A2 June 2024 Paper 2 Q20

AQACurrent spec9 marksIntegration

20 The integral \(I_n\) is defined by

\[I_n = \int_0^{\frac{\pi}{4}} \cos^n x\,\mathrm{d}x \qquad (n \geqslant 0)\]

(a) Show that\[I_n = \left(\frac{n - 1}{n}\right)I_{n-2} + \frac{1}{n\left(2^{\frac{n}{2}}\right)} \qquad (n \geqslant 2)\] [6 marks]

(b) Use the result from part (a) to show that\[\int_0^{\frac{\pi}{4}} \cos^6 x\,\mathrm{d}x = \frac{a\pi + b}{192}\]

where \(a\) and \(b\) are integers to be found. [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Uses integration by parts.	M1	3.1a
Obtains a correct result of integration by parts.	A1	1.1b
Substitutes limits into first expression on RHS and simplifies.	M1	1.1a
Uses a trig identity to obtain an equation involving \(I_n\) and \(I_{n-2}\)	M1	3.1a
Rearranges to make \(I_n\) the subject of a three term equation	M1	1.1a
Completes a rigorous argument to reach the required result. AG	R1	2.1
	(6)

Typical solution

\[I_n = \int_0^{\frac{\pi}{4}} \cos^n x\,\mathrm{d}x = \int_0^{\frac{\pi}{4}} \cos^{n-1} x\cos x\,\mathrm{d}x\]\[u = \cos^{n-1} x \qquad v^{\prime} = \cos x\]\[u^{\prime} = -(n - 1)\cos^{n-2} x\sin x \qquad v = \sin x\]\[I_n = \Big[\sin x\cos^{n-1} x\Big]_0^{\frac{\pi}{4}} + (n - 1)\int_0^{\frac{\pi}{4}} \cos^{n-2} x\sin^2 x\,\mathrm{d}x\]\[I_n = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right)^{n-1} + (n - 1)\int_0^{\frac{\pi}{4}} \cos^{n-2} x\left(1 - \cos^2 x\right)\mathrm{d}x\]\[I_n = \left(\frac{1}{\sqrt{2}}\right)^{n} + (n - 1)\int_0^{\frac{\pi}{4}} \cos^{n-2} x\,\mathrm{d}x - (n - 1)\int_0^{\frac{\pi}{4}} \cos^{n} x\,\mathrm{d}x\]\[I_n = \frac{1}{2^{\frac{n}{2}}} + (n - 1)I_{n-2} - (n - 1)I_n\]\[nI_n = \frac{1}{2^{\frac{n}{2}}} + (n - 1)I_{n-2}\]\[I_n = \left(\frac{n - 1}{n}\right)I_{n-2} + \frac{1}{n\left(2^{\frac{n}{2}}\right)}\]

Mark scheme (b)

Scheme	Marks	AO
Obtains and uses \(I_0 = \dfrac{\pi}{4}\)	B1	1.1b
Uses the formula three times.	M1	1.1a
Completes fully correct working to obtain \(\dfrac{15\pi + 44}{192}\)	R1	2.1
	(3)
	(9 marks)

Typical solution

\[\begin{aligned} I_0 &= \frac{\pi}{4} \\ I_2 &= \frac{1}{2}\left(\frac{\pi}{4}\right) + \frac{1}{2}(2^{-1}) \\ &= \frac{\pi}{8} + \frac{1}{4} \\ I_4 &= \frac{3}{4}\left(\frac{\pi}{8} + \frac{1}{4}\right) + \frac{1}{4}(2^{-2}) \\ &= \frac{3\pi}{32} + \frac{3}{16} + \frac{1}{16} \\ &= \frac{3\pi}{32} + \frac{1}{4} \\ I_6 &= \frac{5}{6}\left(\frac{3\pi}{32} + \frac{1}{4}\right) + \frac{1}{6}(2^{-3}) \\ &= \frac{15\pi}{192} + \frac{5}{24} + \frac{1}{48} \\ &= \frac{15\pi + 44}{192} \end{aligned}\]