<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\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}

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

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

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

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

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

\begin{eqnarray*} J_m &amp; = &amp; [-x^{m-1} \cos
x]_0^{\frac{\pi}{2}}+(m-1)\ds\int_0^{\frac{\pi}{2}}x^{m-2} \cos x
\,dx\\ &amp; = &amp; \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 &amp; = &amp; \ds\int_0^{\frac{\pi}{2}} \cos x
\,dx\\ &amp; = &amp; [\sin x]_0^{\frac{\pi}{2}}\\ &amp; = &amp; 1-0\\ &amp; = &amp; \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 &amp; = &amp;
\left(\ds\frac{\pi}{2}\right)^4-12\left[\left(\ds\frac{\pi}{2}\right)^2-2\right]\\
I_4 &amp; = &amp;
\left(\ds\frac{\pi}{2}\right)^4-12\left(\ds\frac{\pi}{2}\right)^2+24\\
&amp; = &amp; \ds\frac{\pi^4}{16}-3\pi^2+24\\ &amp; = &amp; \un{0.479255...}
\end{eqnarray*}
\end{document}
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