June 2023 Paper 1 Q8

AQACurrent spec6 marksIntegration

8 Show that

\[\int_0^{\frac{\pi}{2}} (x\sin 4x)\,\mathrm{d}x = -\frac{\pi}{8}\]

[6 marks]

Mark scheme

Scheme	Marks	AO
Begins integration by parts by writing \(u = x \qquad v^{\prime} = \sin 4x\) \(u^{\prime} = 1 \qquad v = A\cos 4x\) PI by \(Ax\cos 4x - A\int (\cos 4x)\,\mathrm{d}x\) Or \(u = \sin 4x \qquad v^{\prime} = x\) \(u^{\prime} = A\cos 4x \qquad v = Bx^2\) PI by \(Px^2\sin 4x - Q\int \left(x^2\cos 4x\right)\mathrm{d}x\)	M1	3.1a
Selects the correct method for integration by parts \(u = x \qquad v^{\prime} = \sin 4x\) \(u^{\prime} = 1 \qquad v = A\cos 4x\) PI by \(Ax\cos 4x - A\int (\cos 4x)\,\mathrm{d}x\)	M1	1.1a
Substitutes their \(u, u^{\prime}, v, v^{\prime}\) of either of the above forms into the integration by parts formula. Eg \(Px\cos 4x - P\int (\cos 4x)\,\mathrm{d}x\) \(Px^2\sin 4x - Q\int \left(x^2\cos 4x\right)\mathrm{d}x\) \(\dfrac{x^2}{2}\sin 4x - \int \left(2x^2\cos 4x\right)\mathrm{d}x\) PI by \(-\dfrac{1}{4}x\cos 4x + \dfrac{1}{16}\sin 4x\)	M1	1.1a
Obtains \(-\dfrac{1}{4}x\cos 4x - \dfrac{1}{4}\int (-\cos 4x)\,\mathrm{d}x\) Condone missing d\(x\) PI by \(-\dfrac{1}{4}x\cos 4x + \dfrac{1}{16}\sin 4x\)	A1	1.1b
Completes integration by parts to obtain \(-\dfrac{1}{4}x\cos 4x + B\sin 4x\) with \(B \neq \pm 1\)	M1	1.1a
Completes reasoned argument by explicitly substituting correct limits into \(-\dfrac{1}{4}x\cos 4x + \dfrac{1}{16}\sin 4x\) To obtain \(-\dfrac{\pi}{8}\) Accept \(\left(-\dfrac{\pi}{8}\cos 2\pi + \dfrac{1}{16}\sin 2\pi\right) - 0\) AG	R1	2.1
	(6 marks)

Typical solution

\[u = x \qquad v^{\prime} = \sin 4x\]\[u^{\prime} = 1 \qquad v = -\frac{1}{4}\cos 4x\]\[\left[-\frac{1}{4}x\cos 4x\right]_0^{\frac{\pi}{2}} + \frac{1}{4}\int_0^{\frac{\pi}{2}} (\cos 4x)\,\mathrm{d}x\]\[= \left[-\frac{1}{4}x\cos 4x + \frac{1}{16}\sin 4x\right]_0^{\frac{\pi}{2}}\]\[= \left(-\frac{\pi}{8}\cos\frac{4\pi}{2} + \frac{1}{16}\sin\frac{4\pi}{2}\right) - \left(0 \times \cos 0 + \frac{1}{16}\sin 0\right)\]\[= -\frac{\pi}{8}\]