2007 Paper 2 Q3

Year: 2007
Paper: 2
Question Number: 3

Course: LFM Pure
Section: Integration

Difficulty: 1600.0 Banger: 1575.2

Problem

By writing \(x=a\tan\theta\), show that, for \(a\ne0\), $\displaystyle \int \frac 1 {a^2+x^2}\, \d x =\frac 1 a \arctan \frac x a + \text{constant}\,$.
  1. Let $\displaystyle I=\int_0^{\frac{1}{2}\pi} \frac {\cos x}{1+\sin^2 x} \, \d x\,$.
    1. Evaluate \(I\).
    2. Use the substitution \(t=\tan \frac12 x\) to show that \(\displaystyle \int_0^1 \frac {1-t^2}{1+6t^2+t^4} \, \d t = \tfrac12 I\,\).
  2. Evaluate \(\displaystyle \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \,\).

Solution

\begin{align*} && I &= \int \frac{1}{a^2+x^2} \d x\\ x = a \tan \theta, \d x =a \sec^2 \theta \d \theta &&&= \int \frac{1}{a^2+a^2\tan^2 x} a \sec^2 \theta \d \theta \\ &&&=\int \frac{\sec^2 \theta}{a \sec^2 \theta} \d \theta \\ &&&= \frac1a \theta + C \\ &&&= \frac1a \arctan \frac{x}{a} + C \end{align*}
    1. \(\,\) \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac {\cos x}{1+\sin^2 x} \d x \\ &&&= \left [ \arctan (\sin x) \right]_0^{\pi/2} \\ &&&= \arctan(1) - \arctan(0) = \frac{\pi}{4} \end{align*}
    2. \(\,\) \begin{align*} && t &= \tan \frac{x}{2} \\ \Rightarrow && \sin x &= \frac{2t}{1+t^2} \\ && \cos x &= \frac{1-t^2}{1+t^2} \\ && \d x &= \frac{2}{1+t^2} \d t \\ \Rightarrow && I &= \int_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x } \d x \\ &&&= \int_{t=0}^{t = 1} \frac{\frac{1-t^2}{1+t^2}}{1 + \left (\frac{2t}{1+t^2} \right)^2} \frac{2}{1+t^2} \d t \\ &&&= 2 \int_0^1 \frac{1-t^2}{(1+t^2)^2+(2t)^2} \d t\\ &&&= 2 \int_0^1 \frac{1-t^2}{1+6t^2+t^4} \d t\\ \end{align*} From which the conclusion follows
  1. \(\,\) \begin{align*} && J &= \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{1+14t^2+t^4}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{(t^2+1)^2+3(2t)^2}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \frac12\int_{x=0}^{x=\pi/2} \frac {\cos x}{1+3 \sin^2 x} \d x \\ &&&= \frac{1}{6}\left[ \sqrt{3} \arctan(\sin \sqrt{3}x)\right]_0^{\pi/2} \\ &&&= \frac16 \sqrt{3} \frac{\pi}{3} \\ &&&= \frac{\sqrt{3}\pi}{18} \end{align*}
Examiner's report
— 2007 STEP 2, Question 3
Most Popular Along with Q1, Q2 and Q6 by far the most popular; good source of marks

This was another popular question, and was usually a good source of marks for those candidates who attempted it. The first two parts were usually successfully completed. In part (i) (b), candidates had to employ the t = tan ½ x substitution which seems to have fallen into disuse in recent years (due to modularity!). Having said that, most candidates were able to make some progress and, where they did fall down, it was generally due to a lack of confidence in handling trigonometric identities. One of the advantages of these last two parts to the question was that they could be done in one of two directions, and many candidates were able to spot the connections and exploit them satisfactorily. When errors arose, they were frequently due to a lack of care with constants, and a correct final answer was not often to be found as a result.

