Year: 2016
Paper: 3
Question Number: 1
Course: UFM Additional Further Pure
Section: Reduction Formulae
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let
\[
\displaystyle I_n= \int_{-\infty}^\infty \frac 1 {(x^2+2ax+b)^n} \, \d x
\]
where $a$ and $b$ are constants with $b > a^2$, and $n$ is a positive integer.
\begin{questionparts}
\item By using the substitution $x + a = \sqrt{b- a^2} \, \tan u\,$, or otherwise, show that
\[
I_1 = \dfrac \pi {\sqrt{b-a^2}}\, .
\]
\item Show that $2n(b - a^2)\, I_{n+1} =(2n - 1) \, I_n\,$.
\item Hence prove by induction that
\[
I_n =\frac{\pi}{2^{2n-2}( b - a^2)^{n-\frac12}} \, \binom {2n-2}{n-1}
\]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& I_1 &= \int_{-\infty}^{\infty} \frac{1}{x^2+2ax+b} \d x \\
&&&= \int_{-\infty}^{\infty} \frac{1}{b-a^2 +(x+a)^2} \d x \\
&&&= \left [ \frac{1}{\sqrt{b-a^2}} \tan^{-1} \frac{x+a}{\sqrt{b-a^2}} \right]_{-\infty}^{\infty} \\
&&&= \frac{\pi}{\sqrt{b-a^2}}
\end{align*}
\item $\,$ Here is the corrected LaTeX code for the second part, maintaining your exact styling and notation.
\item $\,$
\begin{align*}
&& I_{n} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+2ax+b)^{n}} \d x \\
&&&= \left[ \frac{x}{(x^2+2ax+b)^n} \right]_{-\infty}^{\infty} - \int_{-\infty}^\infty x \cdot \frac{-n(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x \\
&&&= 0 + n \int_{-\infty}^\infty \frac{2x^2+2ax}{(x^2+2ax+b)^{n+1}} \d x \\
&&&= n \int_{-\infty}^\infty \frac{2(x^2+2ax+b) - (2ax+2b)}{(x^2+2ax+b)^{n+1}} \d x \\
&&&= 2n I_n - n \int_{-\infty}^\infty \frac{2ax+2b}{(x^2+2ax+b)^{n+1}} \d x \\
&&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a) + 2(b-a^2)}{(x^2+2ax+b)^{n+1}} \d x \\
&&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x - 2n(b-a^2) I_{n+1} \\
&&&= 2n I_n - n \left[ \frac{-a}{n(x^2+2ax+b)^n} \right]_{-\infty}^\infty - 2n(b-a^2) I{n+1} \\
&&&= 2n I_n - 0 - 2n(b-a^2) I_{n+1} \\
\Rightarrow && 2n(b-a^2)I_{n+1} &= (2n-1)I_n
\end{align*}
\item $\,$ \begin{align*}
&& I_{n+1} &= \frac{2n-1}{2n(b-a^2)} I_n \\
&&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} I_{n-1} \\
&&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} \cdots I_{1} \\
&&&= \frac{(2n-1)(2n-3) \cdots 1}{2n \cdot 2(n-1) \cdots 2 (b-a^2)^n} \frac{\pi}{\sqrt{b-a^2}} \\
&&&= \frac{(2n-1)(2n-3) \cdots 1}{2^n n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\
&&&= \frac{(2n-1)(2n-3) \cdots 1 \cdot 2n \cdot 2(n-1) \cdots 2}{2^{2n} n!n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\
&&&= \frac{(2n)!}{2^{2n}n!n!}\frac{\pi}{(b-a^2)^{n+\frac12}} \\
&&&= \frac{\pi}{2^{2n}(b-a^2)^{n+\frac12}} \binom{2n}{n} \\
\Rightarrow && I_n &= \frac{\pi}{2^{2n-2}(b-a^2)^{n-\frac12}} \binom{2n-2}{n-1} \\
\end{align*}
\end{questionparts}
This was the most frequently attempted question with more than 93% of candidates attempting it. It was also successfully attempted, the second most in this respect and only by a small margin, earning two thirds marks on average. The majority of candidates completed parts (i) and (iii) with little trouble, with various algebraic mistakes occurring throughout, and a few candidates forgetting to substitute for the limits in part (i). Many got started on part (ii) using the substitution from part (i), and then get stuck faced with the consequent integral.