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

{\bf Question}

\begin{description}
\item[(i)]
$\ds\int x(3x^2+7) \,dx$

\item[(ii)]
$\ds\int \sin^4 x \cos x \,dx$

\item[(iii)]
$\ds\int \cos^5 x \,dx$

\item[(iv)]
$\ds\int \cos^4 x \,dx$

\item[(v)]
$\ds\int \sec x \,dx$

\item[(vi)]
$\ds\int \ds\frac{x \,dx}{\sqrt{x-2}}$

\item[(vii)]
$\ds\int \sin^2 x \cos^3 x \,dx$

\item[(viii)]
$\ds\int_1^2 \ds\frac{8x \,dx}{(2x+1)^3}$

\item[(ix)]
$\ds\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \ds\frac{\cot
x}{\sqrt{\csc^3 x}} \,dx$

\item[(x)]
$\ds\int_0^3 \ds\frac{3 \,dx}{\sqrt{9-x^2}}$

\item[(xi)]
$\ds\int \ds\frac{(x-2) \,dx}{(x^2-4x-5)}$

\item[(xii)]
$\ds\int \ds\frac{\,dx}{(x^2+6x+17)}$

\item[(xiii)]
$\ds\int \ds\frac{(2x+5) \,dx}{(x^2+4x+5)}$

\item[(xiv)]
$\ds\int \ds\frac{x^2 \,dx}{(x+1)}$
\end{description}

\newpage
{\bf Answer}

(i)

$\ds\int x(3x^2+7) \,dx$

either $\ds\int (3x^2+7x)
\,dx=\un{\ds\frac{3x^4}{4}+\ds\frac{7x^2}{2}+c}\ \ \ A$

{\bf{or}} spot that $\ds\frac{d}{dx}[(3x^2+7)^2+\bar{c}]=2 \cdot
6x(3x^2+7) = 12x(3x^2+7)$

so that \lq\lq by inspection",

$x(3x^2+7)=\ds\frac{1}{12}\ds\frac{d}{dx}(3x^2+7)^2$

or $\ds\int x(3x^2+7)=\ds\frac{1}{12}(3x^2+7)^2+c_1\ \ \ B$

\un{Check}: evaluate $B$:

$\ds\frac{1}{12}(3x^2+7)^2+c_1 =
\ds\frac{1}{12}(9x^4+42x^2+49)+c_1 =
\ds\frac{3}{4}x^4+\ds\frac{7x^2}{2}+\left(\ds\frac{49}{12}+c_1\right)$

Since it's an indefinite integral and integration constants are
arbitrary $A=B$ if $c=\ds\frac{49}{12}+c_1$.

Of course with integrands as simple as this you'd do it using
method $A$.  However, if I had asked for $\int x(3x^2+7)^4 \,dx$,
method $B$ might have been quicker on spotting that $\int
x(3x^2+7)^4
\,dx=\ds\frac{d}{dx}\left[\ds\frac{1}{30}(3x^2+7)^5+c\right]....$

\bigskip
(ii)

$\ds\int \sin^4 x \cos x \,dx$: spot that

$\ds\frac{d}{dx} (\sin^5{x})=5\sin^4 x \cos x \Rightarrow \ds\int
\sin^4 x\cos x \,dx = \ds\frac{1}{5}\ds\int \frac{d}{dx} (\sin^5
x) \,dx = \un{\ds\frac{1}{5} \sin^5 x +c}$

(check by differentiation)

NB $\ds\int \ds\frac{d(f(x))}{dx} \,dx=f(x)+c$ Standard result:
the integral undoes the derivative.

\bigskip
(iii)

$\begin{array} {rcl} \ds\int \cos^5 x \,dx & = & \ds\int \cos^4 x
\cos x \,dx\\ & = & \ds\int (1-\sin^2 x)^2 \cos x \,dx\\ & = &
(1-2\sin^2 x + \sin^4x) \cos x \,dx\\ & = & \ds\int(\cos x-2\sin^2
x \cos x + \sin^4 x \cos x) \,dx\\ & = & \un{\sin
x-\ds\frac{2}{3}\sin^3 x+\ds\frac{1}{5}\sin^5 x +c} \end{array}$

