Year: 2012
Paper: 2
Question Number: 3
Course: LFM Pure
Section: Integration
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show that, for any function f (for which the integrals exist),
\[
\int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x = \frac12 \int_1^\infty
\left(1+\frac 1 {t^2}\right) \f(t)\, \d t \,.
\]
Hence evaluate
\[
\int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \, \, \d x
\,,
\]
and, using the substitution $x=\tan\theta$,
\[
\int_0^{\frac12\pi} \frac{1}{(1+\sin\theta)^3}\,\d \theta
\,.
\]\begin{align*}
&& t &= x + \sqrt{1+x^2} \\
&&\frac1t &= \frac{1}{x+\sqrt{1+x^2}} \\
&&&= \frac{\sqrt{1+x^2}-x}{1+x^2-1} \\
&&&= \sqrt{1+x^2}-x \\
\Rightarrow && x &=\frac12 \left ( t - \frac1t\right) \\
\Rightarrow && \d x &=\frac12 \left (1 + \frac1{t^2} \right)\d t \\
\\
\Rightarrow && \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x &= \int_{t=1}^{t = \infty}f(t) \frac12\left (1 + \frac1{t^2} \right)\d t \\
&&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right)f(t) \d t
\end{align*}
\begin{align*}
&& I &= \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \d x \\
&&&= \int_0^\infty \frac1 {(x+\sqrt{x^2+1})^2} \d x \\
&&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right) \frac{1}{t^2} \d t \\
&&&= \frac12 \left [-\frac1t-\frac13\frac1{t^3} \right]_1^{\infty} \\
&&&= \frac12 \cdot \frac43 = \frac23
\end{align*}
\begin{align*}
&& J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\
&&x &= \tan \theta\\
&& \d x &= \sec^2 \theta = (1+x^2) \d \theta\\
&& \tan\theta &= \frac{s}{\sqrt{1-s^2}}\\
\Rightarrow && \tan^2 \theta &= \frac{s^2}{1-s^2} \\
\Rightarrow && \sin \theta &= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\
&& J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\
&&&= \int_0^{\frac12 \pi} \frac{1}{\left (1+ \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \right )^3} \d \theta \\
&&&= \int_{x=0}^{x=\infty} \frac{1}{\left(1 + \frac{x}{\sqrt{1+x^2}} \right)^3} \frac{1}{1+x^2} \d x \\
&&&= \int_0^{\infty} \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2}+x)^3} \d x \\
&&J_a &= \int_0^{\infty} \frac{\sqrt{1+x^2}+x}{(\sqrt{1+x^2}+x)^3} \d x \\
&&&= \frac23 \\
&&J_b &= \int_0^{\infty} \frac{\sqrt{1+x^2}-x}{(\sqrt{1+x^2}+x)^3} \d x \\
&&&= \int_0^{\infty} \frac{1}{(\sqrt{1+x^2}+x)^4} \d x\\
&&&= \frac12\int_1^{\infty} \left (1 +\frac1{t^2} \right)\frac{1}{t^4} \d t \\
&&&= \frac12 \left [-\frac13 t^{-3}-\frac15t^{-5} \right]_1^{\infty} \\
&&&= \frac12 \cdot \frac8{15} = \frac4{15} \\
\Rightarrow && J &= \frac12(J_a+J_b) = \frac7{15}
\end{align*}
This was another popular question, scoring just over half of the marks on average. Although most efforts to establish the given initial result were eventually successful, many made hard work of it, failing to notice the obvious result that if t = x² + 1 − x then 1/t = x² + 1 + x. The first integral could then be found by realising that f(t) = 1/t² was the relevant function here, or by repeating the substitution already used. Quite a few candidates thought that the second integral followed from the first, which was unfortunate, as it didn't. However, most efforts at this second integral were unsuccessful anyhow, with candidates usually getting 3 of the 10 marks for setting up the substitution and then often going round in circles. The main problem lay in using sin and cos instead of tan and sec, or in continuing with x² + 1 − x without identifying a suitable function f(t). It was helpful to find this, but not essential.