| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Definite integral with simple linear/polynomial substitution |
| Difficulty | Challenging +1.2 This is a standard FP1 question on the Weierstrass t-substitution (tan half-angle). Part (i) requires algebraic manipulation using known identities but follows a routine procedure. Part (ii) is a textbook application of the substitution to evaluate a definite integral. While requiring careful algebra and knowledge of the substitution formulas, this represents expected Further Maths content without novel problem-solving demands. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08h Integration by substitution4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\dfrac{\sin x - \cos x + 1}{\sin x + \cos x - 1} = \dfrac{\dfrac{2t}{1+t^2} - \dfrac{1-t^2}{1+t^2} + 1}{\dfrac{2t}{1+t^2} + \dfrac{1-t^2}{1+t^2} - 1} = \ldots\) | M1 | Substitutes half-angle \(t\) substitutions correctly |
| Numerator \(= \dfrac{2t-(1-t^2)+1+t^2}{1+t^2}\), denominator \(= \dfrac{2t+1-t^2-(1+t^2)}{1+t^2}\) and divides | M1 | Combines fractions correctly in numerator and denominator and divides |
| \(= \dfrac{2t^2+2t}{2t-2t^2} \left(= \dfrac{1+t}{1-t}\right)\) | A1 | Correct simplified expression |
| \(= \dfrac{t+1}{1-t} \times \dfrac{1+t}{1+t} = \dfrac{1+t^2}{1-t^2} + \dfrac{2t}{1-t^2}\); Alt: \(\sec x + \tan x = \dfrac{1+t^2}{1-t^2} + \dfrac{2t}{1-t^2} = \dfrac{1+2t+t^2}{1-t^2}\) | M1 | Connects back to \(\sec x + \tan x\) |
| \(= \dfrac{1}{\cos x} + \tan x = \sec x + \tan x\) * ; Alt: \(= \dfrac{(t+1)^2}{(1-t)(1+t)} = \dfrac{1+t}{1-t} = LHS\) hence proved * | A1* | Correct completion to printed result (cso) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\displaystyle\int_{0}^{\pi/2} \dfrac{5}{4+2\cos\theta}\,d\theta = \int_{0}^{1} \dfrac{5}{4+2\cdot\dfrac{1-t^2}{1+t^2}} \times \dfrac{2}{1+t^2}\,dt\) ; Alt: \(\dfrac{dt}{d\theta} = \dfrac{1}{2}\sec^2\dfrac{\theta}{2} = \dfrac{1}{\cos\theta+1}\) leading to equivalent form | M1 | Correct substitution of \(t\) and \(d\theta\) |
| \(= \displaystyle\int_{0}^{1} \dfrac{10}{4(1+t^2)+2(1-t^2)}\,dt = \int_{0}^{1} \dfrac{5}{3+t^2}\,dt\) o.e. | A1 | Correct simplified integrand |
| \(= \left[5 \times \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{t}{\sqrt{3}}\right)\right]_{0}^{1}\) | M1 | Correct integration to arctan form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= \frac{5}{\sqrt{3}}\left(\arctan\left(\frac{1}{\sqrt{3}}\right)-0\right)\) or \(\frac{5}{\sqrt{3}}\left(\arctan\left(\frac{\tan(\pi/4)}{\sqrt{3}}\right)-\arctan\left(\frac{\tan(0)}{\sqrt{3}}\right)\right)\) | M1 | Deduces correct limits and applies them correctly, OR deduces integral in terms of \(\theta\) and applies original limits |
| \(= \frac{5\pi\sqrt{3}}{18}\) oe in surd form e.g. \(\frac{5\pi}{6\sqrt{3}}\) | A1 | Answer must be in exact surd form; note calculator gives \(\approx 1.51149947\) which scores M1 A1 M0 M1 A0 |
