Year: 2017
Paper: 3
Question Number: 6
Course: LFM Pure
Section: Integration
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\textit{In this question, you are not permitted to use any properties of trigonometric functions or inverse trigonometric functions.}
The function $\T$ is defined for $x>0$ by
\[
\T(x) = \int_0^x \! \frac 1 {1+u^2} \, \d u\,,
\]
and $\displaystyle T_\infty =
\int_0^\infty \!\! \frac 1 {1+u^2} \, \d u\,$ (which has a finite value).
\begin{questionparts}
\item
By making an appropriate substitution in the integral for $\T(x)$,
show that \[\T(x) = \T_\infty - \T(x^{-1})\,.\]
\item
Let $v= \dfrac{u+a}{1-au}$, where $a$ is a constant. Verify that, for
$u\ne a^{-1}$,
\[
\frac{\d v}{\d u} = \frac{1+v^2}{1+u^2}
\,.
\]
Hence show that, for $a>0$ and $x< \dfrac1a\,$,
\[
\T(x) = \T\left(\frac{x+a}{1-ax}\right) -\T(a) \,.
\]
Deduce that
\[
\T(x^{-1})
= 2\T_\infty -\T\left(\frac{x+a}{1-ax}\right)
-\T(a^{-1})
\]
and hence that, for
$b>0$ and $y>\dfrac1b\,$,
\[
\T(y) =2\T_\infty - \T\left(\frac{y+b}{by-1}\right) - \T(b) \,.
\]
\item Use the above results to show that
$\T(\sqrt3)= \tfrac23 \T_\infty \,$
and
$\T(\sqrt2 -1)= \frac14 \T_\infty\,$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& T(x) &= \int_0^x \! \frac 1 {1+u^2} \, \d u \\
&&&= \int_0^{\infty} \frac{1}{1+u^2} \d u - \int_x^\infty \frac{1}{1+u^2} \d u \\
&&&= T_\infty - \int_x^\infty \frac{1}{1+u^2} \d u \\
u = 1/v, \d u = -1/v^2 \d v: &&&= T_\infty - \int_{v=x^{-1}}^{v=0} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\
&&&= T_\infty - \int_{0}^{x^{-1}} \frac{1}{1+v^2} \d v \\
&&&= T_\infty - T(x^{-1})
\end{align*}
\item Let $v = \frac{u+a}{1-au}$ then
\begin{align*}
&& \frac{\d v}{\d u} &= \frac{(1-au) \cdot 1 - (u+a)\cdot(-a)}{(1-au)^2} \\
&&&= \frac{1-au+au+a^2}{(1-au)^2} \\
&&&= \frac{1+a^2}{(1-au)^2} \\
\\
&& \frac{1+v^2}{1+u^2} &= \frac{1 + \left ( \frac{u+a}{1-ua} \right)^2}{1+u^2} \\
&&&= \frac{(1-ua)^2+(u+a)^2}{(1-ua)^2(1+u^2)} \\
&&&= \frac{1+u^2a^2+u^2+a^2}{(1-ua)^2(1+u^2)} \\
&&&= \frac{(1+u^2)(1+a^2)}{(1-ua)^2(1+u^2)} \\
&&&= \frac{1+a^2}{(1-ua)^2}
\end{align*}
if $a > 0, x < \frac1a$ then
\begin{align*}
&& T(x) &= \int_0^x \frac{1}{1+u^2} \d u \\
&&&= \int_{v=a}^{v=\frac{a+x}{1-ax}} \frac{1}{1+u^2} \frac{1+u^2}{1+v^2} \d v \\
&&&= T\left ( \frac{x+a}{1-ax} \right) - T(a) \\
\\
\Rightarrow && T(x^{-1}) &= T_\infty - T(x) \\
&&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T(a) \\
&&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T_\infty-T(a^{-1}) \\
&&&= 2T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) -T(a^{-1})
\end{align*}
$b > 0, y > \frac1b$ then $y> 0, b > \frac1y$ (same as letting $x = \frac1y, a = \frac1b$
\begin{align*}
&& T(y) &= 2T_\infty - 2T \left ( \frac{\frac1y+\frac1b}{1-\frac1{by}} \right) + T(b) \\
\Rightarrow && T(y) &= 2T_\infty - 2T \left ( \frac{b+y}{by-1} \right) + T(b) \\
\end{align*}
\item Letting $y = b = \sqrt{3}$ in the final equation
\begin{align*}
&& T(\sqrt{3}) &= 2T_{\infty} - T \left ( \frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}\sqrt{3}-1} \right) -T (\sqrt{3}) \\
&&&= 2T_\infty - 2T(\sqrt{3}) \\
\Rightarrow && T(\sqrt{3}) &= \tfrac23 T_\infty
\end{align*}
Let $x = \sqrt2 - 1, a = 1$ so,
\begin{align*}
&& T(\sqrt2 -1) &= T \left ( \frac{\sqrt2-1+1}{1-\sqrt2+1} \right)-T(1) \\
&&&= T \left ( \frac{\sqrt{2}}{2-\sqrt{2}} \right) - T(1) \\
&&&= T(\frac{\sqrt{2}(2+\sqrt{2})}{2}) - T(1) \\
&&&= T(\sqrt{2}+1) - T(1) \\
&&&= T_\infty - T(\sqrt2-1)-T(1) \\
\Rightarrow && T(\sqrt{2}-1) &= \frac12T_\infty-\frac12T(1) \\
&& T(1) &= T_\infty - T(1) \\
\Rightarrow && T(1) &= \frac12 T_\infty \\
\Rightarrow && T(\sqrt2-1) &= \frac12T_\infty - \frac14T_\infty \\
&&&= \frac14 T_\infty
\end{align*}
\end{questionparts}
The second most popular question attempted by four fifths of the candidates; the success rate was only very slightly less than that of question 4. As every part of the question required obtaining a given result, it had to be marked strictly on how well things were presented. There were surprising problems with changing the variables in the first part, as often candidates did not clearly understand dummy variables, and others integrated with respect to constants. In spite of the ban on the use of trigonometric functions, some still tried to use the tangent function. The two results in (iii), especially the second, were testing but were found very hard, and previous inapplicable results were used, ignoring the conditions given as inequalities.