NB $\cos^2 x=1-\sin^2 x$ and $\ds\int \sin^2 x \cos x =
\ds\frac{1}{3} \sin^3 x+c$

\newpage
(iv)

$\begin{array} {rcl} \ds\int \cos^4 x \,dx & = & \ds\int (\cos^2
x)^2 \,dx\\ & = & \ds\frac{1}{4} \ds\int(1+\cos 2x)^2 \,dx\\ & &
({\rm{since}}\ 2\cos^2 x-1=\cos 2x)\\ & = &
\ds\frac{1}{4}\ds\int(1+2\cos 2x+\cos^2 2x) \,dx\\ & = &
\ds\frac{1}{4} \ds\int\left[1+2\cos 2x+\ds\frac{1}{2}(1+\cos
4x)\right] \,dx\\ & & ({\rm{since}}\ 2\cos^2 (2x)-1=\cos 4x)\\ & =
& \ds\frac{1}{4}\ds\int\left(\ds\frac{3}{2}+2\cos
2x+\ds\frac{1}{2} \cos 4x\right) \,dx\\ & = &
\ds\frac{1}{4}\left(\ds\frac{3}{2}x + \sin 2x +\ds\frac{1}{8} \sin
4x\right)+c\\ & = & \un{\ds\frac{3x}{8}+\ds\frac{1}{4} \sin
2x+\ds\frac{1}{32} \sin 4x+c} \end{array}$

\bigskip
(v)

$\ds\int \sec x \,dx$

This can be written as a function of the type
$\ds\frac{f'(x)}{f(x)}$

How? Consider

$\begin{array} {l} f(x)=\sec x+\tan x\\ f'(x)=\sec x\tan x+\sec^2
x \end{array}$ (standard derivative)

Thus $\ds\frac{f'(x)}{f(x)}=\ds\frac{\sec x \tan x +\sec^2 x}{\sec
x+\tan x}=\ds\frac{\sec x(\sec x+\tan x)}{(\sec x+\tan
x)}=\un{\sec x}$

Now:

$\ds\int \ds\frac{f'(x)}{f(x)} \,dx=\ln[f(x)]+c$ (standard
integral)

So

$\ds\int \sec x \,dx = \ds\int \ds\frac{\sec x(\sec x+\tan
x)}{(\sec x+\tan x)} \,dx=\ds\int \ds\frac{f'(x)}{f(x)} \,dx$

where $f(x)=\sec x+\tan x$

Therefore \un{$\ds\int \sec x \,dx=\ln[\sec x+\tan x]+c$}

\newpage
(vi)

$\ds\int \ds\frac{x \,dx}{\sqrt{x-2}}$

Try substitution

$\begin{array} {l} u=\sqrt{x-2}=(x-2)^{\frac{1}{2}}\\ \Rightarrow
\ds\frac{du}{dx}=\ds\frac{1}{2}(x-2)^{-\frac{1}{2}}\\ \Rightarrow
\ds\frac{dx}{du}=2(x-2)^{\frac{1}{2}} \end{array}$

and also $u^2=x-2 \Rightarrow x=u^2+2$

So

$\begin{array} {rcl} \ds\int \ds\frac{x \,dx}{\sqrt{x-2}} & = &
\ds\int \ds\frac{x}{\sqrt{x-2}} \ds\frac{dx}{du} du\\ & = &
\ds\int \ds\frac{(u^2+2)}{\sqrt{x-2}} 2\sqrt{x-2} du\\ & = &
2\ds\int (u^2+2) \,du\\ & = & \ds\frac{2}{3}u^3+4u+c\\ & = &
\ds\frac{2}{3}u(u^2+6)+c \end{array}$

\un{Now} have to replace $u$ by $(x-2)^{\frac{1}{2}}$

So

\un{$\ds\int \ds\frac{x \,dx}{\sqrt{x-2}} =
\ds\frac{2}{3}(x-2)^{\frac{1}{2}}(x-2+6)+c=\ds\frac{2}{3}(x+4)\sqrt{x-2}+c$}

\newpage
(vii)

$\ds\int \sin^2 x \cos^3 x \,dx$

Try substitution

$\begin{array} {l} u=\sin x\\ \Rightarrow \ds\frac{du}{dx}=\cos
x\\ \Rightarrow \ds\frac{dx}{du}=\ds\frac{1}{\cos x} \end{array}$

