Year: 2014
Paper: 3
Question Number: 2
Course: UFM Pure
Section: Integration using inverse trig and hyperbolic functions
A 10% increase in the number of candidates and the popularity of all questions ensured that all questions had a good number of attempts, though the first two questions were very much the most popular. Every question received at least one absolutely correct solution. In most cases when candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five sixths gave in at least six attempts.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1513.2
Banger Comparisons: 3
\begin{questionparts}
\item Show, by means of the substitution $u=\cosh x\,$, that
\[
\int \frac{\sinh x}{\cosh 2x} \d x
= \frac 1{2\sqrt2}
\ln \left\vert \frac{\sqrt2 \cosh x - 1}{\sqrt2 \cosh x + 1 } \right\vert
+ C
\,.\]
\item Use a similar substitution to find an expression for
\[
\int \frac{\cosh x}{\cosh 2x} \d x
\,.\]
\item Using parts (i) and (ii) above, show that
\[
\int_0^1 \frac 1{1+u^4} \d u = \frac{\pi + 2\ln(\sqrt2 +1)}{4\sqrt2}\,.
\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& \int \frac{\sinh x}{\cosh 2x} \d x &= \int \frac{\sinh x}{2\cosh^2 x -1} \d x \\
u = \cosh x, \d u = \sinh x \d x &&&= \int \frac{1}{2u^2 -1} \d u \\
&&&= \int\frac12 \left ( \frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1} \right) \d u \\
&&&= \frac1{2\sqrt{2}} \left (\ln (\sqrt{2}u-1) - \ln(\sqrt{2}u+1) \right) + C \\
&&&= \frac{1}{2\sqrt{2}} \ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) + C
\end{align*}
\item \begin{align*}
&& \int \frac{\cosh x}{\cosh 2x} \d x &= \int \frac{\cosh x}{1+2\sinh^2 x} \d x \\
u = \sinh x && &= \int \frac{1}{1+2u^2} \d u \\
&&&=\frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}u) + C \\
&&&= \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) + C
\end{align*}
\item \begin{align*}
u = e^x : && \int_0^1 \frac{1}{1+u^4} \d u &= \int_{x=-\infty}^{x=0} \frac{1}{1+e^{4x}}e^{x} \d x \\
&&&= \int_{-\infty}^{0} \frac{e^{-x}}{e^{2x}+e^{-2x}} \d x \\
&&&= \int_{-\infty}^{0} \frac{\cosh x - \sinh x}{2\cosh 2x } \d x \\
&&&= \frac12 \int_{-\infty}^{0} \frac{\cosh x}{\cosh 2x} \d x - \frac12 \int_{-\infty}^{0} \frac{\sinh x}{\cosh 2x} \\
&&&= \frac12 \left [\frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}\sinh x) \right]_{-\infty}^{0}-\frac12 \left [ \frac{1}{2\sqrt{2}}\ln \left ( \frac{\sqrt{2}\cosh x -1}{\sqrt{2}\cosh x +1} \right) \right]_{-\infty}^{0} \\
&&&= 0 - \frac1{2\sqrt{2}} \frac{-\pi}{2}-\left (\frac1{4\sqrt{2}} \ln \left (\frac{\sqrt{2}-1}{\sqrt{2}+1} \right) - 0 \right) \\
&&&= \frac{\pi - \ln((\sqrt{2}-1)^2)}{4\sqrt{2}} \\
&&&= \frac{\pi + 2 \ln(1+\sqrt{2})}{4\sqrt{2}}
\end{align*}
\end{questionparts}
This was only marginally less popular than question 1, but was the most successfully attempted with a mean of two thirds marks. Most that attempted the question were able to do the first two parts easily, but could not find a suitable substitution to do the last part. In about a tenth of the attempts, a helpful substitution was made in part (iii) which then usually resulted in successful completion of the question. Modulus signs were often ignored, or could not be distinguished from usual parentheses, and the arbitrary constant, even though it appeared in the result for part (i), was frequently overlooked. A few did not use the correct formulae for cosh 2x, instead resorting to the trigonometric versions. A handful of candidates attempted partial fractions in the last part having correctly factorised the quartic, but this did not use the previous parts as instructed.