| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Trigonometric substitution: show transformation then evaluate |
| Difficulty | Standard +0.8 This is a structured P4/Further Maths integration question requiring trigonometric substitution with multiple algebraic manipulations. While the substitution is given, students must correctly handle dx transformation, simplify the denominator using trig identities (cos³u), change limits, and integrate sec·tan and sec² functions. The multi-step nature and requirement to work with compound trig expressions places it moderately above average difficulty. |
| Spec | 1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 = 2\sin u \Rightarrow p = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\) | B1 | Allow if seen anywhere, even in (b). \(p=30\) is B0 |
| \(x = 2\sin u \Rightarrow \frac{dx}{du} = 2\cos u\) | M1 | Or any rearrangement; \(\frac{dx}{du} = \pm...\cos u\) |
| Full substitution giving \(\int \frac{6\sin u + 2}{(4\cos^2 u)^{\frac{3}{2}}} \cdot 2\cos u \, du\) | M1 | Full substitution using \(\sin^2 u + \cos^2 u = 1\); do not concern with limits |
| \(= \int_0^{\frac{\pi}{6}} \left(\frac{3}{2}\sec u \tan u + \frac{1}{2}\sec^2 u\right) du\) | A1* | Achieves given answer with no errors and at least one intermediate step; condone missing \(du\) in intermediate lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\left(\frac{3}{2}\sec u \tan u + \frac{1}{2}\sec^2 u\right)du = \frac{3}{2}\sec u + \frac{1}{2}\tan u\) | M1A1 | M1: \(= ...\sec u + ...\tan u\); A1: \(\frac{3}{2}\sec u + \frac{1}{2}\tan u\), ignore constant \(c\) |
| \(\left[\frac{3}{2}\sec u + \frac{1}{2}\tan u\right]_0^{\frac{\pi}{6}}\) evaluated by substituting \(p \neq 1\) and \(0\), subtracting | M1 | Substitution must be seen or clearly implied by correct values for each term |
| \(= \sqrt{3} + \frac{\sqrt{3}}{6} - \frac{3}{2} = \frac{7\sqrt{3}}{6} - \frac{3}{2} \left(= \frac{7\sqrt{3}-9}{6}\right)\) | A1 | Or exact equivalent e.g. \(\frac{7\sqrt{3}-9}{6}\); allow if \(p=30°\) was used |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 = 2\sin u \Rightarrow p = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ | B1 | Allow if seen anywhere, even in (b). $p=30$ is B0 |
| $x = 2\sin u \Rightarrow \frac{dx}{du} = 2\cos u$ | M1 | Or any rearrangement; $\frac{dx}{du} = \pm...\cos u$ |
| Full substitution giving $\int \frac{6\sin u + 2}{(4\cos^2 u)^{\frac{3}{2}}} \cdot 2\cos u \, du$ | M1 | Full substitution using $\sin^2 u + \cos^2 u = 1$; do not concern with limits |
| $= \int_0^{\frac{\pi}{6}} \left(\frac{3}{2}\sec u \tan u + \frac{1}{2}\sec^2 u\right) du$ | A1* | Achieves given answer with no errors and at least one intermediate step; condone missing $du$ in intermediate lines |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\left(\frac{3}{2}\sec u \tan u + \frac{1}{2}\sec^2 u\right)du = \frac{3}{2}\sec u + \frac{1}{2}\tan u$ | M1A1 | M1: $= ...\sec u + ...\tan u$; A1: $\frac{3}{2}\sec u + \frac{1}{2}\tan u$, ignore constant $c$ |
| $\left[\frac{3}{2}\sec u + \frac{1}{2}\tan u\right]_0^{\frac{\pi}{6}}$ evaluated by substituting $p \neq 1$ and $0$, subtracting | M1 | Substitution must be seen or clearly implied by correct values for each term |
| $= \sqrt{3} + \frac{\sqrt{3}}{6} - \frac{3}{2} = \frac{7\sqrt{3}}{6} - \frac{3}{2} \left(= \frac{7\sqrt{3}-9}{6}\right)$ | A1 | Or exact equivalent e.g. $\frac{7\sqrt{3}-9}{6}$; allow if $p=30°$ was used |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying on calculator technology are not acceptable.}
(a) Use the substitution $x = 2 \sin u$ to show that
$$\int _ { 0 } ^ { 1 } \frac { 3 x + 2 } { \left( 4 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } d x = \int _ { 0 } ^ { p } \left( \frac { 3 } { 2 } \operatorname { secutanu } + \frac { 1 } { 2 } \sec ^ { 2 } u \right) d u$$
where $p$ is a constant to be found.\\
(b) Hence find the exact value of
$$\int _ { 0 } ^ { 1 } \frac { 3 x + 2 } { \left( 4 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } d x$$
\hfill \mbox{\textit{Edexcel P4 2022 Q5 [8]}}