| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Multi-part with preliminary simplification |
| Difficulty | Standard +0.8 Part (i) requires the standard double-angle identity cos²θ = (1+cos2θ)/2, which is routine for C4. Part (ii) requires integration by parts with the result from (i), involving careful substitution and evaluation at limits π/12 (which simplifies nicely). This is a solid multi-step question requiring two techniques sequentially, but follows predictable patterns for C4 integration—moderately above average difficulty. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Integration method | ||
| Attempt to change \(\cos^2 6x\) into \(\frac{1}{2}(\cos 12x)\) | M1 | |
| \(\cos^2 6x = \frac{1}{2}(1 + \cos 12x)\) | A1 | with \(\cos^2 6x\) as the subject of the formula |
| \(\int\frac{1}{2}x + \frac{1}{24}\sin 12x + c\) | A1 | AG Accept \(\frac{1}{2}(x + \frac{1}{12}\sin 12x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Differentiate RHS producing \(\frac{1}{2} + \frac{1}{2}\cos 12x\) --- (E) | B1 | |
| Attempt to change \(\cos 12x\) into \(\frac{1}{2}(\cos 6x)\) | M1 | Accept +/- 2 \(\cos^2 6x + l - 1\) |
| Simplify (E) WWW to \(\cos^2 6x\) + satis finish | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Parts with \(u = x\), dv \(= \cos^2 6x\) | *M1 | |
| \(x(\frac{1}{2}x + \frac{1}{24}\sin 12x) - \int(\frac{1}{2}x + \frac{1}{24}\sin 12x)dx\) | A1 | Correct expression only |
| \(\int\sin 12x\,dx = -\frac{1}{12}\cos 12x\) | B1 | Clear indication somewhere in this part |
| Correct use of limits to whole integral | dep*M1 | Accept ( ) (-0) |
| \(\frac{\pi^2}{288}, \frac{\pi^2}{576}, \frac{1}{288}, \frac{1}{288}\) | A1 | AE unsupp exp. Accept 12x24, sin \(\pi\) here |
| \(\frac{\pi^2}{576} - \frac{1}{144}\) | +A1 | 6 Tolerate e.g. \(-\frac{2}{288}\) here |
| S.R. If final marks are A0 + A0, allow SR A1 for | 0.01/0.0100/0.0101/0.0102/0.0101902 |
### (i)
Integration method | |
Attempt to change $\cos^2 6x$ into $\frac{1}{2}(\cos 12x)$ | M1 |
$\cos^2 6x = \frac{1}{2}(1 + \cos 12x)$ | A1 | with $\cos^2 6x$ as the subject of the formula
$\int\frac{1}{2}x + \frac{1}{24}\sin 12x + c$ | A1 | AG Accept $\frac{1}{2}(x + \frac{1}{12}\sin 12x)$
**Differentiation method**
Differentiate RHS producing $\frac{1}{2} + \frac{1}{2}\cos 12x$ --- (E) | B1 |
Attempt to change $\cos 12x$ into $\frac{1}{2}(\cos 6x)$ | M1 | Accept +/- 2 $\cos^2 6x + l - 1$
Simplify (E) WWW to $\cos^2 6x$ + satis finish | A1 | 3
---
### (ii)
Parts with $u = x$, dv $= \cos^2 6x$ | *M1 |
$x(\frac{1}{2}x + \frac{1}{24}\sin 12x) - \int(\frac{1}{2}x + \frac{1}{24}\sin 12x)dx$ | A1 | Correct expression only
$\int\sin 12x\,dx = -\frac{1}{12}\cos 12x$ | B1 | Clear indication somewhere in this part
Correct use of limits to whole integral | dep*M1 | Accept ( ) (-0)
$\frac{\pi^2}{288}, \frac{\pi^2}{576}, \frac{1}{288}, \frac{1}{288}$ | A1 | AE unsupp exp. Accept 12x24, sin $\pi$ here
$\frac{\pi^2}{576} - \frac{1}{144}$ | +A1 | 6 Tolerate e.g. $-\frac{2}{288}$ here
S.R. If final marks are A0 + A0, allow SR A1 for | | 0.01/0.0100/0.0101/0.0102/0.0101902
---\begin{enumerate}[label=(\roman*)]
\item Show that $\int \cos^2 6x dx = \frac{1}{2}x + \frac{1}{24}\sin 12x + c$. [3]
\item Hence find the exact value of $\int_0^{\frac{\pi}{12}} x\cos^2 6x dx$. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2006 Q8 [9]}}