| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 10 |
| Topic | Integration by Substitution |
| Type | Algebraic manipulation before substitution |
| Difficulty | Standard +0.3 This is a standard A-level integration question combining completing the square (routine), a guided trigonometric substitution with explicit instructions, and a final part using logarithmic integration. While multi-part with 10 marks total, each step follows well-established techniques with clear guidance, making it slightly easier than average but requiring competent execution of multiple methods. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution |
| Answer | Marks |
|---|---|
| \(p = 3\) | B1 |
| \(q = 1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x - 3 = \tan u \Rightarrow dx = \sec^2 u\,du\) | M1 | Attempt to connect \(dx\) and \(du\) |
| \(\int_3^4 \frac{1}{x^2 - 6x + 10}\,dx = \int_0^{\frac{\pi}{4}} \frac{\sec^2 u}{\tan^2 u + 1}\,du\) | A1ft | Correct integral and limits; ft their \(p\) |
| \(= \int_0^{\frac{\pi}{4}} du = \frac{1}{4}\pi\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Integral is \(\frac{1}{2}\int_3^4 \frac{2x - 6}{x^2 - 6x + 10}\,dx + \int_3^4 \frac{3}{x^2 - 6x + 10}\,dx\) | B1 | Or with single numerator \(2x - 6 + 6\) or similar; Limits not required |
| Use of \(\int \frac{f'(x)}{f(x)}\,dx\) and their answer to part (ii) | *M1 | |
| \(\frac{1}{2}[\ln(x^2 - 6x + 10)]_3^4 + \frac{3}{4}\pi\) | A1ft | FT \(3 \times\) their answer to part (ii) |
| Correct use of limits and correct use of logs | dep*M1 | |
| \(\frac{3}{4}\pi + \ln\sqrt{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\frac{\pi}{4}} \frac{3 + \tan u}{\tan^2 u + 1}\sec^2 u\,du\) | *M1 | Attempt use of previous substitution; Limits not required |
| \(\int_0^{\frac{\pi}{4}} (3 + \tan u)\,du\) | A1 | Correct integral and limits |
| \(= [3u - \ln(\cos u)]_0^{\frac{\pi}{4}}\) | A1 | |
| Correct use of limits and correct use of logs | dep*M1 | |
| \(\frac{3}{4}\pi + \ln\sqrt{2}\) | A1 |
## Part (i):
$p = 3$ | B1 |
$q = 1$ | B1 |
## Part (ii):
$x - 3 = \tan u \Rightarrow dx = \sec^2 u\,du$ | M1 | Attempt to connect $dx$ and $du$
$\int_3^4 \frac{1}{x^2 - 6x + 10}\,dx = \int_0^{\frac{\pi}{4}} \frac{\sec^2 u}{\tan^2 u + 1}\,du$ | A1ft | Correct integral and limits; ft their $p$
$= \int_0^{\frac{\pi}{4}} du = \frac{1}{4}\pi$ | A1 |
## Part (iii):
Integral is $\frac{1}{2}\int_3^4 \frac{2x - 6}{x^2 - 6x + 10}\,dx + \int_3^4 \frac{3}{x^2 - 6x + 10}\,dx$ | B1 | Or with single numerator $2x - 6 + 6$ or similar; Limits not required
Use of $\int \frac{f'(x)}{f(x)}\,dx$ and their answer to part (ii) | *M1 |
$\frac{1}{2}[\ln(x^2 - 6x + 10)]_3^4 + \frac{3}{4}\pi$ | A1ft | FT $3 \times$ their answer to part (ii)
Correct use of limits and correct use of logs | dep*M1 |
$\frac{3}{4}\pi + \ln\sqrt{2}$ | A1 |
**Alternative solution:**
$\int_0^{\frac{\pi}{4}} \frac{3 + \tan u}{\tan^2 u + 1}\sec^2 u\,du$ | *M1 | Attempt use of previous substitution; Limits not required
$\int_0^{\frac{\pi}{4}} (3 + \tan u)\,du$ | A1 | Correct integral and limits
$= [3u - \ln(\cos u)]_0^{\frac{\pi}{4}}$ | A1 |
Correct use of limits and correct use of logs | dep*M1 |
$\frac{3}{4}\pi + \ln\sqrt{2}$ | A1 |\begin{enumerate}[label=(\roman*)]
\item Determine the values of $p$ and $q$ for which
$$x^2 - 6x + 10 \equiv (x - p)^2 + q.$$ [2]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Use the substitution $x - p = \tan u$, where $p$ takes the value found in part (i), to evaluate
$$\int_3^4 \frac{1}{x^2 - 6x + 10} \, dx.$$ [3]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Determine the value of
$$\int_3^4 \frac{x}{x^2 - 6x + 10} \, dx,$$
giving your answer in the form $a + \ln b$, where $a$ and $b$ are constants to be determined. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q6 [10]}}