| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.8 This is a standard Further Maths reduction formula question requiring integration by parts with a specific choice of u and dv, followed by algebraic manipulation and recursive application. While technically demanding with multiple steps, it follows a well-established pattern that FP2 students practice extensively. The definite integral evaluation in part (b) is computational but straightforward once the reduction formula is established. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int \text{cosec}^{n-2}x\,\text{cosec}^2x\,dx\); set \(u=\text{cosec}^{n-2}x\), \(\frac{dv}{dx}=\text{cosec}^2x\); apply \(\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx\) | M1 | Splits integrand as shown and begins integration by parts |
| \(I_n = -\text{cosec}^{n-2}x\cot x - (n-2)\int\text{cosec}^{n-3}x(-\text{cosec}\,x\cot x)(-\cot x)\,dx\) | A1 | Correct expression |
| \(I_n = -\text{cosec}^{n-2}x\cot x - (n-2)\int\text{cosec}^{n-2}x\cot^2x\,dx\) | ||
| Applies \(\cot^2x = \pm1\pm\text{cosec}^2x\), introduces \(I_n\) and \(I_{n-2}\) | dM1 | Depends on previous M |
| \(I_n = -\text{cosec}^{n-2}x\cot x - (n-2)I_n + (n-2)I_{n-2}\) | ||
| \((n-1)I_n = -\text{cosec}^{n-2}x\cot x + (n-2)I_{n-2}\) | ||
| \(I_n = \dfrac{n-2}{n-1}I_{n-2} - \dfrac{\text{cosec}^{n-2}x\cot x}{n-1}\) | A1* | Completes proof making \(I_n\) subject with no errors |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_6 = \dfrac{4}{5}I_4 - \dfrac{\text{cosec}^4x\cot x}{5}\) or \(\left[I_6\right]_{\pi/3}^{\pi/2} = \dfrac{4}{5}\left[I_4\right]_{\pi/3}^{\pi/2} - \left[\dfrac{\text{cosec}^4x\cot x}{5}\right]_{\pi/3}^{\pi/2}\) | M1 | Begins applying reduction formula to find \(I_6\) in terms of \(I_4\) |
| \(= \dfrac{4}{5}\!\left(\dfrac{2}{3}I_2 - \dfrac{\text{cosec}^2x\cot x}{3}\right) - \dfrac{\text{cosec}^4x\cot x}{5}\) or with limits | M1 | Uses reduction formula to find \(I_4\) in terms of \(I_2\) |
| \(\left[I_6\right]_{\pi/3}^{\pi/2} = \dfrac{8}{15}\left[-\cot x\right]_{\pi/3}^{\pi/2} - \left[\dfrac{4\text{cosec}^2x\cot x}{15}\right]_{\pi/3}^{\pi/2} - \left[\dfrac{\text{cosec}^4x\cot x}{5}\right]_{\pi/3}^{\pi/2}\) | M1 | Fully correct method using reduction formula to reach value for \(I_6\); substitutions shown for non-zero terms |
| \(= \dfrac{8}{15}\!\left(\dfrac{\sqrt{3}}{3}\right) + \dfrac{16\sqrt{3}}{135} + \dfrac{16\sqrt{3}}{135} = \dfrac{56}{135}\sqrt{3}\) | A1* | Correct exact value |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(I_n = \int \cosec^{n+1}x \sin x \, dx\) (Allow with \(n \pm 1\) in power for M's) | ||
| \(u = \cosec^{n+1}x, \frac{dv}{dx} = \sin x, \int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx\) | M1 | 2.1 |
| \(I_n = -\cosec^{n+1}x\cos x - (n+1)\int \cosec^n x(-\cosec x \cot x)(-\cos x)dx\) | A1 | 1.1b |
| \(I_n = -\cosec^n x \cot x - (n+1)\int \cosec^n x \cot^2 x \, dx\) | ||
| \(I_n = -\cosec^n x \cot x - (n+1)\int \cosec^n x(\cosec^2 x - 1)dx\) | dM1 | 1.1b |
| \(I_n = -\cosec^n x \cot x - (n+1)I_{n+2} + (n+1)I_n\) | ||
| \((n+1)I_{n+2} = -\cosec^n x \cot x + nI_n\) | ||
