Year: 2013
Paper: 1
Question Number: 4
Course: LFM Pure
Section: Integration
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were good with candidates finding much to occupy them profitably during the three hours of the examination. In hindsight, two or three of the questions lacked sufficient 'punch' in their later parts, but at least most candidates showed sufficient skill to identify them and work on them as part of their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6 questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-attempts, most or all of which were fantastically accomplished mathematically, and such excellence is very heart-warming.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item
Show that, for $n> 0$,
\[
\int_0^{\frac14\pi} \tan^n x \,\sec^2 x \, \d x =
\frac 1 {n+1} \;
\text{ and }
\int_0^{\frac14\pi} \!\! \sec ^n\! x \, \tan x \,
\d x = \frac{(\sqrt 2)^n - 1}n \,.
\]
\item Evaluate the following integrals:
\[
\displaystyle
\int_0^{\frac14\pi}
\!\! x\, \sec ^4 \! x\, \tan x \, \d x \,
\text{ and }
\int_0^{\frac14\pi}
\!\! x^2 \sec ^2 \! x\, \tan x \, \d x \,.
\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
u = \tan x, \d u = \sec^2 x \d x: &&\int_0^{\pi/4} \tan^n x \sec^2 x \d x &= \int_0^1 u^n \d u \\
&&&= \frac{1}{n+1}
\end{align*}
\begin{align*}
u = \sec x, \d u = \sec x \tan x \d x: &&\int_0^{\pi/4} \sec^n x \tan x \d x &= \int_{u=1}^{u=\sqrt{2}} u^{n-1} \d u \\
&&&= \left [ \frac{u^n}{n}\right] \\
&&&= \frac{(\sqrt{2})^n - 1}n
\end{align*}
\item \begin{align*}
&&\int_0^{\frac14\pi}
x \sec ^4 x \tan x \d x &= \left [x \frac{1}{4} \sec^4 x \right]_0^{\frac14\pi} - \frac14 \int_0^{\frac14\pi} \sec^4 x \d x \\
&&&= \frac{\pi}{4} - \frac14 \int_0^{\frac14\pi} \sec^2 x(1+ \tan^2 x) \d x \\
&&&= \frac{\pi}{4} - \frac14 \left [ \tan x+ \frac13 \tan^3 x \right] _0^{\frac14\pi} \\
&&&= \frac{\pi}{4} - \frac{1}{3}
\end{align*}
\begin{align*}
\int_0^{\frac14\pi}
\!\! x^2 \sec ^2 \! x\, \tan x \, \d x &= \left [x^2 \frac12 \tan^2 x \right]_0^{\frac14\pi} - \int_0^{\frac14\pi} x \tan^2 x \d x\\
&= \frac{\pi^2}{32} - \int_0^{\frac14\pi} x (\sec^2 x - 1) \d x\\
&= \frac{\pi^2}{32} - \left [x (-x + \tan x) \right]_0^{\frac14\pi} + \int_0^{\frac14\pi}-x + \tan x \d x \\
&= \frac{\pi^2}{32} - \frac{\pi}{4} (-\frac{\pi}{4} + 1) -\frac{\pi^2}{32} + \left [ -\ln \cos x \right]_0^{\pi/4} \\
&= \frac{\pi^2}{16}- \frac{\pi}{4} + \frac12 \ln 2
\end{align*}
\end{questionparts}
This was another very popular and high-scoring question (attracting over 1200 attempts and with a mean mark of more than 10/20). The first part to this question involved two integrals which can be integrated immediately by "recognition", although many students took a lot of time and trouble to establish the given results by substitution and surprising amounts of working. Those candidates who had found these easy introductory parts especially troubling usually did not proceed far, if at all, into part (ii). Those who did venture further usually picked up quite a lot of marks. One of the great advantages to continued progress in the question is that the two integrals in part (ii) can be approached in so many different ways – the examiners worked out more than 25 slightly different approaches, depending upon how, and when, one used the identity sec2x = 1 + tan2x; how one split the "parts" in the process of "integrating by parts"; and even whether one approached the various secondary integrals that arose as a function of sec x or tan x. This meant that, with care, most of the marks were accessible, although many candidates clearly got into a considerable tangle at some stage of proceedings. The most common "howler" was the mix-up between the definite integrals (i.e. numerical values) given in (i) and their associated unevaluated indefinite integrals (i.e. functions) which formed part of a subsequent integral.