Year: 2016
Paper: 2
Question Number: 7
Course: LFM Pure
Section: Integration
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show that
\[
\int_0^a \f(x) \d x= \int _0^a \f(a-x) \d x\,,
\tag{$*$}
\]
where f is any function for which the integrals exist.
\begin{questionparts}
\item Use ($*$) to evaluate
\[
\int_0^{\frac12\pi} \frac{\sin x}{\cos x + \sin x} \, \d x
\,.
\]
\item Evaluate
\[
\int_0^{\frac14\pi} \frac{\sin x}{\cos x + \sin x} \, \d x
\,.
\]
\item Evaluate
\[
\int_0^{\frac14\pi} \ln (1+\tan x) \, \d x
\,.
\]
\item Evaluate
\[
\int_0^{\frac14 \pi}
\frac x {\cos x \, (\cos x + \sin x)}\, \d x
\,.
\]
\end{questionparts}\begin{align*}
u = a-x, \d u = - \d x: && \int_0^a f(x) \d x &= \int_{u=a}^{u=0} f(a-u) (-1) \d u \\
&&&= \int_0^a f(a-u) \d u \\
&&&= \int_0^a f(a-x) \d x
\end{align*}
\begin{questionparts}
\item \begin{align*}
&& I &= \int_0^{\frac12 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\
&&&= \int_0^{\frac12 \pi} \frac{\sin (\frac12 \pi - x)}{\cos (\frac12 \pi-x) + \sin (\frac12 \pi-x) } \d x\\
&&&= \int_0^{\frac12 \pi} \frac{\cos x}{\sin x + \cos x } \d x\\
\Rightarrow && 2I &= \int_0^{\frac12 \pi} 1 \d x \\
\Rightarrow && I &= \frac{\pi}{4}
\end{align*}
\item \begin{align*}
&& I &= \int_0^{\frac14 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\
&&&= \int_0^{\frac14 \pi} \frac{\sin (\frac14 \pi - x)}{\cos (\frac14 \pi-x) + \sin (\frac14 \pi-x) } \d x\\
&&&= \int_0^{\frac14 \pi} \frac{\frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}{\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x + \frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x} \d x \\
&&&= \int_0^{\frac14 \pi} \frac{\cos x - \sin x}{2 \cos x} \d x \\
&&&= \left [\frac12 x + \ln(\cos x) \right]_0^{\pi/4} \\
&&&= \frac{\pi}{8} -\frac12\ln2 - 1
\end{align*}
\item \begin{align*}
&& I &= \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \\
&&&= \int_0^{\frac14 \pi} \ln \left (1 + \tan \left(\frac{\pi}{4} - x\right) \right) \, \d x\\
&&&= \int_0^{\frac14 \pi} \ln \left (1 +\frac{1 - \tan x}{1+ \tan x} \right) \, \d x\\
&&&= \int_0^{\frac14 \pi} \ln \left (\frac{2}{1+ \tan x} \right) \, \d x\\
&&&= \frac{\pi}{4} \ln 2 - I \\
\Rightarrow && I &= \frac{\pi}{8} \ln 2
\end{align*}
\item \begin{align*}
&& I &= \int_0^{\frac14 \pi}
\frac x {\cos x \, (\cos x + \sin x)}\, \d x \\
&&&= \int_0^{\frac14 \pi}
\frac {\frac14 \pi - x} {(\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) \, (\frac{2}{\sqrt{2}}\cos x)}\, \d x \\
&&&= \int_0^{\frac14 \pi}
\frac {\frac14 \pi - x} {\cos x \, (\cos x + \sin x)}\, \d x \\ \\
\Rightarrow && I &= \frac{\pi}{8} \int_0^{\pi/4} \frac{\sec^2 x}{1 + \tan x} \d x\\
&&&= \frac{\pi}{8} \left [\ln (1 + \tan x) \right]_0^{\pi/4} \\
&&&= \frac{\pi}{8} \ln 2
\end{align*}
\end{questionparts}
This was the second most popular and one of the best-answered questions on the paper, with many candidates scoring very high marks. In many cases the initial result was explained clearly and then applied successfully to the first example. Candidates were generally able to follow through the calculations where they were able to see the way in which the result could be achieved, and so most of the attempts that followed a correct method only lost marks through occasional errors in calculation.