Edexcel C4 2012 January — Question 2 6 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2012
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeSequential multi-part (building on previous)
DifficultyStandard +0.8 Part (a) is a standard integration by parts application, but part (b) requires recognizing that differentiating the result from (a) leads to the desired integral—a non-routine insight that goes beyond mechanical application. The multi-step nature and need for strategic thinking elevates this above average difficulty.
Spec1.08i Integration by parts
2. (a) Use integration by parts to find \(\int x \sin 3 x \mathrm {~d} x\).
(b) Using your answer to part (a), find \(\int x ^ { 2 } \cos 3 x \mathrm {~d} x\).
Question 2:
Part (a): \(\int x\sin 3x \, dx\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(-\frac{1}{3}x\cos 3x - \int -\frac{1}{3}\cos 3x \{dx\}\)M1 A1 Use of IBP formula \(uv - \int vu'\) in correct direction; \(u=x \Rightarrow u'=1\), \(v'=\sin 3x \Rightarrow v=k\cos 3x\). Must achieve \(x(k\cos 3x) - \int(k\cos 3x)\)
\(= -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x \{+c\}\)A1 With/without \(+c\)
Part (b): \(\int x^2\cos 3x \, dx\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{3}x^2\sin 3x - \int \frac{2}{3}x\sin 3x \{dx\}\)M1 A1 IBP in correct direction; \(u=x^2 \Rightarrow u'=2x\), \(v'=\cos 3x \Rightarrow v=\lambda\sin 3x\). Must achieve \(x^2(\lambda\sin 3x) - \int 2x(\lambda\sin 3x)\)
\(= \frac{1}{3}x^2\sin 3x - \frac{2}{3}\left(-\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x\right)\{+c\}\)A1 isw Substituting part (a) result
\(= \frac{1}{3}x^2\sin 3x + \frac{2}{9}x\cos 3x - \frac{2}{27}\sin 3x \{+c\}\) Ignore subsequent working 
# Question 2:

## Part (a): $\int x\sin 3x \, dx$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $-\frac{1}{3}x\cos 3x - \int -\frac{1}{3}\cos 3x \{dx\}$ | M1 A1 | Use of IBP formula $uv - \int vu'$ in correct direction; $u=x \Rightarrow u'=1$, $v'=\sin 3x \Rightarrow v=k\cos 3x$. Must achieve $x(k\cos 3x) - \int(k\cos 3x)$ |
| $= -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x \{+c\}$ | A1 | With/without $+c$ |

## Part (b): $\int x^2\cos 3x \, dx$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{3}x^2\sin 3x - \int \frac{2}{3}x\sin 3x \{dx\}$ | M1 A1 | IBP in correct direction; $u=x^2 \Rightarrow u'=2x$, $v'=\cos 3x \Rightarrow v=\lambda\sin 3x$. Must achieve $x^2(\lambda\sin 3x) - \int 2x(\lambda\sin 3x)$ |
| $= \frac{1}{3}x^2\sin 3x - \frac{2}{3}\left(-\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x\right)\{+c\}$ | A1 isw | Substituting part (a) result |
| $= \frac{1}{3}x^2\sin 3x + \frac{2}{9}x\cos 3x - \frac{2}{27}\sin 3x \{+c\}$ | | Ignore subsequent working |

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2. (a) Use integration by parts to find $\int x \sin 3 x \mathrm {~d} x$.\\
(b) Using your answer to part (a), find $\int x ^ { 2 } \cos 3 x \mathrm {~d} x$.\\

\hfill \mbox{\textit{Edexcel C4 2012 Q2 [6]}}
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