Year: 2014
Paper: 2
Question Number: 4
Course: LFM Pure
Section: Integration
There were good solutions presented to all of the questions, although there was generally less success in those questions that required explanations of results or the use of diagrams and graphs to reach the solution. Algebraic manipulation was generally well done by many of the candidates although a range of common errors such as confusing differentiation and integration and simple arithmetic slips were evident. Candidates should also be advised to use the methods that are asked for in questions unless it is clear that other methods will be accepted (such as by the use of the phrase "or otherwise").
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item
By using the substitution $u=1/x$,
show that for
$b>0$
\[
\int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1)} \d x =0 \,.
\]
\item
By using the substitution $u=1/x$,
show that for $b>0$,
\[
\int_{1/b}^b \frac{\arctan x}{x} \d x = \frac{\pi \ln b} 2\,.
\]
\item
By using the result $ \displaystyle \int_0^\infty \frac 1 {a^2+x^2} \d x = \frac {\pi}{2 a} $ (where $a > 0$),and a substitution of the form $u=k/x$, for suitable $k$, show that
\[
\int_0^\infty \frac 1 {(a^2+x^2)^2} \d x = \frac {\pi}{4a^3 }
\, \ \ \ \ \ \ (a > 0).
\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& I &= \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1} \d x \\
u = 1/x, \d u = -1/x^2 \d x: &&&= \int_{u=b}^{u=1/b} \frac{1/u \ln(1/u)}{(a^2+u^{-2})(a^2u^{-2}+1)} (- \frac{1}{u^2}) \d u \\
&&&= \int_{1/b}^b \frac{-u\ln u}{(a^2u^2+1)(a^2+u^2)} \d u \\
&&&= -I \\
\Rightarrow && I &= 0
\end{align*}
\item $\,$ \begin{align*}
&& I &= \int_{1/b}^b \frac{\arctan x}{x} \d x \\
u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=b}^{u=1/b} \frac{\arctan \frac1u}{\frac1u} \frac{-1}{u^2} \d u \\
&&&= \int_{1/b}^b \frac{\arctan \frac1u}{u} \d u \\
\Rightarrow && 2I &= \int_{1/b}^b \frac{\arctan x + \arctan \frac1x}{x} \d x \\
&&&= \int_{1/b}^b \frac{\frac{\pi}2}{x} \d x \\
&&&= \pi \ln b \\
\Rightarrow && I &= \frac{\pi}{2} \ln b
\end{align*}
\item $\,$ \begin{align*}
&& I_a &= \int_0^{\infty} \frac{1}{(a^2+x^2)^2} \d x \\
u = a/x, \d x = -a/u^2 \d u:&&&= \int_{u=0}^{u=\infty} \frac{1}{\left (a^2+\frac{a^2}{u^2} \right)^2} \frac{a}{u^2} \d u \\
&&&= \frac1{a^3}\int_0^{\infty} \frac{1}{(u+1/u)^2} \d u \\
&&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2}{(u^2+1)^2} \d u \\
&&&= \frac{1}{a^3} \int_0^{\infty} \frac{u^2+1-1}{(u^2+1)^2} \d u \\
&&&= \frac{1}{a^3} \int_0^{\infty} \frac{1}{(u^2+1)} - \frac{1}{(u^2+1)^2} \d u \\
&&&= \frac1{a^3} \frac{\pi}{2} - \frac{1}{a^3} I_1 \\
\Rightarrow && 2I_1 &= \frac{\pi}{2} \\
\Rightarrow && I_1 &= \frac{\pi}{4} \\
\Rightarrow && I_a &= \frac{\pi}{4a^3}
\end{align*}
\end{questionparts}
Many candidates were able to perform the given substitution correctly and then correctly explain how this demonstrates that the integral is equal to 1. The second part caused more difficulty, particularly with candidates not able to state the relationship between arctan(x) and arctan(1/x). Attempts to integrate with the substitution u = arctan(x) often resulted in an incorrect application of the chain rule when finding du. In the final part of the question many candidates attempted to use integration by parts to reach the given answer.