C4 January 2012 Q2

Notes

M1: Use of ‘integration by parts’ formula \(uv - \displaystyle\int vu'\) (whether stated or not stated) in the correct direction, where \(u = x \to u' = 1\) and \(v' = \sin 3x \to v = k\cos 3x\) (seen or implied), where \(k\) is a positive or negative constant. (Allow \(k = 1\)).
This means that the candidate must achieve \(x(k\cos 3x) - \displaystyle\int (k\cos 3x)\), where \(k\) is a consistent constant.
If \(x^2\) appears after the integral, this would imply that the candidate is applying integration by parts in the wrong direction, so M0.

A1: \(-\dfrac{1}{3}x\cos 3x - \displaystyle\int -\frac{1}{3}\cos 3x\ \{\mathrm{d}x\}\). Can be un-simplified. Ignore the \(\{\mathrm{d}x\}\).

A1: \(-\dfrac{1}{3}x\cos 3x + \dfrac{1}{9}\sin 3x\) with/without \(+\,c\). Can be un-simplified.

Notes

M1: Use of ‘integration by parts’ formula \(uv - \displaystyle\int vu'\) (whether stated or not stated) in the correct direction, where \(u = x^2 \to u' = 2x\) or \(x\) and \(v' = \cos 3x \to v = \lambda\sin 3x\) (seen or implied), where \(\lambda\) is a positive or negative constant. (Allow \(\lambda = 1\)).
This means that the candidate must achieve \(x^2(\lambda\sin 3x) - \displaystyle\int 2x(\lambda\sin 3x)\), where \(u' = 2x\)
or \(x^2(\lambda\sin 3x) - \displaystyle\int x(\lambda\sin 3x)\), where \(u' = x\).
If \(x^3\) appears after the integral, this would imply that the candidate is applying integration by parts in the wrong direction, so M0.

A1: \(\dfrac{1}{3}x^2\sin 3x - \displaystyle\int \frac{2}{3}x\sin 3x\ \{\mathrm{d}x\}\). Can be un-simplified. Ignore the \(\{\mathrm{d}x\}\).

A1: \(\dfrac{1}{3}x^2\sin 3x - \dfrac{2}{3}\left(-\dfrac{1}{3}x\cos 3x + \dfrac{1}{9}\sin 3x\right)\) with/without \(+\,c\), can be un-simplified.
You can ignore subsequent working here.

Special Case: If the candidate scores the first two marks of M1A1 in part (b), then you can award the final A1 as a follow through for \(\dfrac{1}{3}x^2\sin 3x - \dfrac{2}{3}(\text{their follow through part(a) answer})\).