\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Given that

$$I_m=\ds\int_0^{\frac{\pi}{2}} x^m \cos x \,dx\ \ ,$$

integrate by parts twice to derive the reduction formula

$$I_m=\left(\ds\frac{\pi}{2}\right)^m-m(m-1)I_{m-2},\ \ m \geq
2.$$

Show that $I_0=1$ and calculate $I_4$.

\medskip

{\bf Answer}

$I_m=\ds\int_0^{\frac{\pi}{2}} x^m \cos x \,dx$

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

$I_m=[x^m \sin x]_0^{\frac{\pi}{2}}-m\ds\int_0^{\frac{\pi}{2}}
x^{m-1} \sin x \,dx$

$J_{m-1}=\ds\int_0^{\frac{\pi}{2}} x^{m-1} \sin x \,dx$

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

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

Therefore $I_m=\left[\left(\ds\frac{\pi}{2}\right)^m \sin
\ds\frac{\pi}{2}-0^m \sin 0\right]-m(m-1)I_{m-2}$

$\Rightarrow
\un{I_m=\left(\ds\frac{\pi}{2}\right)^m-m(m-1)I_{m-2}}$

\begin{eqnarray*} I_0 & = & \ds\int_0^{\frac{\pi}{2}} \cos x
\,dx\\ & = & [\sin x]_0^{\frac{\pi}{2}}\\ & = & 1-0\\ & = & \un{1}
\end{eqnarray*}

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

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

$I_0=1$

$\Rightarrow$ \begin{eqnarray*} I_4 & = &
\left(\ds\frac{\pi}{2}\right)^4-12\left[\left(\ds\frac{\pi}{2}\right)^2-2\right]\\
I_4 & = &
\left(\ds\frac{\pi}{2}\right)^4-12\left(\ds\frac{\pi}{2}\right)^2+24\\
& = & \ds\frac{\pi^4}{16}-3\pi^2+24\\ & = & \un{0.479255...}
\end{eqnarray*}
\end{document}