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

{\bf Question}

Evaluate
\begin{description}
\item[(i)]
$\ds\int_0^{\frac{\pi}{2}} \sin^3 x \cos^4 x \,dx$

\item[(ii)]
$\ds\int_0^{\frac{\pi}{2}} \sin^5 x \cos ^7 x \,dx$

\item[(iii)]
$\ds\int_0^{\frac{\pi}{2}} \sin^ x \cos^4 x \,dx$

\item[(iv)]
$\ds\int_0^1 x^691-x^2)^{\frac{3}{2}} \,dx$

\end{description}
\medskip

{\bf Answer} %note ref to question 6%
\begin{description}
\item[(i)]
This is $I_{3,4}$. Use formula of Q6.

$I_{m,n}=\ds\frac{(m-1)}{(m+n)}I_{m-2,n}$ valid for $m,n>0$

$I_{3,4}=\ds\frac{(3-1)}{(3+4)}I_{3-2,4}=\ds\frac{2}{7}I_{1,4}$

What's $I_{1,4}$? $\begin{array} {rcl} I_{1,4} & = &
\ds\int_0^{\frac{\pi}{2}}\sin x\cos^4 x \,dx\\ & = &
\left[-\ds\frac{\cos^5
x}{5}\right]_0^{\frac{\pi}{2}}=-0+\ds\frac{1}{5}=\ds\frac{1}{5}\end{array}$

Therefore $I_{3,4}=\ds\frac{2}{7} \times
\ds\frac{1}{5}=\un{\ds\frac{2}{35}}$.

\item[(ii)]
This is $I_{5,7}$. Use formula of Q6.

$I_{5,7}=\ds\frac{(5-1)}{(5+7)}I_{5-2,7}=\ds\frac{4}{12}I_{3,7}$

$I_{3,7}=\ds\frac{(3-1)}{(3+7)}I_{3-2,7}=\ds\frac{2}{10}I_{1,7}$


What's $I_{1,7}$?

$I_{1,7}=\ds\int_0^{\frac{\pi}{2}}\sin x\cos^7 x
\,dx=\left[-\ds\frac{\cos^8
x}{8}\right]_0^{\frac{\pi}{2}}=\ds\frac{1}{8}$

So

$I_{3,7}=\ds\frac{2}{10} \times \ds\frac{1}{8}$

$I_{5,7}=\ds\frac{4}{12} \times \ds\frac{2}{10} \times
\ds\frac{1}{8} = \ds\frac{1}{10 \times 12} =
\un{\ds\frac{1}{120}}$

\item[(iii)]
This is $I_{6,4}$. Use formula of Q6.

$I_{6,4}=\ds\frac{(6-1)}{(6+4)}I_{4,4}$

$I_{4,4}=\ds\frac{(4-1)}{(4+4)}I_{2,4}$

$I_{2,4}=\ds\frac{(2-1)}{(2+4)}I_{0,4}$

What's $I_{0,4}$?

$I_{0,4}=\ds\int_0^{\frac{\pi}{2}}\sin^0 x\cos^4 x
\,dx=\ds\int_0^{\frac{\pi}{2}} \cos^4 x \,dx$

Use result from lecture notes: equations (1.38)

$\Rightarrow \begin{array} {rcl} I_{0,4} & = &
\left(\ds\frac{4-1}{4}\right)I_{0,2}\\ & & \\ I_{0,2} & = &
\left(\ds\frac{2-1}{2}\right)I_{0,0}\\ & & \\ & = &
\left(\ds\frac{2-1}{2}\right)\ds\int_0^{\frac{\pi}{2}} \,dx\\ & &
\\ & = & \ds\frac{\pi}{4} \end{array}$

So $I_{0,4}=\ds\frac{3}{4}
\times\ds\frac{1}{2}\times\ds\frac{\pi}{2}=\ds\frac{3\pi}{16}$

Hence $\begin{array} {rcl} I_{2,4} & = &
\ds\frac{1}{6}\times\ds\frac{3\pi}{16}\\ & & \\ I_{4,4} & = &
\ds\frac{3}{8}\times\ds\frac{1}{6}\times\ds\frac{3\pi}{16}\\ & &
\\ I_{6,4} & = & \ds\frac{5}{10} \times \ds\frac{3}{8} \times
\ds\frac{1}{6} \times \ds\frac{3\pi}{16} =
\un{\ds\frac{3\pi}{512}} \end{array}$

\newpage
\item[(iv)]
$\ds\int_0^1 x^6(1-x^2)^{\frac{3}{2}} \,dx$

set $\begin{array} {rcl} x & = & \sin u\\ \Rightarrow \,dx & = &
\cos u \,du\\ \rm{and}\ x=1 & \rightarrow & u=\ds\frac{\pi}{2}\\
x=0 & \rightarrow & u=0 \end{array}$

So integral becomes

$\begin{array} {rcl} & & \ds\int_0^{\frac{\pi}{2}} \,du \cos u
(\sin u)^6(1-\sin^2 u)^\frac{3}{2}\\ & & \\ & = &
\ds\int_0^{\frac{\pi}{2}} \,du (\cos u) \sin^6 x(\cos^2
u)^{\frac{3}{2}}\\ & & \\ & = & \ds\int_0^{\frac{\pi}{2}} \,du
\cos u \sin^6 x \cos^3 u\\ & & \\ & = & \ds\int_0^{\frac{\pi}{2}}
\,du \cos^4 u \sin^6 u = I_{6,4} \end{array}$

We've already done this in part (iii), so the answer is
\un{$\ds\frac{3\pi}{512}$}.
\end{description}

\end{document}
