Year: 2011
Paper: 3
Question Number: 6
Course: UFM Pure
Section: Integration using inverse trig and hyperbolic functions
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
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Banger Rating: 1536.7
Banger Comparisons: 7
The definite integrals $T$, $U$, $V$ and $X$ are defined by
\begin{align*}
T&= \int_{\frac13}^{\frac12} \frac{{\rm artanh}\, t}t \,\d t\,, &
U&= \int _{\ln 2 }^{\ln 3 } \frac{u}{2\sinh u}\, \d u \,, \\[3mm]
V&= - \int_{\frac13}^{\frac12} \frac{\ln v}{1-v^2} \,\d v \,, &
X&= \int _{\frac12\ln2}^{\frac12\ln3} \ln ({\coth x})\, \d x\,.
\end{align*}
Show, without evaluating any of them, that $T$, $U$, $V$ and $X$ are
all equal.\begin{align*}
&& T &= \int_{\frac13}^{\frac12} \frac{{\rm artanh}\, t}t \,\d t \\
&& &=\int_{\frac13}^{\frac12} \frac{1}{2t}\ln \left ( \frac{1+t}{1-t} \right) \,\d t \\
u = \tfrac{1+t}{1-t}, t= \tfrac{u-1}{u+1}, \d t = \tfrac{2}{(u+1)^2} \d t &&&= \int_{u=2}^{u=3} \frac{1}{2t} \ln u \frac{2}{(u+1)^2} \d u \\
&&&= \int_2^3 \frac{u+1}{u-1} \ln u \frac{1}{(u+1)^2} \d u \\
&&&= \int_2^3 \frac{1}{u^2-1} \ln u \d u
\end{align*}
\begin{align*}
&& U&= \int _{\ln 2 }^{\ln 3 } \frac{u}{2\sinh u}\, \d u \\
v = e^u, \d v = e^u \d u &&&= \int_{v=2}^{v=3} \frac{\ln v}{v - \frac{1}{v}} \frac{1}{v} \d v \\
&&&= \int_2^3 \frac{1}{v^2-1} \ln v \d v
\end{align*}
\begin{align*}
&&V &= - \int_{\frac13}^{\frac12} \frac{\ln v}{1-v^2} \,\d v \\
u = \tfrac1v, \d u = -\tfrac1{v^2} \d v &&&= -\int_{u=3}^{u=2} \frac{-\ln u}{1 - \frac{1}{u^2}} \frac{-1}{u^2} \d u \\
&&&= -\int_3^2 \frac{\ln u}{u^2-1} \d u \\
&&&= \int_2^3 \frac{1}{u^2-1} \ln u \d u
\end{align*}
\begin{align*}
&&X&= \int _{\frac12\ln2}^{\frac12\ln3} \ln ({\coth x})\, \d x \\
u = \coth x, \d u =(1-u^2) \d x &&&= \int_{u = 3}^{u=2} \ln u \frac{1}{1-u^2} \d u \\
&&&= \int_2^3 \frac{\ln u}{u^2-1} \d u
\end{align*}
Therefore all integrals are equal to the same integral, namely $\displaystyle \int_2^3 \frac{\ln u}{u^2-1} \d u$
This was quite popular, with attempts from three quarters of the candidates, and slightly more success than questions like 2 and 3. Needing to prove three equalities, many got close to doing two well and, with the others splitting half and half between getting close to all three or just one. A small number of candidates made several attempts without always having any sense of direction and often proved a particular pair equal both ways round. The other weaknesses were in dealing with the limits when changing variable and evaluating the definite term (which was zero!) when employing integration by parts.