OCR C4 2016 June — Question 2 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2016
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeUse trig identity before definite integration
DifficultyStandard +0.3 This question requires using the double angle identity cos²(4x) = (1 + cos(8x))/2 to simplify before integrating, then applying the reverse chain rule to integrate cos(8x). While it involves multiple steps (identity application, integration, substitution of limits), these are standard C4 techniques with no novel problem-solving required. The question is slightly easier than average because the identity application is straightforward and the integration is routine once simplified.
Spec1.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

2 Use integration to find the exact value of \(\int _ { \frac { 1 } { 16 } \pi } ^ { \frac { 1 } { 8 } \pi } \left( 9 - 6 \cos ^ { 2 } 4 x \right) \mathrm { d } x\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\cos 8x\) seen in integrandM1
\(F[x] = Ax + B\sin 8x\)M1* \(A\) and \(B\) are non-zero constants
\(F[x] = 6x - \frac{3}{8}\sin 8x\)A1
\(F[\frac{1}{8}\pi] - F[\frac{1}{16}\pi]\)M1*dep
\(\frac{3}{8}\pi + \frac{3}{8}\)A1 allow e.g. \(0.375\pi + 0.375\) or fractions not in lowest terms
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos 8x$ seen in integrand | M1 | |
| $F[x] = Ax + B\sin 8x$ | M1* | $A$ and $B$ are non-zero constants |
| $F[x] = 6x - \frac{3}{8}\sin 8x$ | A1 | |
| $F[\frac{1}{8}\pi] - F[\frac{1}{16}\pi]$ | M1*dep | |
| $\frac{3}{8}\pi + \frac{3}{8}$ | A1 | allow e.g. $0.375\pi + 0.375$ or fractions not in lowest terms |

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2 Use integration to find the exact value of $\int _ { \frac { 1 } { 16 } \pi } ^ { \frac { 1 } { 8 } \pi } \left( 9 - 6 \cos ^ { 2 } 4 x \right) \mathrm { d } x$.

\hfill \mbox{\textit{OCR C4 2016 Q2 [5]}}