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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}
\begin{description}
\item[(i)]
Show that if,

$$I_n=\ds\int_0^{\frac{\pi}{2}}x^n \sin x \,dx,\ \ n \geq 1,$$

then by integrating by parts twice,

$$I_n=n\left(\ds\frac{\pi}{2}\right)^{n-1}-n(n-1)I_{n-2}$$

{\bf{Hint:}} remember you want to try to reduce the power of $n$.

\item[(ii)]
Evaluate $I_0$ and hence calculate $I_6$.

\item[(iii)]
Given that when $n=\ds\frac{1}{2},\ I_{\frac{1}{2}}=0.977451...$,
calculate $I_\frac{5}{2}$ to six decimal places.

\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$I_n=\ds\int_0^{\frac{\pi}{2}} x^n \sin x \,dx\ \ \ n \geq 1$

$\begin{array} {ll} u=x^n & \ds\frac{dv}{dx}=\sin x\\ du=nx^{n-1}
& v=-\cos x \end{array}$

$\Rightarrow$

\begin{eqnarray*} I_n & = &
\left[-x^n\cos
x\right]_0^{\frac{\pi}{2}}+n\ds\int_0^{\frac{\pi}{2}}x^{n-1}\cos x
\,dx\\ & = & -\left(\ds\frac{\pi}{2}\right)^n
\cos\left(\ds\frac{\pi}{2}\right)-0^n\cos
0+n\ds\int_0^{\frac{\pi}{2}}x^{n-1}\cos x \,dx\\ & = &
n\ds\int_0^{\frac{\pi}{2}} x^{n-1} \cos x \,dx \end{eqnarray*}

$\begin{array} {ll} u=x^{n-1} & \ds\frac{dv}{dx}=\cos x\\
du=(n-1)x^{n-2} & v=\sin x \end{array}$

$\Rightarrow$

\begin{eqnarray*} I_n & = & n\left\{[x^{n-1} \sin
x]_0^{\frac{\pi}{2}}-(n-1)\ds\int_0^{\frac{\pi}{2}} x^{n-2} \sin
x\,dx\right\}\\ & = & n\left(\left(\ds\frac{\pi}{2}\right)^{n-1}
\sin \ds\frac{\pi}{2}-0^{n-1} \sin
0\right]-n(n-1)\ds\int_0^{\frac{\pi}{2}} x^{n-1} \sin x \,dx
\end{eqnarray*}

Therefore $I_n=n\left(\ds\frac{\pi}{2}\right)^{n-1}-n(n-1)I_{n-1}$

\item[(ii)]
$I_0=\ds\int_0^{\frac{\pi}{2}}\sin x \,dx =[-\cos
x]_0^{\frac{\pi}{2}}=1$

$I_6=6\left(\ds\frac{\pi}{2}\right)^5-6\times 5\times I_4$

$I_4=4\left(\ds\frac{\pi}{2}\right)^3-4\times 3\times I_2$

$I_2=2\left(\ds\frac{\pi}{2}\right)^1-2\times 1\times
I_0=2\ds\frac{\pi}{2}-2\times 1=\pi-2$ Back substitute

$I_4=\ds\frac{4}{8}\pi^3-12(\pi-2)$

$I_6=\ds\frac{6}{32}\pi^5-30\left(\ds\frac{\pi^3}{2}-12(\pi-2)\right)$

  $\ds\frac{3\pi^5}{16}-15\pi^3+360\pi-720 = 3.257896...$

\item[(iii)]
$I_{\frac{5}{2}}=\ds\frac{5}{2}\left(\ds\frac{\pi}{2}\right)^{\frac{3}{2}}-\ds\frac{5}{2}
\cdot \ds\frac{3}{2} \cdot I_{\frac{1}{2}}=1.256311858...$

\end{description}

\end{document}