Therefore

$\begin{array} {rcl} \ds\int \sin^2 x\cos^3 x \,dx & = & \ds\int
\sin^2x \cos^3x \ds\frac{dx}{du} \,du\\ & = & \ds\int \sin^2 x
\cos^2 x \cos x \ds\frac{dx}{du} \,du\\ & = & \ds\int \sin^2
x(1-\sin^2 x) \cos x \ds\frac{dx}{du} \,du\\ & = & \ds\int
u^2(1-u^2) \cos x \ds\frac{1}{\cos x} \,du\\ & = & \ds\int
(u^2-u^4) \,du\\ & = & \ds\frac{u^3}{3}-\ds\frac{u^5}{5}+c\\ & = &
\ds\frac{u^3}{15}(5-3u^2)+c \end{array}$

Replace $u$ by $\sin x$ to get:

\un{$\ds\int \sin^2 x \cos^3 x \,dx=\ds\frac{1}{15} \sin^3
x(5-3\sin^2 x)+c$}

\newpage
(viii)

$\ds\int_1^2 \ds\frac{8x \,dx}{(2x+1)^3}$

Try substitution

$u=2x+1 \Rightarrow x=\ds\frac{1}{2}(u-1)$

$\ds\frac{du}{dx}=2 \Rightarrow \ds\frac{dx}{du}=\ds\frac{1}{2}$

Under this substitution the interval $1 \leq x \leq 2$ maps onto
$3 \leq u \leq 5$ (when $x=1,\ u=(2 \times 1)+1=3$, when $x=2,\
u=(2 \times 2)+1=5$)

Hence,

$\begin{array} {rcl} & & \ds\int \ds\frac{8x}{(2x+1)^3}\\ & = &
\ds\int_{x=1}^{x=2} \ds\frac{8x}{(2x+1)^4}\ds\frac{dx}{du} \,du\\
& = & \ds\int_{u=3}^{u=5} \ds\frac{8 \times \frac{1}{2}(u-1)}{u^3}
\times \ds\frac{1}{2}\,du\\ & = & 2 \ds\int_3^5
\left(\ds\frac{1}{u^2}-\ds\frac{1}{u^3}\right) \,du\\ & = &
2\ds\int_3^5(u^{-2}-u^{-3}) \,du\\ & = &
2\left[-u^{-1}+\ds\frac{1}{2}u^{-2}\right]_3^5\ \\ & & {\rm{NB\
keep\ the}}\ u\ {\rm{ranges\ in\ the\ \un{limits}\ since\ we\
have\ an\ answer\ in}}\ u\\ & = &
2\left\{\left(-\ds\frac{1}{5}+\ds\frac{1}{50}\right)-
\left(-\ds\frac{1}{3}+\ds\frac{1}{18}\right)\right\}\\ & = &
2\left(-\ds\frac{9}{50}+\ds\frac{5}{18}\right)\\ & = &
\un{\ds\frac{44}{225}} \end{array}$

\newpage
(ix)

$\ds\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \ds\frac{\cot
x}{\sqrt{\csc^3 x}} \,dx$

Let $u=\csc x$

$\ds\frac{du}{dx}=-\csc x \cot x \Rightarrow
\ds\frac{dx}{du}=\ds\frac{-1}{\csc x \cot x}$

When $x=\ds\frac{\pi}{6},\ u=\csc
\frac{\pi}{6}=\ds\frac{1}{\sin\frac{\pi}{6}}=\ds\frac{1}{(\frac{1}{2})}=2$

When $x=\ds\frac{\pi}{2},\ u=\csc
\frac{\pi}{2}=\ds\frac{1}{\sin\frac{\pi}{2}}=\ds\frac{1}{1}=1$

So $\frac{\pi}{6} \leq x \leq \frac{\pi}{2}$ maps to $2 \leq u
\leq 1$.

