| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show definite integral equals specific value (requiring partial fractions or complex algebra) |
| Difficulty | Challenging +1.2 This is a guided substitution question where the substitution is provided and students must execute it systematically. While it requires careful algebraic manipulation (using sin 2x = 2sin x cos x, finding du/dx, expressing cos x in terms of u) and partial fractions or algebraic simplification, the path is clearly signposted. The multi-step nature and need to handle both algebraic and logarithmic terms elevates it slightly above average, but it remains a standard P4/Further Maths technique question without requiring novel insight. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.06f Laws of logarithms: addition, subtraction, power rules1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = 3 + 4\sin x \Rightarrow \frac{du}{dx} = 4\cos x\) | B1 | States or uses \(\frac{du}{dx} = 4\cos x\). May be seen within the integrand |
| \(\int \frac{16\sin 2x}{(3+4\sin x)^2}\,dx = \int \frac{32\sin x\cos x}{(3+4\sin x)^2}\,dx = \int \frac{2(u-3)}{u^2}\,du\) | M1 A1 | M1: Attempts to write all in terms of \(u\). Look for \(\int \frac{\ldots\sin x\cos x}{(3+4\sin x)^2}\,dx = \pm\int \frac{\ldots(u\pm3)}{u^2}\,du\). Common error: losing the square on \(u^2\) would be M0 |
| \(= \int \frac{2}{u} - \frac{6}{u^2}\,du = 2\ln u + \frac{6}{u}\) | dM1 A1 | dM1: Integrates \(\int \frac{\ldots(u\pm3)}{u^2}\,du\) to \(\ldots\ln u \pm \frac{\ldots}{u}\). A1: Correct \(2\ln u + \frac{6}{u}\). Note \(\ln u^2 + \frac{6}{u}\) also correct |
| Uses limits 5 and 7: \(\Rightarrow 2\ln 7 + \frac{6}{7} - 2\ln 5 - \frac{6}{5} = -\frac{12}{35} + \ln\frac{49}{25}\) | M1 A1 | M1: Uses limits 5 and 7 within attempted integral and subtracts. Condone poor attempts at integration, or converting back to \(x\) using correct substitution. A1: \(-\frac{12}{35} + \ln\frac{49}{25}\) or equivalent such as \(\ln 1.96 - \frac{12}{35}\). ISW after a correct answer |
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = 3 + 4\sin x \Rightarrow \frac{du}{dx} = 4\cos x$ | B1 | States or uses $\frac{du}{dx} = 4\cos x$. May be seen within the integrand |
| $\int \frac{16\sin 2x}{(3+4\sin x)^2}\,dx = \int \frac{32\sin x\cos x}{(3+4\sin x)^2}\,dx = \int \frac{2(u-3)}{u^2}\,du$ | M1 A1 | M1: Attempts to write all in terms of $u$. Look for $\int \frac{\ldots\sin x\cos x}{(3+4\sin x)^2}\,dx = \pm\int \frac{\ldots(u\pm3)}{u^2}\,du$. Common error: losing the square on $u^2$ would be M0 |
| $= \int \frac{2}{u} - \frac{6}{u^2}\,du = 2\ln u + \frac{6}{u}$ | dM1 A1 | dM1: Integrates $\int \frac{\ldots(u\pm3)}{u^2}\,du$ to $\ldots\ln u \pm \frac{\ldots}{u}$. A1: Correct $2\ln u + \frac{6}{u}$. Note $\ln u^2 + \frac{6}{u}$ also correct |
| Uses limits 5 and 7: $\Rightarrow 2\ln 7 + \frac{6}{7} - 2\ln 5 - \frac{6}{5} = -\frac{12}{35} + \ln\frac{49}{25}$ | M1 A1 | M1: Uses limits 5 and 7 within attempted integral and subtracts. Condone poor attempts at integration, or converting back to $x$ using correct substitution. A1: $-\frac{12}{35} + \ln\frac{49}{25}$ or equivalent such as $\ln 1.96 - \frac{12}{35}$. **ISW after a correct answer** |
---6. In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08756c4b-6619-42da-ac8a-2bf065c01de8-18_650_938_413_504}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with equation
$$y = \frac { 16 \sin 2 x } { ( 3 + 4 \sin x ) ^ { 2 } } \quad 0 \leqslant x \leqslant \frac { \pi } { 2 }$$
The region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis and the line with equation $x = \frac { \pi } { 6 }$
Using the substitution $u = 3 + 4 \sin x$, show that the area of $R$ can be written in the form $a + \ln b$, where $a$ and $b$ are rational constants to be found.
\hfill \mbox{\textit{Edexcel P4 2021 Q6 [7]}}