Year: 2015
Paper: 2
Question Number: 6
Course: LFM Pure
Section: Integration
As in previous years the Pure questions were the most popular of the paper with questions 1, 2 and 6 the most popular. The least popular questions on the paper were questions 8, 11 and 13 with fewer than 250 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: 1504.2
Banger Comparisons: 2
\begin{questionparts}
\item Show that
\[ \mathrm{sec}^2\left(\tfrac14\pi-\tfrac12 x\right)=\frac{2}{1+\sin x} \,.
\]
Hence integrate $\dfrac{1}{1+\sin x}$ with respect to $x$.
\item By means of the substitution $y=\pi -x$, show that
\[
\int_0^\pi x \f (\sin x)\, \d x = \frac \pi 2 \int_0^\pi \f(\sin x) \, \d x
,\]
where $\mathrm{f}$ is any function for which these integrals exist.
Hence evaluate
\[
\int_0^\pi \frac x {1+\sin x} \, \d x
\,.
\]
\item Evaluate
\[
\int_0^\pi\frac{ 2x^3 -3\pi x^2}{(1+\sin x)^2}\, \d x
.\]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& \sec^2\left(\tfrac14\pi-\tfrac12 x\right) &= \frac{1}{\cos^2 \left(\tfrac14\pi-\tfrac12 x\right)} \\
&&&= \frac{1}{\frac{1+\cos 2\left(\tfrac14\pi-\tfrac12 x\right)}{2}} \\
&&&= \frac{2}{1 + \cos \left(\tfrac12\pi- x\right)} \\
&&&= \frac{2}{1+\sin x} \\
\\
&& \int \frac{1}{1+\sin x} \d x &= \int \tfrac12\sec^2\left(\tfrac14\pi-\tfrac12 x\right) \d x\\
&&&= - \tan\left(\tfrac14\pi-\tfrac12 x\right) + C
\end{align*}
\item $\,$ \begin{align*}
&& I &= \int_0^{\pi} x f(\sin x) \d x \\
y = \pi - x, \d y = - \d x: &&&= \int_{y=\pi}^{y = 0} (\pi - y) f(\sin(\pi - y))(-1) \d y \\
&&&= \int_0^\pi (\pi - y) f(\sin y) \d y \\
&&&= \pi \int_0^\pi f(\sin y) \d y - I \\
\Rightarrow && I &= \frac{\pi}{2} \int_0^\pi f(\sin x) \d x \\
\\
\Rightarrow && \int_0^{\pi} \frac{x}{1 + \sin x} \d x &= \frac{\pi}{2} \int_0^{\pi} \frac{1}{1 + \sin x} \d x\\
&&&=\frac{\pi}{2} \left [- \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\
&&&= \frac{\pi}{2} \left (-\tan (-\tfrac{\pi}{4}) + \tan \tfrac{\pi}{4} \right) \\
&&&= \pi
\end{align*}
\item $\,$ \begin{align*}
&& J &= \int_0^{\pi} \frac{2x^3-3\pi x^2}{(1+\sin x)^2} \d x \\
y = \pi - x: &&&= \int_0^{\pi} \frac{2(\pi-y)^3-3\pi (\pi - y)^2}{(1+\sin x)^2 } \d y \\
&&&= \int_0^{\pi} \frac{-2 y^3 + 3 \pi y^2 - \pi^3}{(1+ \sin x)^2}\\
&&&= -\pi^3 \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x -J \\
\Rightarrow && J &= -\frac{\pi^3}{2} \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x\\
&&&= -\frac{\pi^3}{2} \int_0^\pi \tfrac14 \sec^4\left(\tfrac14\pi-\tfrac12 x\right) \d x \\
&&&= -\frac{\pi^3}{8} \int_0^\pi \sec^2\left(\tfrac14\pi-\tfrac12 x\right)\left (1 + \tan^2\left(\tfrac14\pi-\tfrac12 x\right) \right) \d x \\
&&&= -\frac{\pi^3}{8} \left [-\frac23 \tan^3\left(\tfrac14\pi-\tfrac12 x\right) - 2 \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\
&&&= -\frac{\pi^3}{8} \left (\frac43+4 \right) \\
&&&= -\frac{2\pi^3}{3}
\end{align*}
\end{questionparts}
This was the most popular question on the paper with over 1000 attempts made. The first section did not present significant difficulties to candidates and the integration was generally well completed, although occasionally with an error in the factor. The second part proved difficult for a number of candidates who failed to change the variable in the integral correctly, or in some cases did not change the variable in every position that it occurred. Other candidates did not apply a correct result for dealing with the trigonometric functions involved or did not clearly show how the required result was reached as the solution jumped through several steps to a statement of the result asked for in the question. There were very few successful attempts at the final part of the question, but they did include a variety of methods for evaluating the integral once the substitution had been made.