C4 June 2014 Q6

Notes

M1: Integration by parts is applied in the form \(\pm\alpha x\mathrm{e}^{4x} - \displaystyle\int \beta\mathrm{e}^{4x}\{\mathrm{d}x\}\), where \(\alpha \neq 0,\ \beta > 0\).
(must be in this form).

A1: \(\dfrac{1}{4}x\mathrm{e}^{4x} - \displaystyle\int \dfrac{1}{4}\mathrm{e}^{4x}\{\mathrm{d}x\}\) or equivalent.

A1: \(\dfrac{1}{4}x\mathrm{e}^{4x} - \dfrac{1}{16}\mathrm{e}^{4x}\) with/without \(+\,c\). Can be un-simplified.

isw: You can ignore subsequent working following on from a correct solution.

SC: SPECIAL CASE: A candidate who uses \(u = x\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{4x}\), writes down the correct “by parts” formula,
but makes only one error when applying it can be awarded Special Case M1.

Notes

M1: \(\pm\lambda(2x - 1)^{-2},\ \lambda \neq 0\). Note that \(\lambda\) can be 1.

A1: \(\dfrac{8(2x - 1)^{-2}}{(2)(-2)}\) or \(-2(2x - 1)^{-2}\) or \(\dfrac{-2}{(2x - 1)^2}\) with/without \(+\,c\). Can be un-simplified.

Note: You can ignore subsequent working which follows from a correct answer.

Notes

B1: Separates variables as shown. \(\mathrm{d}y\) and \(\mathrm{d}x\) should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs.

Note: Allow B1 for \(\displaystyle\int \dfrac{1}{\operatorname{cosec} 2y\,\operatorname{cosec} y} = \displaystyle\int \mathrm{e}^x\) or \(\displaystyle\int \sin 2y\sin y = \displaystyle\int \mathrm{e}^x\)

M1: \(\dfrac{1}{\operatorname{cosec} 2y} \to 2\sin y\cos y\) or \(\sin 2y \to 2\sin y\cos y\) or \(\sin 2y\sin y \to \pm\lambda\cos 3y \pm \lambda\cos y\)
seen anywhere in the candidate’s working to (iii).

M1: Integrates to give \(\pm\mu\sin^3 y,\ \mu \neq 0\) or \(\pm\alpha\sin 3y \pm \beta\sin y,\ \alpha \neq 0,\ \beta \neq 0\)

A1: \(2\sin^2 y\cos y \to \dfrac{2}{3}\sin^3 y\) (with no extra terms) or integrates to give \(-\dfrac{1}{2}\left(\dfrac{1}{3}\sin 3y - \sin y\right)\)

B1: Evidence that \(\mathrm{e}^x\) has been integrated to give \(\mathrm{e}^x\) as part of solving their DE.

M1: Some evidence of using both \(y = \dfrac{\pi}{6}\) and \(x = 0\) in an integrated or changed equation containing \(c\).

Note: that is mark can be implied by the correct value of \(c\).

A1: \(\dfrac{2}{3}\sin^3 y = \mathrm{e}^x - \dfrac{11}{12}\) or \(-\dfrac{1}{6}\sin 3y + \dfrac{1}{2}\sin y = \mathrm{e}^x - \dfrac{11}{12}\) or any equivalent correct answer.

Note: You can ignore subsequent working which follows from a correct answer.

Alternative Method 1

Alternative Method 2 (Using integration by parts twice)