Year: 2010
Paper: 3
Question Number: 2
Course: UFM Pure
Section: Integration using inverse trig and hyperbolic functions
About 80% of candidates attempted at least five questions, and well less than 20% made genuine attempts at more than six. Those attempting more than six questions fell into three camps which were those weak candidates who made very little progress on any question, those with four or five fair solutions casting about for a sixth, and those strong candidates that either attempted 7th or even 8th questions as an "insurance policy" against a solution that seemed strong but wasn't, or else for entertainment!
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1485.5
Banger Comparisons: 1
In this question, $a$ is a positive constant.
\begin{questionparts}
\item Express $\cosh a$ in terms of exponentials.
By using partial fractions, prove that
\[
\int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x = \frac a {2\sinh a}\,.
\]
\item Find, expressing your answers in terms of hyperbolic functions,
\[
\int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x
\,
\]
and
\[
\int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x
\,.\]
\end{questionparts}\begin{questionparts}
\item $\cosh a = \frac12 (e^a + e^{-a})$
\begin{align*}
\int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x &= \int_0^1 \frac{1}{x^2+(e^a+e^{-a})x+e^ae^{-a}} \d x \\
&= \int_0^1 \frac{1}{e^a-e^{-a}}\left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\
&= \frac{1}{2 \sinh a} \int_0^1 \left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\
&= \frac{1}{2 \sinh a}\left [\ln(x+e^{-a})-\ln(x+e^a) \right]_0^1 \\
&= \frac{1}{2 \sinh a} \left (\ln(1+e^a)-\ln(1+e^{-a}) - (\ln e^{-a}-\ln e^a) \right) \\
&= \frac{1}{2\sinh a}\left (2a + \ln \frac{1+e^a}{1+e^{-a}}\right) \\
&= \frac1{2\sinh a} \left ( 2a -a \right) \\
&= \frac{a}{2 \sinh a}
\end{align*}
\item
\begin{align*}
\int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x &= \int_1^{\infty} \frac{1}{(x+e^a)(x-e^{-a})} \d x \\
&= \int_1^{\infty} \frac{1}{e^a+e^{-a}} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\
&= \frac{1}{2\cosh a} \int_1^{\infty} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\
&= \frac{1}{2\cosh a} \left [\ln(x-e^{-a}) - \ln (x + e^{a} ) \right]_1^{\infty} \\
&= \frac1{2\cosh a} \left [ \ln \frac{x-e^{-a}}{x+e^{a}} \right]_1^{\infty} \\
&= \frac{1}{2\cosh a} \left ( 0 - \ln \frac{1-e^{-a}}{1+e^a}{}\right) \\
&= \frac{1}{2\cosh a} \ln \frac{1+e^a}{1-e^{-a}}\\
&= \frac{1}{2\cosh a} \left ( a + \ln \coth \frac{a}{2} \right)
\end{align*}
and
\begin{align*}
\int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x &= \int_0^\infty\frac{1}{(x^2+e^a)(x^2+e^{-a})} \d x \\
&= \int_0^\infty \frac{1}{e^a-e^{-a}} \left ( \frac{1}{x^2+e^{-a}} - \frac{1}{x^2+e^{a}} \right) \d x \\
&= \frac{1}{2\sinh a} \left [ \frac{1}{e^{-a/2}} \tan^{-1} \frac{x}{e^{-a/2}} - \frac{1}{e^{a/2}}\tan^{-1} \frac{x}{e^{a/2}} \right]_0^{\infty} \\
&= \frac{1}{2\sinh a} \left (e^{a/2}\frac{\pi}{2}-e^{-a/2}\frac{\pi}{2} - 0 \right) \\
&= \frac{1}{2\sinh a} \pi \sinh \frac{a}{2} \\
&= \frac{\pi \sinh \tfrac{a}{2}}{2\sinh a} \\
&= \frac{\pi \sinh \tfrac{a}{2}}{4\sinh \tfrac{a}{2} \cosh \tfrac{a}{2}} \\
&= \frac{\pi}{4\cosh \tfrac{a}{2}}
\end{align*}
\end{questionparts}
The most popular question, the scoring rate was very similar to the first. Quite a few candidates did not take the hint provided in part (i) to express cosh a in terms of exponentials in order to perform the integration. However, apart from those that did not correctly substantiate the given result, many handled the partial fractions and exponentials well, and quite a number dealt with the infinite limit impressively. Problems arose later in the question with manipulating logarithms and the instruction to express answers in terms of hyperbolic functions was either overlooked or beyond their capacity.