| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show integral transforms via substitution then evaluate (trigonometric/Weierstrass) |
| Difficulty | Challenging +1.2 This is a standard Further Maths FP1 integration question using the Weierstrass substitution (t = tan(θ/2)) with well-known formulas. Part (a) requires systematic algebraic manipulation to complete the square, and part (b) involves a standard arctangent integral with definite limits. While it requires multiple steps and careful algebra, the techniques are routine for FP1 students with no novel problem-solving insight needed. The 'show that' format provides clear targets, making it slightly above average difficulty but not challenging by Further Maths standards. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.06f Laws of logarithms: addition, subtraction, power rules4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{2\sin\theta + \cos\theta + 2}\,\mathrm{d}\theta = \int \frac{1}{2\left(\frac{2t}{1+t^2}\right)+\left(\frac{1-t^2}{1+t^2}\right)+2} \times \frac{2}{1+t^2}\,\mathrm{d}t\) | M1 | Applies correct substitutions \(\sin\theta = \frac{2t}{1+t^2}\), \(\cos\theta = \frac{1-t^2}{1+t^2}\), \(\mathrm{d}\theta = \frac{2}{1+t^2}\,\mathrm{d}t\). The \(\mathrm{d}t\) may be missing. |
| \(\int \frac{2}{4t+(1-t^2)+2(1+t^2)}\,\mathrm{d}t = \int \frac{2}{t^2+4t+3}\,\mathrm{d}t\) | M1 | Simplifies to form \(\frac{a}{pt^2+qt+r}\), allowing slips in coefficients but "\(1+t^2\)" must be correctly handled. |
| \(\int \frac{2}{(t+2)^2-1}\,\mathrm{d}t\) | A1 | Correct answer including the \(\int\,\mathrm{d}t\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{a}{(t+b)^2-c}\,\mathrm{d}t = \frac{a}{2\sqrt{c}}\ln\left\lvert\frac{(t+b)-\sqrt{c}}{(t+b)+\sqrt{c}}\right\rvert, c>0\) giving \(\int \frac{2}{(t+2)^2-1}\,\mathrm{d}t = \frac{2}{2(1)}\ln\left\lvert\frac{(t+2)-1}{(t+2)+1}\right\rvert\) | M1 | Correct method for integration by standard formula. Modulus signs not necessary. |
| Correctly uses limits \(t=1\) and \(t=\sqrt{3}\): \(\ln\left\lvert\frac{\sqrt{3}+1}{\sqrt{3}+3}\right\rvert - \ln\left\lvert\frac{2}{4}\right\rvert\) | M1 | Correctly uses limits \(t=1\) and \(t=\sqrt{3}\) to produce exact expression. |
| \(= \ln\left(\frac{\sqrt{3}+1}{\sqrt{3}+3} \div \frac{1}{2}\right)\) | M1 | Correctly combines ln terms with at least one step showing substitution of limits before final answer. |
| \(= \ln\left(2 \times \frac{\sqrt{3}+1}{\sqrt{3}+3} \times \frac{\sqrt{3}-3}{\sqrt{3}-3}\right) = \ln\left(\frac{2\sqrt{3}}{3}\right)\)* | A1\* | Shows rationalisation of denominator and completes to given answer. Note \(\sqrt{3}+3 = \sqrt{3}(1+\sqrt{3})\) may be used. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{2}{(t+2)^2-1}\,\mathrm{d}t = \int \frac{2}{(t+1)(t+3)}\,\mathrm{d}t = \int\frac{1}{t+1}-\frac{1}{t+3}\,\mathrm{d}t = \ln | t+1 | -\ln |
| \(\ln(\sqrt{3}+1)-\ln(\sqrt{3}+3)-\ln 2+\ln 4\) | M1 | Correctly uses limits \(t=1\) and \(t=\sqrt{3}\). |
| \(= \ln\left(2\times\frac{\sqrt{3}+1}{\sqrt{3}+3}\right)\) | M1 | Correctly combines ln terms. |
