Year: 2017
Paper: 1
Question Number: 1
Course: LFM Pure
Section: Integration
The pure questions were again the most popular of the paper with questions 1 and 3 being attempted by almost all candidates. The least popular questions on the paper were questions 10, 11, 12 and 13 and a significant proportion of attempts at these were brief, attracting few or no marks. Candidates generally demonstrated a high level of competence when completing the standard processes and there were many good attempts made when questions required explanations to be given, particularly within the pure questions. A common feature of the stronger responses to questions was the inclusion of diagrams.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item Use the substitution $u= x\sin x +\cos x$ to find
\[
\int \frac{x }{x\tan x +1 } \, \d x
\,.
\]
Find by means of a similar substitution, or otherwise,
\[
\int \frac{x }{x\cot x -1 } \, \d x
\,.
\]
\item Use a substitution to find
\[
\int \frac{x\sec^2 x \, \tan x}{x\sec^2 x -\tan x} \,\d x
\,
\]
and
\[
\int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \d x
\,.
\]
\end{questionparts}\begin{questionparts}
\item $\,$
\begin{align*}
&& I &= \int \frac{x}{x \tan x + 1} \d x \\
&&&= \int \frac{x \cos x}{x \sin x + \cos x} \d x \\
u = x \sin x + \cos x , \d u = x \cos x \d x: &&&= \int \frac{\d u}{u} \\
&&&= \ln u + C \\
&&&= \ln (x \sin x + \cos x) + C \\
\\
&& J &= \int \frac{x}{x \cot x - 1} \d x \\
&&&= \int \frac{x \sin x }{x \cos x - \sin x} \d x \\
u = x \cos x - \sin x, \d u = x \sin x \d x: &&&= \int \frac{1}{u} \d u \\
&&&= \ln u + K \\
&&&= \ln (x \cos x -\sin x) + K
\end{align*}
\item $\,$
\begin{align*}
&& I &= \int \frac{x \sec^2 x \tan x}{x \sec^2 x - \tan x} \d x \\
u = x\sec^2 x-\tan x, \d u = 2x \sec^2 x \tan x&&&= \frac12 \int \frac{1}{u} \d u \\
&&&= \frac12 \ln (x \sec^2 x - \tan x) + C \\
\\
&& J &= \int \frac{x \sin x \cos x}{(x - \sin x \cos x)^2} \d x \\
u = x \sec^2 x -\tan x, \d u=2x \frac{\sin x}{\cos^3 x} &&&= \int \frac{x \sin x \cos x}{\cos^4x(x\sec^2 x -\tan x)^2} \d x \\
&&&= \frac12 \int \frac{1}{u^2} \d u \\
&&&= -\frac12u^{-1} + K \\
&&&= \frac{1}{2(\tan x - x \sec^2 x)} + K
\end{align*}
\end{questionparts}
This was a very popular question, attempted by 94% of the candidates. The substitution was often correctly made for the first part and a large number of candidates were able to identify the similar substitution required for the second integration. The substitution for the first integration in the second part of the question caused some difficulties, but was again completed successfully by many candidates; however it sometimes required a number of attempts before the correct substitution was completed. A small proportion of the candidates were then able to complete the final integration successfully.