| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Parametric or substitution with partial fractions |
| Difficulty | Standard +0.3 This is a standard C4 integration question following a predictable three-part structure: verify a substitution (routine differentiation), perform partial fractions decomposition (standard technique with three linear factors), and evaluate a definite integral. While it requires multiple techniques and careful algebra, each step follows well-practiced procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08h Integration by substitution1.08j Integration using partial fractions |
| Answer | Marks |
|---|---|
| \(u = \sin x \Rightarrow \frac{du}{dx} = \cos x\) | B1 |
| \(I = \int \frac{6\cos x}{\cos^2 x(2 - \sin x)} dx = \int \frac{6\cos x}{(1 - \sin^2 x)(2 - \sin x)} dx\) | M1 |
| \(= \int \frac{6}{(1 - u^2)(2 - u)} du\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\frac{6}{(1 + u)(1 - u)(2 - u)} \equiv \frac{A}{1 + u} + \frac{B}{1 - u} + \frac{C}{2 - u}\) | M1 |
| \(6 = A(1 - u)(2 - u) + B(1 + u)(2 - u) + C(1 + u)(1 - u)\) | |
| \(u = -1 \quad \Rightarrow 6 = 6A \quad \Rightarrow A = 1\) | A1 |
| \(u = 1 \quad \Rightarrow 6 = 2B \quad \Rightarrow B = 3\) | A1 |
| \(u = 2 \quad \Rightarrow 6 = -3C \quad \Rightarrow C = -2\) | A1 |
| \(\therefore \frac{6}{(1 - u^2)(2 - u)} \equiv \frac{1}{1 + u} + \frac{3}{1 - u} - \frac{2}{2 - u}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 0 \Rightarrow u = 0, \quad x = \frac{\pi}{6} \Rightarrow u = \frac{1}{2}\) | M1 | |
| \(I = \int_0^{1/2} \left(\frac{1}{1 + u} + \frac{3}{1 - u} - \frac{2}{2 - u}\right) du\) | ||
| \(= [\ln | 1 + u | - 3\ln |
| \(= (\ln\frac{3}{2} - 3\ln\frac{1}{2} + 2\ln\frac{3}{2}) - (0 + 0 + 2\ln 2)\) | M1 | |
| \(= 3\ln\frac{3}{2} + 3\ln 2 - 2\ln 2\) | ||
| \(= 3\ln 3 - 3\ln 2 + \ln 2 = 3\ln 3 - 2\ln 2\) | M1 A1 | (14) |
## (i)
$u = \sin x \Rightarrow \frac{du}{dx} = \cos x$ | B1 |
$I = \int \frac{6\cos x}{\cos^2 x(2 - \sin x)} dx = \int \frac{6\cos x}{(1 - \sin^2 x)(2 - \sin x)} dx$ | M1 |
$= \int \frac{6}{(1 - u^2)(2 - u)} du$ | M1 A1 |
## (ii)
$\frac{6}{(1 + u)(1 - u)(2 - u)} \equiv \frac{A}{1 + u} + \frac{B}{1 - u} + \frac{C}{2 - u}$ | M1 |
$6 = A(1 - u)(2 - u) + B(1 + u)(2 - u) + C(1 + u)(1 - u)$ | |
$u = -1 \quad \Rightarrow 6 = 6A \quad \Rightarrow A = 1$ | A1 |
$u = 1 \quad \Rightarrow 6 = 2B \quad \Rightarrow B = 3$ | A1 |
$u = 2 \quad \Rightarrow 6 = -3C \quad \Rightarrow C = -2$ | A1 |
$\therefore \frac{6}{(1 - u^2)(2 - u)} \equiv \frac{1}{1 + u} + \frac{3}{1 - u} - \frac{2}{2 - u}$ | |
## (iii)
$x = 0 \Rightarrow u = 0, \quad x = \frac{\pi}{6} \Rightarrow u = \frac{1}{2}$ | M1 |
$I = \int_0^{1/2} \left(\frac{1}{1 + u} + \frac{3}{1 - u} - \frac{2}{2 - u}\right) du$ | |
$= [\ln|1 + u| - 3\ln|1 - u| + 2\ln|2 - u|]_0^{1/2}$ | M1 A1 |
$= (\ln\frac{3}{2} - 3\ln\frac{1}{2} + 2\ln\frac{3}{2}) - (0 + 0 + 2\ln 2)$ | M1 |
$= 3\ln\frac{3}{2} + 3\ln 2 - 2\ln 2$ | |
$= 3\ln 3 - 3\ln 2 + \ln 2 = 3\ln 3 - 2\ln 2$ | M1 A1 | (14) |
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**Total: (72)**\begin{enumerate}[label=(\roman*)]
\item Show that the substitution $u = \sin x$ transforms the integral
$$\int \frac{6}{\cos x(2 - \sin x)} dx$$
into the integral
$$\int \frac{6}{(1-u^2)(2-u)} du.$$ [4]
\item Express $\frac{6}{(1-u^2)(2-u)}$ in partial fractions. [4]
\item Hence, evaluate
$$\int_0^{\pi/6} \frac{6}{\cos x(2 - \sin x)} dx,$$
giving your answer in the form $a \ln 2 + b \ln 3$, where $a$ and $b$ are integers. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q9 [14]}}