Exam Board OCR Module C4 (Core Mathematics 4) Year 2012 Session June Marks 7 Paper Download PDF ↗ Mark scheme Download PDF ↗ Topic Standard Integrals and Reverse Chain Rule Type Definite integral with trigonometric functions Difficulty Standard +0.3 This question requires expanding the squared bracket, integrating standard trigonometric functions, and applying the reverse chain rule for sin 3x terms. While it involves multiple steps (expand, integrate, substitute limits), each step uses routine C4 techniques with no novel insight required. The exact value requirement and the specific limits add minor complexity, but this remains a straightforward application of standard methods, making it slightly easier than average. Spec 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)
7 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) ^ { 2 } \mathrm {~d} x\).
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Question 7:
Answer Marks
Guidance
Answer Marks
Guidance
Satisfactory start: attempt to square \((1-\sin 3x)\) M1
Not e.g. \(\frac{(1-\sin 3x)^3}{3}\)
Obtain \(\int -2\sin 3x\,dx = \frac{2}{3}\cos 3x\) *A1
Obtain \(-\frac{2}{3}\) or \((\ldots+0\ldots)-(\ldots+\frac{2}{3}\ldots)\) A1dep*
Use \(\sin^2 3x = k(+/-1+/-\cos 6x)\) M1
or for integrating \(\sin^2 ax\) where \(a=6\) or \(9\) only: \(\sin^2 ax = k(+/-1+/-\cos 2ax)\)
Correct version \(= \frac{1}{2}(1-\cos 6x)\) A1
Correct \(= \frac{1}{2}(1-\cos 2ax)\)
\(\int\cos 6x\,dx = \frac{1}{6}\sin 6x\), seen anywhere, indep B1
or \(\int\cos 2ax\,dx = \frac{1}{2a}\sin 2ax\)
Final answer \(= \frac{1}{4}\pi + \text{their}\, {-\frac{2}{3}}\) A1
Check that \(\frac{1}{4}\pi\) is from \(\left[\frac{3}{2}x - \frac{1}{12}\sin 6x\right]_0^{\frac{1}{6}\pi}\)
Total: [7]
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# Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Satisfactory start: attempt to square $(1-\sin 3x)$ | M1 | Not e.g. $\frac{(1-\sin 3x)^3}{3}$ |
| Obtain $\int -2\sin 3x\,dx = \frac{2}{3}\cos 3x$ | *A1 | |
| Obtain $-\frac{2}{3}$ or $(\ldots+0\ldots)-(\ldots+\frac{2}{3}\ldots)$ | A1dep* | |
| Use $\sin^2 3x = k(+/-1+/-\cos 6x)$ | M1 | or for integrating $\sin^2 ax$ where $a=6$ or $9$ only: $\sin^2 ax = k(+/-1+/-\cos 2ax)$ |
| Correct version $= \frac{1}{2}(1-\cos 6x)$ | A1 | Correct $= \frac{1}{2}(1-\cos 2ax)$ |
| $\int\cos 6x\,dx = \frac{1}{6}\sin 6x$, seen anywhere, indep | B1 | or $\int\cos 2ax\,dx = \frac{1}{2a}\sin 2ax$ |
| Final answer $= \frac{1}{4}\pi + \text{their}\, {-\frac{2}{3}}$ | A1 | Check that $\frac{1}{4}\pi$ is from $\left[\frac{3}{2}x - \frac{1}{12}\sin 6x\right]_0^{\frac{1}{6}\pi}$ |
| **Total: [7]** | | |
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7 Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) ^ { 2 } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR C4 2012 Q7 [7]}}