| Replacing \(n\) by \(n-2\): \(I_n = \frac{n-2}{n-1}I_{n-2} - \frac{\cosec^{n-2}x\cot x}{n-1}\) | A1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(I_n = \int \cosec^{n-2}x \cosec^2 x \, dx = \int \cosec^{n-2}x(1+\cot^2 x)dx\) | ||
| \(= \int \cosec^{n-2}x \, dx + \left(\int \cosec^{n-2}x \cot x\right)(\cot x)dx\) | M1 | 2.1 |
| \(u = \cot x, \frac{dv}{dx} = \cosec^{n-2}x\cot x, \int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx\) | ||
| \(I_n = I_{n-2} + \cot x\left(-\frac{\cosec^{n-2}x}{n-2}\right) - \int\left(-\frac{\cosec^{n-2}x}{n-2}\right)(-\cosec^2 x)dx\) | A1 | 1.1b |
| \((n-2)I_n = (n-2)I_{n-2} - \cosec^{n-2}x\cot x - \int \cosec^n x \, dx\) | ||
| \((n-2)I_n = (n-2)I_{n-2} - \cosec^{n-2}x\cot x - I_n\) | dM1 | 1.1b |
| \((n-1)I_n = -\cosec^{n-2}x\cot x + (n-2)I_{n-2}\) | ||
| \(I_n = \frac{n-2}{n-1}I_{n-2} - \frac{\cosec^{n-2}x\cot x}{n-1}\) | A1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(I_2 = \int_{\frac{\pi}{5}}^{\frac{\pi}{3}}\cosec^2 x \, dx = [-\cot x]_{\frac{\pi}{5}}^{\frac{\pi}{3}} = \frac{\sqrt{3}}{3}\) | M1 | 2.2a |
| \(I_4 = \frac{2}{3}I_2 - \left[\frac{\cosec^2 x\cot x}{3}\right]_{\frac{\pi}{5}}^{\frac{\pi}{3}} = \frac{2}{9}\sqrt{3} + \frac{4}{27}\sqrt{3}\) | M1 | 1.1b |
| \(I_6 = \frac{4}{5}I_4 - \left[\frac{\cosec^4 x\cot x}{5}\right]_{\frac{\pi}{5}}^{\frac{\pi}{3}} = \frac{4}{5}\left(\frac{4}{27}\sqrt{3}+\frac{2}{9}\sqrt{3}\right) + \frac{16}{135}\sqrt{3}\) | M1 | 2.1 |
| \(= \frac{56}{135}\sqrt{3}\) | A1* | 1.1b |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int \text{cosec}^{n-2}x\,\text{cosec}^2x\,dx$; set $u=\text{cosec}^{n-2}x$, $\frac{dv}{dx}=\text{cosec}^2x$; apply $\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$ | M1 | Splits integrand as shown and begins integration by parts |
| $I_n = -\text{cosec}^{n-2}x\cot x - (n-2)\int\text{cosec}^{n-3}x(-\text{cosec}\,x\cot x)(-\cot x)\,dx$ | A1 | Correct expression |
| $I_n = -\text{cosec}^{n-2}x\cot x - (n-2)\int\text{cosec}^{n-2}x\cot^2x\,dx$ | | |
| Applies $\cot^2x = \pm1\pm\text{cosec}^2x$, introduces $I_n$ and $I_{n-2}$ | dM1 | Depends on previous M |
| $I_n = -\text{cosec}^{n-2}x\cot x - (n-2)I_n + (n-2)I_{n-2}$ | | |
| $(n-1)I_n = -\text{cosec}^{n-2}x\cot x + (n-2)I_{n-2}$ | | |
| $I_n = \dfrac{n-2}{n-1}I_{n-2} - \dfrac{\text{cosec}^{n-2}x\cot x}{n-1}$ | A1* | Completes proof making $I_n$ subject with no errors |
| | **(4)** | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_6 = \dfrac{4}{5}I_4 - \dfrac{\text{cosec}^4x\cot x}{5}$ or $\left[I_6\right]_{\pi/3}^{\pi/2} = \dfrac{4}{5}\left[I_4\right]_{\pi/3}^{\pi/2} - \left[\dfrac{\text{cosec}^4x\cot x}{5}\right]_{\pi/3}^{\pi/2}$ | M1 | Begins applying reduction formula to find $I_6$ in terms of $I_4$ |
| $= \dfrac{4}{5}\!\left(\dfrac{2}{3}I_2 - \dfrac{\text{cosec}^2x\cot x}{3}\right) - \dfrac{\text{cosec}^4x\cot x}{5}$ or with limits | M1 | Uses reduction formula to find $I_4$ in terms of $I_2$ |
