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

{\bf Question} Let $I_{m,n} \ds\int_0^{\frac{\pi}{2}} \sin^m x
\cos ^n x \,dx$. Prove that $I_{m,n}=I_{n,m}$. By differentiating
$\sin^{m-1} x \cos^{n+1} x$, prove that
$I_{m,n}=\left(\ds\frac{m-1}{m+n}\right)I_{m-2,n}$


\medskip

{\bf Answer} $I_{m,n}=\ds\int_0^{\frac{\pi}{2}} \sin^m x \cos ^n x
\,dx$

Now $I_{n,m}=\ds\int_0^{\frac{\pi}{2}}\sin^n x\cos^m x \,dx$
(reverse $n$ and $m$)

Now $\left. \begin{array} {l}
\cos\left(\ds\frac{\pi}{2}-\alpha\right)=\sin \alpha\\
\sin\left(\ds\frac{\pi}{2}-\alpha\right)=\cos \alpha \end{array}
\right\}$ standard trig. results

So in $I_{n,m}$ substitute $x=\ds\frac{\pi}{2}-u$.

$\Rightarrow \left\{ \begin{array} {rcl} \,dx & = & -\,du\\
x=\ds\frac{\pi}{2} & \rightarrow & u=0\\ x=0 & \rightarrow &
u=\ds\frac{\pi}{2}\end{array}\right.$

Then

$I_{n,m} = -\ds\int_{\frac{\pi}{2}}^0 \,du
\sin^n\left(\ds\frac{\pi}{2}-u\right) \cos ^m
\left(\ds\frac{\pi}{2}-u\right)$

$\Rightarrow I_{n,m} = -\ds\int_{\frac{\pi}{2}}^0 \,du \cos^n u
\sin ^m u$

$\Rightarrow I_{n,m} = +\ds\int^{\frac{\pi}{2}}_0 \,du \sin^m u
\cos^n u$ (reverse sign by integral property $\int_a^b=-\int_b^a$)

$\Rightarrow \un{I_{n,m} = I_{m,n}}$ (as defined above) as
required.

Differentiate $sin^{m-1}x \cos ^{n+1}x$

$\begin{array} {rcl} \ds\frac{d}{dx}(\sin^{m-1}x \cos^{n+1}x) & =
&  (m-1)\sin^{m-2}x \cos x \cos^{n+1} x\\ & - & (n+1) \cos^n x
\sin x \sin^{m-1}x \\ & = & (m-1)\sin^{m-2} x \cos^{n+1}x\\ & - &
(n+1)\underbrace{\sin^m x \cos ^n x} \end{array}$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ look familiar?

In other words, rearranging for $\sin^m x \cos^n x$ we have:

$(n+1)\sin^m x\cos^n x=(m-1)\sin^{m-2} x \cos^{n+2} x
-\ds\frac{d}{dx}(\sin^{m-1} x \cos ^{n+1} x)$

\newpage
or, integrating both sides:


$\begin{array} {rcl} \ds\int_0^{\frac{\pi}{2}} \sin^m x \cos ^n x
\,dx & = & \ds\frac{(m-1)}{(n+1)}\ds\int_0^{\frac{\pi}{2}}
\sin^{m-2}x \cos^{n+2} x \,dx\\ & - &
\ds\frac{1}{n+1}\ds\int_0^{\frac{\pi}{2}}
\ds\frac{d}{dx}(\sin^{m-1}x \cos ^{n+1} x) \,dx \end{array}$

$\Rightarrow$

$\begin{array} {rcl} \ds\int_0^{\frac{\pi}{2}} \sin^m x \cos ^n x
\,dx & = & \ds\frac{(m-1)}{(n+1)}\ds\int_0^{\frac{\pi}{2}}
\sin^{m-2}x \cos^{n+2} x \,dx\\ & - & \ds\frac{1}{n+1} -\sin^{m-2}
x \cos^{n+1}x]_0^{\frac{\pi}{2}} \end{array}$

(since $\int_a^b \,dx(\frac{df}{dx})=[f]_a^b$)

i.e. $I_{m,n}=\ds\frac{(m-1)}{(n+1)}\ds\int_0^{\frac{\pi}{2}}
\sin^{m-2}x \cos^{n+2} x \,dx-0$

Now write $\begin{array} {rcl} \sin^{m-2}x \cos^{n+2} x & = &
\sin^{m-2}x \cos^{n} x \cos^2 x\\ & = & \sin^{m-2}x \cos^{n} x
(1-\sin^2 x)\\ & = & \sin^{m-2}x \cos^{n} x-\sin^m x \cos^n x
\end{array}$

Thus, we have

$\begin{array} {rcl} I_{m,n} & = &
\ds\frac{(m-1)}{(n+1)}\ds\int_0^{\frac{\pi}{2}} \sin^{m-2}x
\cos^{n} x \,dx\\ & - &
\ds\frac{(m-1)}{(n+1)}\ds\int_0^{\frac{\pi}{2}} \sin^{m}x \cos^{n}
x \,dx\\ \Rightarrow I_{m,n} & = &
\ds\frac{(m-1)}{(n+1)}I_{m-2,n}-\ds\frac{(m-1)}{(n+1)}I_{m,n}
\\ & \Rightarrow & \un{I_{m,n}=\ds\frac{(m-1)}{(m+n)}I_{m-2,n}}
\end{array}$

\end{document}
