Year: 2013
Paper: 3
Question Number: 1
Course: LFM Pure
Section: Integration
With the number of candidates submitting scripts up by some 8% from last year, and whilst inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a lesser extent, all questions on the paper were attempted by a significant number of candidates. About a sixth of candidates gave in answers to more than six questions, but the extra questions were invariably scoring negligible marks. Two fifths of the candidates gave in answers to six questions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Given that $t= \tan \frac12 x$, show that
$\dfrac {\d t}{\d x} = \frac12(1+t^2)$ and $ \sin x = \dfrac {2t}{1+t^2}\,$.
Hence show that
\[
\int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x =
\frac2 {\sqrt{1-a^2}} \arctan \frac{\sqrt{1-a}}{\sqrt{1+a}}\,
\qquad \quad (0 < a < 1).
\]
Let
\[
I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x
\qquad \quad (n\ge0).
\]
By considering $I_{n+1}+2I_{n}\,$, or otherwise, evaluate $I_3$.Let $t = \tan \frac12 x$, then
\begin{align*}
\frac{\d t}{\d x} &= \tfrac12 \sec^2 \tfrac12 t \\
&= \tfrac12 (1 + \tan^2 \tfrac12 ) \\
&= \tfrac12 (1 + t^2) \\
\\
\sin x &= 2 \sin \tfrac12 x \cos \tfrac12 \\
&= \frac{2 \frac{\sin \tfrac12 x}{ \cos \tfrac12x}}{\frac{1}{\cos^2 \tfrac12 x}} \\
&= \frac{2 \tan \tfrac12 x}{\sec^2 \tfrac12 } \\
&= \frac{2t }{1+t^2}
\end{align*}
Now consider
\begin{align*}
t = \tan \tfrac12 x: && \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x &= \int_{t=0}^{t = 1} \frac{1}{1 + a \frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\
&&&= \int_0^1 \frac{2}{1+2at+t^2} \d t \\
&&&= \int_0^1 \frac{2}{(t+a)^2 + 1-a^2} \d t \\
(1-a^2) > 0: &&&= \left [ \frac{2}{\sqrt{1-a^2}} \arctan \frac{t+a}{\sqrt{1-a^2}} \right]_0^1 \\
&&&= \frac{2}{\sqrt{1-a^2}} \left ( \arctan \frac{1+a}{\sqrt{1-a^2}} - \arctan \frac{a}{\sqrt{1-a^2}} \right) \\
&&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1+a}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{1+a}{\sqrt{1-a^2}}\frac{a}{\sqrt{1-a^2}}} \right) \\
&&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\
&&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\sqrt{1-a}}{\sqrt{1+a}} \right)
\end{align*}
as required.
Let
\[
I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x
\qquad \quad (n\ge0).
\]
and consider
\begin{align*}
I_{n+1} + 2I_n &= \int_0^{\frac12\pi} \frac{ \sin ^{n+1}x+2\sin^{n} x}{2+\sin x} \, \d x \\
&= \int_0^{\frac12\pi} \frac{ \sin^n x (2 + \sin x)}{2+\sin x} \, \d x \\
&= \int_0^{\frac12\pi} \sin^n x \d x
\end{align*}
Therefore we can compute
\begin{align*}
I_0 &= \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\
&= \frac12 \int_0^{\pi/2} \frac{1}{1 + \frac12 \sin x} \d x \\
&= \frac{1}{\sqrt{3/4}} \arctan \frac{\sqrt{1/2}}{\sqrt{3/2}} \\
&= \frac{2}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}} \\
&= \frac{\pi}{3\sqrt{3}} \\
\\
I_1 &= \int_0^{\pi/2} 1 \d x - 2 I_0 \\
&= \frac{\pi}{2} - \frac{2\pi}{3\sqrt{3}} \\
I_2 &= \int_0^{\pi/2} \sin x \d x - 2I_1 \\
&= 1 - \pi + \frac{4\pi}{3\sqrt{3}} \\
I_3 &= \int_0^{\pi/2} \sin^2 x \d x - 2I_2 \\
&= \frac12 \int_0^{\pi/2} \sin^2 + \cos^2 x \d x - 2I_2 \\
&= \frac{\pi}{4} - 2 + 2\pi - \frac{8\pi}{3\sqrt{3}} \\
&= -2 + \frac{9\pi}{4} - \frac{8\pi}{3\sqrt{3}}
\end{align*}
Most candidates attempted this question, making it the most popular and it was also the most successful with a mean score of about two thirds marks. The first two standard results caused few problems, nor did the integration, but some struggled to simplify to the single inverse tan form. In the final part, common errors were failure to reduce to the θ = 0 case, confusion with the index e.g. instead of the correct result, or for those that were more successful, algebraic inaccuracies let them down. Some attempted a recursive formula to evaluate with varying success. Most attempting the last part saw the connection between and the main result of the question.