C4 June 2007 Q3
3.
| Scheme | Marks |
|---|---|
| \(\left\{\begin{aligned} u &= x &&\Rightarrow\ \tfrac{\mathrm{d}u}{\mathrm{d}x} = 1\\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= \cos 2x &&\Rightarrow\ v = \tfrac{1}{2}\sin 2x\end{aligned}\right\}\) (see note below) | |
| \(\text{Int} = \displaystyle\int x\cos 2x\,\mathrm{d}x = \tfrac{1}{2}x\sin 2x - \int \tfrac{1}{2}\sin 2x.1\,\mathrm{d}x\) | M1 A1 |
| \(= \tfrac{1}{2}x\sin 2x - \tfrac{1}{2}\left(-\tfrac{1}{2}\cos 2x\right) + c\) | dM1 |
| \(= \tfrac{1}{2}x\sin 2x + \tfrac{1}{4}\cos 2x + c\) | A1 |
| (4) |
Notes
M1: Use of ‘integration by parts’ formula in the correct direction. A1: Correct expression.
dM1: \(\sin 2x \to -\tfrac{1}{2}\cos 2x\) or \(\sin kx \to -\tfrac{1}{k}\cos kx\) with \(k \ne 1,\ k \gt 0\)
A1: Correct expression with \(+c\)
| Scheme | Marks |
|---|---|
| \(\text{Int} = \displaystyle\int x\cos 2x\,\mathrm{d}x = \tfrac{1}{2}x\sin 2x \pm \int \tfrac{1}{2}\sin 2x.1\,\mathrm{d}x\) This is acceptable for M1 | M1 |
| \(\left\{\begin{aligned} u &= x &&\Rightarrow\ \tfrac{\mathrm{d}u}{\mathrm{d}x} = 1\\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= \cos 2x &&\Rightarrow\ v = \lambda\sin 2x\end{aligned}\right\}\) | |
| \(\text{Int} = \displaystyle\int x\cos 2x\,\mathrm{d}x = \lambda x\sin 2x \pm \int \lambda\sin 2x.1\,\mathrm{d}x\) This is also acceptable for M1 | M1 |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int x\cos^2 x\,\mathrm{d}x = \int x\left(\tfrac{\cos 2x + 1}{2}\right)\mathrm{d}x\) | M1 |
| \(= \dfrac{1}{2}\displaystyle\int x\cos 2x\,\mathrm{d}x + \frac{1}{2}\int x\,\mathrm{d}x\) | |
| \(= \underline{\dfrac{1}{2}\left(\dfrac{1}{2}x\sin 2x + \dfrac{1}{4}\cos 2x\right)};\ + \dfrac{1}{2}\displaystyle\int x\,\mathrm{d}x\) | A1ft |
| \(= \dfrac{1}{4}x\sin 2x + \dfrac{1}{8}\cos 2x + \dfrac{1}{4}x^2\ (+c)\) | A1 |
| (3) | |
| (7 marks) |
Notes
M1: Substitutes correctly for \(\cos^2 x\) in the given integral
A1ft: \(\tfrac{1}{2}\)(their answer to (a)); or underlined expression
A1: Completely correct expression with/without \(+c\)
Aliter (b) Way 2
| Scheme | Marks |
|---|---|
| \(\displaystyle\int x\cos^2 x\,\mathrm{d}x = \int x\left(\tfrac{\cos 2x + 1}{2}\right)\mathrm{d}x\) | M1 |
| \(\left\{\begin{aligned} u &= x &&\Rightarrow\ \tfrac{\mathrm{d}u}{\mathrm{d}x} = 1\\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= \tfrac{1}{2}\cos 2x + \tfrac{1}{2} &&\Rightarrow\ v = \tfrac{1}{4}\sin 2x + \tfrac{1}{2}x\end{aligned}\right\}\) | |
| \(= \tfrac{1}{4}x\sin 2x + \tfrac{1}{2}x^2 - \displaystyle\int\left(\tfrac{1}{4}\sin 2x + \tfrac{1}{2}x\right)\mathrm{d}x\) | |
| \(= \underline{\tfrac{1}{4}x\sin 2x} + \tfrac{1}{2}x^2 + \underline{\tfrac{1}{8}\cos 2x} - \tfrac{1}{4}x^2 + c\) | A1ft |
| \(= \dfrac{1}{4}x\sin 2x + \dfrac{1}{8}\cos 2x + \dfrac{1}{4}x^2\ (+c)\) | A1 |
| (3) |
M1: Substitutes correctly for \(\cos^2 x\) in the given integral … or \(u = x\) and \(\tfrac{\mathrm{d}v}{\mathrm{d}x} = \tfrac{1}{2}\cos 2x + \tfrac{1}{2}\)
A1ft: \(\tfrac{1}{2}\)(their answer to (a)); or underlined expression. A1: Completely correct expression with/without \(+c\)
Aliter (b) Way 3
| Scheme | Marks |
|---|---|
| \(\displaystyle\int x\cos 2x\,\mathrm{d}x = \int x\left(2\cos^2 x - 1\right)\mathrm{d}x\) | M1 |
| \(\Rightarrow 2\displaystyle\int x\cos^2 x\,\mathrm{d}x - \int x\,\mathrm{d}x = \tfrac{1}{2}x\sin 2x + \tfrac{1}{4}\cos 2x + c\) | |
| \(\Rightarrow \displaystyle\int x\cos^2 x\,\mathrm{d}x = \underline{\frac{1}{2}\left(\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x\right)};\ + \frac{1}{2}\int x\,\mathrm{d}x\) | A1ft |
| \(= \dfrac{1}{4}x\sin 2x + \dfrac{1}{8}\cos 2x + \dfrac{1}{4}x^2\ (+c)\) | A1 |
| (3) |
M1: Substitutes correctly for \(\cos 2x\) in \(\displaystyle\int x\cos 2x\,\mathrm{d}x\)
A1ft: \(\tfrac{1}{2}\)(their answer to (a)); or underlined expression. A1: Completely correct expression with/without \(+c\)