# Question 2:
## Part (i)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\dfrac{\sin x - \cos x + 1}{\sin x + \cos x - 1} = \dfrac{\dfrac{2t}{1+t^2} - \dfrac{1-t^2}{1+t^2} + 1}{\dfrac{2t}{1+t^2} + \dfrac{1-t^2}{1+t^2} - 1} = \ldots$ | M1 | Substitutes half-angle $t$ substitutions correctly |
| Numerator $= \dfrac{2t-(1-t^2)+1+t^2}{1+t^2}$, denominator $= \dfrac{2t+1-t^2-(1+t^2)}{1+t^2}$ and divides | M1 | Combines fractions correctly in numerator and denominator and divides |
| $= \dfrac{2t^2+2t}{2t-2t^2} \left(= \dfrac{1+t}{1-t}\right)$ | A1 | Correct simplified expression |
| $= \dfrac{t+1}{1-t} \times \dfrac{1+t}{1+t} = \dfrac{1+t^2}{1-t^2} + \dfrac{2t}{1-t^2}$; Alt: $\sec x + \tan x = \dfrac{1+t^2}{1-t^2} + \dfrac{2t}{1-t^2} = \dfrac{1+2t+t^2}{1-t^2}$ | M1 | Connects back to $\sec x + \tan x$ |
| $= \dfrac{1}{\cos x} + \tan x = \sec x + \tan x$ * ; Alt: $= \dfrac{(t+1)^2}{(1-t)(1+t)} = \dfrac{1+t}{1-t} = LHS$ hence proved * | A1* | Correct completion to printed result (cso) |
## Part (ii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\displaystyle\int_{0}^{\pi/2} \dfrac{5}{4+2\cos\theta}\,d\theta = \int_{0}^{1} \dfrac{5}{4+2\cdot\dfrac{1-t^2}{1+t^2}} \times \dfrac{2}{1+t^2}\,dt$ ; Alt: $\dfrac{dt}{d\theta} = \dfrac{1}{2}\sec^2\dfrac{\theta}{2} = \dfrac{1}{\cos\theta+1}$ leading to equivalent form | M1 | Correct substitution of $t$ and $d\theta$ |
| $= \displaystyle\int_{0}^{1} \dfrac{10}{4(1+t^2)+2(1-t^2)}\,dt = \int_{0}^{1} \dfrac{5}{3+t^2}\,dt$ o.e. | A1 | Correct simplified integrand |
| $= \left[5 \times \dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{t}{\sqrt{3}}\right)\right]_{0}^{1}$ | M1 | Correct integration to arctan form |
# Question 2 (Part ii continued):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \frac{5}{\sqrt{3}}\left(\arctan\left(\frac{1}{\sqrt{3}}\right)-0\right)$ or $\frac{5}{\sqrt{3}}\left(\arctan\left(\frac{\tan(\pi/4)}{\sqrt{3}}\right)-\arctan\left(\frac{\tan(0)}{\sqrt{3}}\right)\right)$ | M1 | Deduces correct limits and applies them correctly, OR deduces integral in terms of $\theta$ and applies original limits |
| $= \frac{5\pi\sqrt{3}}{18}$ oe in surd form e.g. $\frac{5\pi}{6\sqrt{3}}$ | A1 | Answer must be in exact surd form; note calculator gives $\approx 1.51149947$ which scores M1 A1 M0 M1 A0 |
---\begin{enumerate}
\item (i) Use the substitution $t = \tan \frac { X } { 2 }$ to prove the identity
\end{enumerate}
$$\frac { \sin x - \cos x + 1 } { \sin x + \cos x - 1 } \equiv \sec x + \tan x \quad x \neq \frac { n \pi } { 2 } \quad n \in \mathbb { Z }$$
(ii) Use the substitution $t = \tan \frac { \theta } { 2 }$ to determine the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { 5 } { 4 + 2 \cos \theta } d \theta$$
giving your answer in simplest form.
\hfill \mbox{\textit{Edexcel FP1 2021 Q2 [10]}}