8. (i) Find \(\int x \sin x d x\)
(ii) (a) Use the substitution \(x = \sec \theta\) to show that
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Question 8:
Part (i):
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\int x\sin x\,dx = -x\cos x + \int \cos x\,dx\) M1
Attempts IBP the correct way: \(\int x\sin x\,dx = \pm x\cos x \pm \int \cos x\,dx\)
\(= -x\cos x + \sin x\,(+c)\) dM1A1
dM1: integrates again; A1: correct answer. Also accept \(-\cos x\cdot x + \sin x\,(+c)\)
Part (ii)(a):
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(dx = \sec\theta\tan\theta\,d\theta\) B1
Or equivalent, e.g. \(\frac{dx}{d\theta}=\sec\theta\tan\theta\)
\(\int_1^2\sqrt{1-\frac{1}{x^2}}\,dx = \int_0^{\pi/3}\sqrt{1-\cos^2\theta}\,\sec\theta\tan\theta\,d\theta\) M1
Substitutes \(x=\sec\theta\), simplifies to \(\sin\theta\), \(\frac{1}{\csc\theta}\), \(\frac{\tan\theta}{\sec\theta}\) or \(\sqrt{\sin^2\theta}\,\sqrt{\frac{1}{\csc^2\theta}}\)
\(= \int_0^{\pi/3}\sqrt{\sin^2\theta}\times\frac{1}{\cos\theta}\tan\theta\,d\theta = \int_0^{\pi/3}\tan^2\theta\,d\theta\) A1*
Completes proof; \(\sec\theta\) replaced by \(\frac{1}{\cos\theta}\) or cancelled; correct limits and notation
Part (ii)(b):
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\int\sqrt{1-\frac{1}{x^2}}\,dx = \int\tan^2\theta\,d\theta = \int(\sec^2\theta-1)\,d\theta = \tan\theta - \theta\) M1A1
M1: \(\int\tan^2\theta\,d\theta = \int(\pm\sec^2\theta \pm 1)\,d\theta = \pm\tan\theta\pm\theta\); A1: \(\tan\theta-\theta\)
\(\int_1^2\sqrt{1-\frac{1}{x^2}}\,dx = \left[\tan\theta-\theta\right]_{\theta=0}^{\theta=\pi/3} = \tan\!\left(\frac{\pi}{3}\right)-\frac{\pi}{3} = \sqrt{3}-\frac{\pi}{3}\) dM1A1
dM1: uses limits \(\frac{\pi}{3}\) and \(0\); A1: \(\sqrt{3}-\frac{\pi}{3}\)
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## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x\sin x\,dx = -x\cos x + \int \cos x\,dx$ | M1 | Attempts IBP the correct way: $\int x\sin x\,dx = \pm x\cos x \pm \int \cos x\,dx$ |
| $= -x\cos x + \sin x\,(+c)$ | dM1A1 | dM1: integrates again; A1: correct answer. Also accept $-\cos x\cdot x + \sin x\,(+c)$ |
### Part (ii)(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $dx = \sec\theta\tan\theta\,d\theta$ | B1 | Or equivalent, e.g. $\frac{dx}{d\theta}=\sec\theta\tan\theta$ |
| $\int_1^2\sqrt{1-\frac{1}{x^2}}\,dx = \int_0^{\pi/3}\sqrt{1-\cos^2\theta}\,\sec\theta\tan\theta\,d\theta$ | M1 | Substitutes $x=\sec\theta$, simplifies to $\sin\theta$, $\frac{1}{\csc\theta}$, $\frac{\tan\theta}{\sec\theta}$ or $\sqrt{\sin^2\theta}\,\sqrt{\frac{1}{\csc^2\theta}}$ |
| $= \int_0^{\pi/3}\sqrt{\sin^2\theta}\times\frac{1}{\cos\theta}\tan\theta\,d\theta = \int_0^{\pi/3}\tan^2\theta\,d\theta$ | A1* | Completes proof; $\sec\theta$ replaced by $\frac{1}{\cos\theta}$ or cancelled; correct limits and notation |
### Part (ii)(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\sqrt{1-\frac{1}{x^2}}\,dx = \int\tan^2\theta\,d\theta = \int(\sec^2\theta-1)\,d\theta = \tan\theta - \theta$ | M1A1 | M1: $\int\tan^2\theta\,d\theta = \int(\pm\sec^2\theta \pm 1)\,d\theta = \pm\tan\theta\pm\theta$; A1: $\tan\theta-\theta$ |
| $\int_1^2\sqrt{1-\frac{1}{x^2}}\,dx = \left[\tan\theta-\theta\right]_{\theta=0}^{\theta=\pi/3} = \tan\!\left(\frac{\pi}{3}\right)-\frac{\pi}{3} = \sqrt{3}-\frac{\pi}{3}$ | dM1A1 | dM1: uses limits $\frac{\pi}{3}$ and $0$; A1: $\sqrt{3}-\frac{\pi}{3}$ |
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8. (i) Find $\int x \sin x d x$\\
(ii) (a) Use the substitution $x = \sec \theta$ to show that\\
(b) Hence find the exact value of
$$\int _ { 1 } ^ { 2 } \sqrt { 1 - \frac { 1 } { x ^ { 2 } } } \mathrm {~d} x = \int _ { 0 } ^ { \frac { \pi } { 3 } } \tan ^ { 2 } \theta \mathrm {~d} \theta$$
Hence find the exact value of
$$\int _ { 1 } ^ { 2 } \sqrt { 1 - \frac { 1 } { x ^ { 2 } } } \mathrm {~d} x$$
\hfill \mbox{\textit{Edexcel C34 2018 Q8 [10]}}