| $\left[I_6\right]_{\pi/3}^{\pi/2} = \dfrac{8}{15}\left[-\cot x\right]_{\pi/3}^{\pi/2} - \left[\dfrac{4\text{cosec}^2x\cot x}{15}\right]_{\pi/3}^{\pi/2} - \left[\dfrac{\text{cosec}^4x\cot x}{5}\right]_{\pi/3}^{\pi/2}$ | M1 | Fully correct method using reduction formula to reach value for $I_6$; substitutions shown for non-zero terms |
| $= \dfrac{8}{15}\!\left(\dfrac{\sqrt{3}}{3}\right) + \dfrac{16\sqrt{3}}{135} + \dfrac{16\sqrt{3}}{135} = \dfrac{56}{135}\sqrt{3}$ | A1* | Correct exact value |
| | **(4)** | |
# Question 5(a) ALT 1:
| Working | Mark | Guidance |
|---------|------|----------|
| $I_n = \int \cosec^{n+1}x \sin x \, dx$ (Allow with $n \pm 1$ in power for M's) | | |
| $u = \cosec^{n+1}x, \frac{dv}{dx} = \sin x, \int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$ | M1 | 2.1 |
| $I_n = -\cosec^{n+1}x\cos x - (n+1)\int \cosec^n x(-\cosec x \cot x)(-\cos x)dx$ | A1 | 1.1b |
| $I_n = -\cosec^n x \cot x - (n+1)\int \cosec^n x \cot^2 x \, dx$ | | |
| $I_n = -\cosec^n x \cot x - (n+1)\int \cosec^n x(\cosec^2 x - 1)dx$ | dM1 | 1.1b |
| $I_n = -\cosec^n x \cot x - (n+1)I_{n+2} + (n+1)I_n$ | | |
| $(n+1)I_{n+2} = -\cosec^n x \cot x + nI_n$ | | |
| Replacing $n$ by $n-2$: $I_n = \frac{n-2}{n-1}I_{n-2} - \frac{\cosec^{n-2}x\cot x}{n-1}$ | A1* | 2.1 |
---
# Question 5(a) ALT 2:
| Working | Mark | Guidance |
|---------|------|----------|
| $I_n = \int \cosec^{n-2}x \cosec^2 x \, dx = \int \cosec^{n-2}x(1+\cot^2 x)dx$ | | |
| $= \int \cosec^{n-2}x \, dx + \left(\int \cosec^{n-2}x \cot x\right)(\cot x)dx$ | M1 | 2.1 |
| $u = \cot x, \frac{dv}{dx} = \cosec^{n-2}x\cot x, \int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$ | | |
| $I_n = I_{n-2} + \cot x\left(-\frac{\cosec^{n-2}x}{n-2}\right) - \int\left(-\frac{\cosec^{n-2}x}{n-2}\right)(-\cosec^2 x)dx$ | A1 | 1.1b |
| $(n-2)I_n = (n-2)I_{n-2} - \cosec^{n-2}x\cot x - \int \cosec^n x \, dx$ | | |
| $(n-2)I_n = (n-2)I_{n-2} - \cosec^{n-2}x\cot x - I_n$ | dM1 | 1.1b |
| $(n-1)I_n = -\cosec^{n-2}x\cot x + (n-2)I_{n-2}$ | | |
| $I_n = \frac{n-2}{n-1}I_{n-2} - \frac{\cosec^{n-2}x\cot x}{n-1}$ | A1* | 2.1 |
---
# Question 5(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $I_2 = \int_{\frac{\pi}{5}}^{\frac{\pi}{3}}\cosec^2 x \, dx = [-\cot x]_{\frac{\pi}{5}}^{\frac{\pi}{3}} = \frac{\sqrt{3}}{3}$ | M1 | 2.2a |
| $I_4 = \frac{2}{3}I_2 - \left[\frac{\cosec^2 x\cot x}{3}\right]_{\frac{\pi}{5}}^{\frac{\pi}{3}} = \frac{2}{9}\sqrt{3} + \frac{4}{27}\sqrt{3}$ | M1 | 1.1b |
| $I_6 = \frac{4}{5}I_4 - \left[\frac{\cosec^4 x\cot x}{5}\right]_{\frac{\pi}{5}}^{\frac{\pi}{3}} = \frac{4}{5}\left(\frac{4}{27}\sqrt{3}+\frac{2}{9}\sqrt{3}\right) + \frac{16}{135}\sqrt{3}$ | M1 | 2.1 |
| $= \frac{56}{135}\sqrt{3}$ | A1* | 1.1b |
---5.
$$I _ { n } = \int \operatorname { cosec } ^ { n } x \mathrm {~d} x \quad n \in \mathbb { Z }$$
\begin{enumerate}[label=(\alph*)]
\item Prove that, for $n \geqslant 2$
$$I _ { n } = \frac { n - 2 } { n - 1 } I _ { n - 2 } - \frac { \operatorname { cosec } ^ { n - 2 } x \cot x } { n - 1 }$$
\item Hence show that
$$\int _ { \frac { \pi } { 3 } } ^ { \frac { \pi } { 2 } } \operatorname { cosec } ^ { 6 } x \mathrm {~d} x = \frac { 56 } { 135 } \sqrt { 3 }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2019 Q5 [8]}}