3. (a) Find \(\int x \cos 2 x d x\).
Show mark scheme
Show mark scheme source
Question 3:
Part (a)
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(u=x \Rightarrow \frac{du}{dx}=1\); \(\frac{dv}{dx}=\cos 2x \Rightarrow v=\frac{1}{2}\sin 2x\) See note below
\(\int x\cos 2x\,dx = \frac{1}{2}x\sin 2x - \int \frac{1}{2}\sin 2x \cdot 1\,dx\) M1
Use of integration by parts formula in the correct direction. Correct expression.
A1 \(= \frac{1}{2}x\sin 2x - \frac{1}{2}\left(-\frac{1}{2}\cos 2x\right)+c\) dM1
\(\sin 2x \to -\frac{1}{2}\cos 2x\) or \(\sin kx \to -\frac{1}{k}\cos kx\) with \(k\neq 1\), \(k>0\)
\(= \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + c\) A1
Correct expression with \(+c\)
Total: 4 marks [4]
Part (b)
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\int x\cos^2 x\,dx = \int x\left(\frac{\cos 2x+1}{2}\right)dx\) M1
Substitutes correctly for \(\cos^2 x\) in the given integral
\(= \frac{1}{2}\int x\cos 2x\,dx + \frac{1}{2}\int x\,dx\) \(= \frac{1}{2}\left(\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x\right) + \frac{1}{2}\int x\,dx\) A1\(\checkmark\)
\(\frac{1}{2}\)(their answer to (a)); or underlined expression
\(= \frac{1}{4}x\sin 2x + \frac{1}{8}\cos 2x + \frac{1}{4}x^2\ (+c)\) A1
Completely correct expression with/without \(+c\)
Total: 3 marks [3]
Question 3 Total: 7 marks
Copy
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u=x \Rightarrow \frac{du}{dx}=1$; $\frac{dv}{dx}=\cos 2x \Rightarrow v=\frac{1}{2}\sin 2x$ | | See note below |
| $\int x\cos 2x\,dx = \frac{1}{2}x\sin 2x - \int \frac{1}{2}\sin 2x \cdot 1\,dx$ | M1 | Use of integration by parts formula in the correct direction. Correct expression. |
| | A1 | |
| $= \frac{1}{2}x\sin 2x - \frac{1}{2}\left(-\frac{1}{2}\cos 2x\right)+c$ | dM1 | $\sin 2x \to -\frac{1}{2}\cos 2x$ or $\sin kx \to -\frac{1}{k}\cos kx$ with $k\neq 1$, $k>0$ |
| $= \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + c$ | A1 | Correct expression with $+c$ |
| **Total: 4 marks** | [4] | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x\cos^2 x\,dx = \int x\left(\frac{\cos 2x+1}{2}\right)dx$ | M1 | Substitutes correctly for $\cos^2 x$ in the given integral |
| $= \frac{1}{2}\int x\cos 2x\,dx + \frac{1}{2}\int x\,dx$ | | |
| $= \frac{1}{2}\left(\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x\right) + \frac{1}{2}\int x\,dx$ | A1$\checkmark$ | $\frac{1}{2}$(their answer to (a)); or underlined expression |
| $= \frac{1}{4}x\sin 2x + \frac{1}{8}\cos 2x + \frac{1}{4}x^2\ (+c)$ | A1 | Completely correct expression with/without $+c$ |
| **Total: 3 marks** | [3] | |
| **Question 3 Total: 7 marks** | | |
Show LaTeX source
Copy
3. (a) Find $\int x \cos 2 x d x$.\\
(b) Hence, using the identity $\cos 2 x = 2 \cos ^ { 2 } x - 1$, deduce $\int x \cos ^ { 2 } x \mathrm {~d} x$.\\
\hfill \mbox{\textit{Edexcel C4 2007 Q3 [7]}}