Hence:

$\begin{array} {rcl} & & \ds\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}
\ds\frac{\cot x \,dx}{\sqrt{\csc^3 x}}\\ & = &
\ds\int_{x=\frac{\pi}{6}}^{x=\frac{\pi}{2}} \ds\frac{\cot
x}{\sqrt{\csc^3 x}} \ds\frac{dx}{du} \,du\\ & = &
\ds\int_{u=2}^{u=1} \ds\frac{1}{u^{\frac{3}{2}}}\cot x
\left(\ds\frac{-1}{\csc x \cot x}\right) \,du\\ & = &
-\ds\int_{u=2}^{u=1} \ds\frac{\,du}{u^{\frac{3}{2}}\csc x}\\ & = &
-\ds\int_{u=2}^{u=1}\ds\frac{\,du}{u^{\frac{3}{2}} \cdot u}\\ & =
& -\ds\int_2^1 \ds\frac{\,du}{u^{\frac{5}{2}}}\\ & = &
+\ds\int_1^2\ds\frac{\,du}{u^{\frac{5}{2}}}\\ & & {\rm{since}}\
-\ds\int_a^b f(x)\,dx=+\ds\int_b^a f(x)\,dx\ {\rm{check\ from\
definite\ integral\ limits}}\\ & = & \ds\int_1^2 u^{-\frac{5}{2}}
\,du\\ & = & \left[\ds\frac{-2}{3}u^{-\frac{3}{2}}\right]_1^2\\ &
= & \ds\frac{-2}{3}(2^{-\frac{3}{2}}-1^{-\frac{3}{2}})\\ & = &
-\ds\frac{2}{3}\left(\ds\frac{1}{2^{\frac{3}{2}}}-1\right)\\ & = &
-\ds\frac{2}{3}\left(\ds\frac{1}{2\sqrt{2}}-1\right)\\ & = &
-\ds\frac{2}{3}\left(\ds\frac{\sqrt{2}}{4}-1\right)\\ & = &
\un{\ds\frac{1}{6}(4-\sqrt{2})} \end{array}$

\newpage
(x)

$\ds\int_0^3 \ds\frac{3 \,dx}{\sqrt{9-x^2}}$

Could use standard integrals \un{or} if none available, let

$x=3\sin u,\ \ \ \ 0 \leq x \leq 3 \rightarrow 0 \leq u \leq
\frac{\pi}{2}$

$\ds\frac{dx}{du}=3\cos u$

Hence

$\begin{array} {rcl} \ds\int_0^3 \ds\frac{3 \,dx}{\sqrt{9-x^2}}
\,dx & = & \ds\int_0^3 \ds\frac{3}{\sqrt{9-x^2}}\ds\frac{dx}{du}
\,du\\ & = & \ds\int_0^{\frac{\pi}{2}} \ds\frac{3}{\sqrt{9-9\sin^2
u}} 3 \cos u \,du\\ & = & \ds\int_0^{\frac{\pi}{2}}\ds\frac{3
\cdot 3 \cos u}{3 \cos u} \,du\\ & = & 3\ds\int_0^{\frac{\pi}{2}}
\,du\\ & = & 3[u]_0^{\frac{\pi}{2}}\\ & = & \un{\ds\frac{3\pi}{2}}
\end{array}$

\bigskip
(xi)

$\ds\int \ds\frac{(x-2) \,dx}{(x^2-4x-5)}$

Use partial fractions since it's of form $\ds\frac{\alpha x +
\beta}{\gamma x^2+\delta x +\epsilon}$

First we have $x^2-4x-5 \equiv (x+1)(x-5)$ factors: $(x+1),(x-5)$

So we seek $A,\ B$ such that

$\begin{array} {rcl} \ds\frac{x-2}{(x+1)(x-5)} & \equiv &
\ds\frac{A}{(x+1)}+\ds\frac{B}{(x-5)}\\ & \Rightarrow & (x-2)
\equiv A(x-5)+B(x+1) \end{array}$

Equating coefficients of $x$ and numbers on LHS and RHS
$\Rightarrow 1=A+B, -2=-5A+B$