Although the paper was by no means an easy one, it was generally found a more accessible paper than last year's, with most questions clearly offering candidates an attackable starting-point. The candidature represented the usual range of mathematical talents, with a pleasingly high number of truly outstanding students; many more who were able to demonstrate a thorough grasp of the material in at least three questions; and the few whose three-hour long experience was unlikely to have been a particularly pleasant one. However, even for these candidates, many were able to make some progress on at least two of the questions chosen. Really able candidates generally produced solid attempts at five or six questions, and quite a few produced outstanding efforts at up to eight questions. In general, it would be best if centres persuaded candidates not to spend valuable time needlessly in this way – it is a practice that is not to be encouraged, as it uses valuable examination time to little or no avail. Weaker brethren were often to be found scratching around at bits and pieces of several questions, with little of substance being produced on more than a couple. It is an important examination skill – now more so than ever, with most candidates now not having to employ such a skill on the modular papers which constitute the bulk of their examination experience – for candidates to spend a few minutes at some stage of the examination deciding upon their optimal selection of questions to attempt. As a rule, question 1 is intended to be accessible to all takers, with question 2 usually similarly constructed. In the event, at least one – and usually both – of these two questions were among candidates' chosen questions. These, along with questions 3 and 6, were by far the most popularly chosen questions to attempt. The majority of candidates only attempted questions in Section A (Pure Maths), and there were relatively few attempts at the Applied Maths questions in Sections B & C, with Mechanics proving the more popular of the two options. It struck me that, generally, the working produced on the scripts this year was rather better set-out, with a greater logical coherence to it, and this certainly helps the markers identify what each candidate thinks they are doing. Sadly, this general remark doesn't apply to the working produced on the Mechanics questions, such as they were. As last year, the presentation was usually appalling, with poorly labelled diagrams, often with forces missing from them altogether, and little or no attempt to state the principles that the candidates were attempting to apply.

Source: Cambridge STEP 2007 Examiner's Report · 2007-full.pdf
Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1575.2

Banger Comparisons: 9

Show LaTeX source
Problem source
By writing $x=a\tan\theta$, show that, for $a\ne0$, 
 $\displaystyle \int \frac 1 {a^2+x^2}\,
\d x =\frac 1 a \arctan \frac x a + \text{constant}\,$.
\begin{questionparts}
\item Let $\displaystyle I=\int_0^{\frac{1}{2}\pi} 
\frac {\cos x}{1+\sin^2 x} \, \d x\,$.
\begin{enumerate}
\item Evaluate $I$.
\item Use the substitution $t=\tan \frac12 x$ to show that $\displaystyle \int_0^1 \frac {1-t^2}{1+6t^2+t^4} \, \d t = \tfrac12 I\,$.
\end{enumerate}
\item  Evaluate $\displaystyle \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \,$.
\end{questionparts}
Solution source
\begin{align*}
&& I &= \int \frac{1}{a^2+x^2} \d x\\
x = a \tan \theta, \d x =a \sec^2 \theta \d \theta &&&= \int \frac{1}{a^2+a^2\tan^2 x} a \sec^2 \theta \d \theta \\
&&&=\int \frac{\sec^2 \theta}{a \sec^2 \theta}  \d \theta \\
&&&= \frac1a \theta + C \\
&&&= \frac1a \arctan \frac{x}{a} + C
\end{align*}

\begin{questionparts}
\item \begin{enumerate}
\item $\,$ \begin{align*}
&& I &= \int_0^{\frac{1}{2}\pi} 
\frac {\cos x}{1+\sin^2 x} \d x \\
&&&= \left [ \arctan (\sin x) \right]_0^{\pi/2} \\
&&&= \arctan(1) - \arctan(0) = \frac{\pi}{4}
\end{align*}
\item $\,$ \begin{align*}
&& t &= \tan \frac{x}{2} \\
\Rightarrow && \sin x &= \frac{2t}{1+t^2} \\
&& \cos x &= \frac{1-t^2}{1+t^2} \\
&& \d x &= \frac{2}{1+t^2} \d t \\
\Rightarrow && I &= \int_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x } \d x \\
&&&= \int_{t=0}^{t = 1} \frac{\frac{1-t^2}{1+t^2}}{1 + \left (\frac{2t}{1+t^2} \right)^2} \frac{2}{1+t^2} \d t \\
&&&= 2 \int_0^1 \frac{1-t^2}{(1+t^2)^2+(2t)^2} \d t\\
&&&= 2 \int_0^1 \frac{1-t^2}{1+6t^2+t^4} \d t\\
\end{align*}

From which the conclusion follows
\end{enumerate}

\item $\,$ \begin{align*}
&& J &= \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t  \\
&&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{1+14t^2+t^4}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t  \\
&&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{(t^2+1)^2+3(2t)^2}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t  \\
&&&= \frac12\int_{x=0}^{x=\pi/2} \frac {\cos x}{1+3 \sin^2 x} \d x  \\
&&&= \frac{1}{6}\left[ \sqrt{3} \arctan(\sin \sqrt{3}x)\right]_0^{\pi/2} \\
&&&= \frac16 \sqrt{3} \frac{\pi}{3} \\
&&&= \frac{\sqrt{3}\pi}{18}
\end{align*}

\end{questionparts}