| \(= \ln\left(2\times\frac{\sqrt{3}+1}{\sqrt{3}(1+\sqrt{3})}\right) = \ln\left(\frac{2\sqrt{3}}{3}\right)\)* | A1\* | Rationalises denominator and completes to given answer. |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{2\sin\theta + \cos\theta + 2}\,\mathrm{d}\theta = \int \frac{1}{2\left(\frac{2t}{1+t^2}\right)+\left(\frac{1-t^2}{1+t^2}\right)+2} \times \frac{2}{1+t^2}\,\mathrm{d}t$ | **M1** | Applies correct substitutions $\sin\theta = \frac{2t}{1+t^2}$, $\cos\theta = \frac{1-t^2}{1+t^2}$, $\mathrm{d}\theta = \frac{2}{1+t^2}\,\mathrm{d}t$. The $\mathrm{d}t$ may be missing. |
| $\int \frac{2}{4t+(1-t^2)+2(1+t^2)}\,\mathrm{d}t = \int \frac{2}{t^2+4t+3}\,\mathrm{d}t$ | **M1** | Simplifies to form $\frac{a}{pt^2+qt+r}$, allowing slips in coefficients but "$1+t^2$" must be correctly handled. |
| $\int \frac{2}{(t+2)^2-1}\,\mathrm{d}t$ | **A1** | Correct answer including the $\int\,\mathrm{d}t$. |
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# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{a}{(t+b)^2-c}\,\mathrm{d}t = \frac{a}{2\sqrt{c}}\ln\left\lvert\frac{(t+b)-\sqrt{c}}{(t+b)+\sqrt{c}}\right\rvert, c>0$ giving $\int \frac{2}{(t+2)^2-1}\,\mathrm{d}t = \frac{2}{2(1)}\ln\left\lvert\frac{(t+2)-1}{(t+2)+1}\right\rvert$ | **M1** | Correct method for integration by standard formula. Modulus signs not necessary. |
| Correctly uses limits $t=1$ and $t=\sqrt{3}$: $\ln\left\lvert\frac{\sqrt{3}+1}{\sqrt{3}+3}\right\rvert - \ln\left\lvert\frac{2}{4}\right\rvert$ | **M1** | Correctly uses limits $t=1$ and $t=\sqrt{3}$ to produce exact expression. |
| $= \ln\left(\frac{\sqrt{3}+1}{\sqrt{3}+3} \div \frac{1}{2}\right)$ | **M1** | Correctly combines ln terms with at least one step showing substitution of limits before final answer. |
| $= \ln\left(2 \times \frac{\sqrt{3}+1}{\sqrt{3}+3} \times \frac{\sqrt{3}-3}{\sqrt{3}-3}\right) = \ln\left(\frac{2\sqrt{3}}{3}\right)$* | **A1\*** | Shows rationalisation of denominator and completes to given answer. Note $\sqrt{3}+3 = \sqrt{3}(1+\sqrt{3})$ may be used. |
**Alternative (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{2}{(t+2)^2-1}\,\mathrm{d}t = \int \frac{2}{(t+1)(t+3)}\,\mathrm{d}t = \int\frac{1}{t+1}-\frac{1}{t+3}\,\mathrm{d}t = \ln|t+1|-\ln|t+3|$ | **M1** | Correct method: partial fractions and integrating to ln terms. |
| $\ln(\sqrt{3}+1)-\ln(\sqrt{3}+3)-\ln 2+\ln 4$ | **M1** | Correctly uses limits $t=1$ and $t=\sqrt{3}$. |
| $= \ln\left(2\times\frac{\sqrt{3}+1}{\sqrt{3}+3}\right)$ | **M1** | Correctly combines ln terms. |
| $= \ln\left(2\times\frac{\sqrt{3}+1}{\sqrt{3}(1+\sqrt{3})}\right) = \ln\left(\frac{2\sqrt{3}}{3}\right)$* | **A1\*** | Rationalises denominator and completes to given answer. |
---\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 $t = \tan \left( \frac { \theta } { 2 } \right)$ to show that
$$\int \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \int \frac { a } { ( t + b ) ^ { 2 } + c } d t$$
where $a$, $b$ and $c$ are constants to be determined.\\
(b) Hence show that
$$\int _ { \frac { \pi } { 2 } } ^ { \frac { 2 \pi } { 3 } } \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \ln \left( \frac { 2 \sqrt { 3 } } { 3 } \right)$$
\hfill \mbox{\textit{Edexcel FP1 2024 Q7 [7]}}