Solving gives $A=\ds\frac{1}{2}, B=\ds\frac{1}{2}$

So $\begin{array} {rcl} \ds\int\ds\frac{(x-2) \,dx}{(x^2-4x-5)} &
= &
\ds\frac{1}{2}\ds\int\ds\frac{\,dx}{(x+1}+\ds\frac{1}{2}\ds\int{\,dx}{(x-5)}\\
& = & \ds\frac{1}{2}\log(x+1)+\ds\frac{1}{2}\log(x-5)+c\\ & = &
\ds\frac{1}{2}\log[(x+1)(x-5)]+c\\ & = &
\un{\ds\frac{1}{2}\log(x^2-4x-5) +c}\\ \rm{or} & = &
\log(\sqrt{x^2-4x-5})+c\\ \rm{or} & = & \log(k\sqrt{x^2-4x-5})+c
\end{array}$

$k$ = arbitrary constant

\un{NB} could also do this using the $\ds\frac{f'(x)}{f(x)}$
method since

$\ds\int \ds\frac{x-2}{(x^2-4x-5)}=\ds\frac{1}{2}\ds\int
\ds\frac{2x-4}{x^2-4x-5}=\ds\frac{1}{2}\ds\int\ds\frac{f'(x)}{f(x)}$
where $f(x)=x^2-4x-5$

Check for yourself that the answers are identical (up to arbitrary
constants).

\bigskip
(xii)

$\ds\int \ds\frac{\,dx}{(x^2+6x+17)}$

Now $x^2+6x+17 \equiv (x+3)^2+8$

So

$\begin{array} {rcl} \ds\int\ds\frac{\,dx}{(x^2+6x+17)} & = &
\ds\int\ds\frac{\,dx}{(x+3)^2+(\sqrt{3})^2}\\ & & \rm{Now\ set}\
u=x+3,\ du=dx\\ & = & \ds\int\ds\frac{\,du}{u^2+a^2}\ \rm{where}\
a=\sqrt{3}\\ & = & \ds\frac{1}{a}
\arctan\left(\ds\frac{u}{a}\right)+c\ \rm{by\ standard\
integral}\\ & = &
\un{\ds\frac{1}{\sqrt{8}}\arctan\left(\ds\frac{x+3}{\sqrt{8}}\right)+c}
\end{array}$

NB Could you have used partial fractions here?

\bigskip
(xiii)

$\ds\int \ds\frac{(2x+5) \,dx}{(x^2+4x+5)}$

Could try partial fractions or spot type $\ds\frac{f'(x)}{f(x)}$:
do this way.

$\ds\frac{d}{dx}(x^2+4x+5)=2x+4$

Hence

$\begin{array} {rcl} & & \ds\int\ds\frac{2x+5}{x^2+4x+5} \,dx\\ &
= & \ds\int \ds\frac{(2x+4)\, dx}{x^2+4x+5}+\ds\int \ds\frac{1
\,dx}{x^2+4x+5}\\ & = &
\log(x^2+4x+5)+\ds\int\ds\frac{\,dx}{x^2+4x+5}\\ & &
\ln\left(\ds\frac{f'}{f}\right)\ \rm{type\ integral\ and\
integral\ of\ type\ (xxiv)}\\ & = &
\log(x^2+4x+5)+\ds\int\ds\frac{\,dx}{(x+2)^2+1^2}\\ & = &
\un{\log(x^2+4x+5)+\arctan(x+2)+c} \end{array}$ (standard
integrals)

\newpage
(xiv)

$\ds\int \ds\frac{x^2 \,dx}{(x+1)}$

Partial fractions won't work: divide out.

$$\ds\frac{x^2}{x+1} \equiv x-1+\ds\frac{1}{x+1}$$

This comes either from

PICTURE \vspace{1in}

$\Rightarrow
\ds\frac{x^2}{x+1}=(x-1)+\left(\ds\frac{1}{x+1}\right)$

\un{or} $\begin{array} {rcl}
\ds\frac{x^2}{x+1}=\ds\frac{x^2-1+1}{x+1} & = &
\ds\frac{x^2-1}{x+1}+\ds\frac{1}{x+1}\\ & = &
\ds\frac{(x-1)(x+1)}{(x+1)}+\ds\frac{1}{x+1}\\ & = &
(x-1)+\ds\frac{1}{x+1}\end{array}$

Thus $\ds\int\ds\frac{x^2}{x+1} \,dx = \ds\int9x-1) \,dx + \ds\int
\ds\frac{\,dx}{(x+1)} = \un{\ds\frac{x^2}{2}-x+\log(x+1)+c}$


\end{document}