\right)I_{n} &= \left(n+\frac12\right)aI_{n-1}\\
\Rightarrow && (2n+4)I_n &= (2n+1)aI_{n-1} \\
\\
\Rightarrow && I_n &= \frac{2n+1}{2n+4}a I_{n-1} \\
&&&=\frac{2n+1}{2n+4}\frac{2n-1}{2n+2}a^2 I_{n-2} \\
&&&= \frac{(2n+1)!!}{(2n+4)!!}  \pi a^{n+2}
\end{align*}