| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | September |
| Marks | 5 |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with trigonometric functions |
| Difficulty | Challenging +1.2 This question tests understanding of improper integrals and discontinuities, requiring students to recognize that tan x has a discontinuity at π/2 within the integration interval. Part (i) involves symmetric integration around a discontinuity (requiring careful limit analysis or recognition of odd function properties), while part (ii) requires explaining why the integral is undefined due to the asymptote. This goes beyond routine integration to test conceptual understanding of when integrals exist, making it moderately harder than average but still within standard Further Maths scope. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int 2\tan xdx = -2\ln(\cos x)\) or \(2\ln(\sec x)\) | B1 | Limits can be ignored for M1A1 |
| \(-2[\ln(\cos x)]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -2(\ln \cos \frac{\pi}{3} - \ln \cos \frac{\pi}{4})\) | M1 | Correct use of limits in integral of the form \(a\ln f(x)\) where \(f(x)\) is trigonometric |
| \(= -2(\ln \frac{1}{2} - \ln \frac{1}{\sqrt{2}}) = -2 \ln 2^{-\frac{1}{2}} = \ln 2\) cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^z 2\tan xdx = -2\ln \cos z\) | M1 | Using a variable to denote the upper limit and finding the integral. Could be their incorrect integral from (i) or \(\ln 0\) is undefined but argument must be complete |
| \(\text{So } \int_0^z 2\tan xdx = \lim(−2\ln \cos z)\) but \(z → \frac{\pi}{2}^−\) \(z \to \frac{\pi}{2} \Rightarrow \cos z \to 0 \Rightarrow \ln \cos z \to -\infty\) so the integral is undefined. | A1 |
| Answer | Marks |
|---|---|
| Or \(\int_0^z 2\tan xdx = 2\ln \sec z\) Or \(\int_0^z 2\tan xdx = \lim(2\ln \sec z)\) \(z \to \frac{\pi}{2}^−\) but \(z \to \frac{\pi}{2} \Rightarrow \sec z \to \infty\) \(\Rightarrow \ln \sec z \to \infty\) so the integral is undefined. | (alternative answer format) |
## (i)
$\int 2\tan xdx = -2\ln(\cos x)$ or $2\ln(\sec x)$ | B1 | Limits can be ignored for M1A1
$-2[\ln(\cos x)]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -2(\ln \cos \frac{\pi}{3} - \ln \cos \frac{\pi}{4})$ | M1 | Correct use of limits in integral of the form $a\ln f(x)$ where $f(x)$ is trigonometric
$= -2(\ln \frac{1}{2} - \ln \frac{1}{\sqrt{2}}) = -2 \ln 2^{-\frac{1}{2}} = \ln 2$ cao | A1 |
[3]
## (ii)
$\int_0^z 2\tan xdx = -2\ln \cos z$ | M1 | Using a variable to denote the upper limit and finding the integral. Could be their incorrect integral from (i) or $\ln 0$ is undefined but argument must be complete
$\text{So } \int_0^z 2\tan xdx = \lim(−2\ln \cos z)$ but $z → \frac{\pi}{2}^−$ $z \to \frac{\pi}{2} \Rightarrow \cos z \to 0 \Rightarrow \ln \cos z \to -\infty$ so the integral is undefined. | A1 |
[2]
Or $\int_0^z 2\tan xdx = 2\ln \sec z$ Or $\int_0^z 2\tan xdx = \lim(2\ln \sec z)$ $z \to \frac{\pi}{2}^−$ but $z \to \frac{\pi}{2} \Rightarrow \sec z \to \infty$ $\Rightarrow \ln \sec z \to \infty$ so the integral is undefined. | (alternative answer format) |
---In this question you must show detailed reasoning.
\begin{enumerate}[label=(\roman*)]
\item Find $\int_{-\frac{3\pi}{4}}^{\frac{3\pi}{4}} 2\tan x \, dx$ giving your answer in the form $\ln p$. [3]
\item Show that $\int_0^{\frac{3\pi}{4}} 2\tan x \, dx$ is undefined explaining your reasoning. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q2 [5]}}