9 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x\).
Given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x\), prove that, for \(n > 1\),
$$I _ { n } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 }$$
By first using the substitution \(x = \cos ^ { - 1 } u\), find the value of
$$\int _ { 0 } ^ { 1 } \left( \cos ^ { - 1 } u \right) ^ { 3 } \mathrm {~d} u$$
giving your answer in an exact form.
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Question 9:
Answer Marks
Guidance
Answer Marks
Guidance
\(\int_0^{\pi/2} x\sin x\,dx = \left[-x\cos x\right]_0^{\pi/2} + \int_0^{\pi/2}\cos x\,dx = 0 + \left[\sin x\right]_0^{\pi/2} = 1\) M1A1
\(I_n = \left[-x^n\cos x\right]_0^{\pi/2} + \int_0^{\pi/2} nx^{n-1}\cos x\,dx\) M1A1
\(= \left[-x^n\cos x + nx^{n-1}\sin x\right]_0^{\pi/2} - \int_0^{\pi/2} n(n-1)x^{n-2}\sin x\,dx\) A1
\(= n\left(\dfrac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2}\) A1
\(x = \cos^{-1}u \Rightarrow u = \cos x \Rightarrow \dfrac{du}{dx} = -\sin x\); \(u=0,1 \Rightarrow x = \pi/2, 0\) B1B1
\(\int_0^1 \left([\cos^{-1}u]^3\right)du = -\int_{\pi/2}^{0} x^3\sin x\,dx = \int_0^{\pi/2} x^3\sin x\,dx = I_3\) B1
\(I_3 = 3\left(\dfrac{\pi}{2}\right)^2 - 3\times 2\times I_1 = \dfrac{3}{4}\pi^2 - 6\) M1A1
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## Question 9:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\pi/2} x\sin x\,dx = \left[-x\cos x\right]_0^{\pi/2} + \int_0^{\pi/2}\cos x\,dx = 0 + \left[\sin x\right]_0^{\pi/2} = 1$ | M1A1 | |
| $I_n = \left[-x^n\cos x\right]_0^{\pi/2} + \int_0^{\pi/2} nx^{n-1}\cos x\,dx$ | M1A1 | |
| $= \left[-x^n\cos x + nx^{n-1}\sin x\right]_0^{\pi/2} - \int_0^{\pi/2} n(n-1)x^{n-2}\sin x\,dx$ | A1 | |
| $= n\left(\dfrac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2}$ | A1 | |
| $x = \cos^{-1}u \Rightarrow u = \cos x \Rightarrow \dfrac{du}{dx} = -\sin x$; $u=0,1 \Rightarrow x = \pi/2, 0$ | B1B1 | |
| $\int_0^1 \left([\cos^{-1}u]^3\right)du = -\int_{\pi/2}^{0} x^3\sin x\,dx = \int_0^{\pi/2} x^3\sin x\,dx = I_3$ | B1 | |
| $I_3 = 3\left(\dfrac{\pi}{2}\right)^2 - 3\times 2\times I_1 = \dfrac{3}{4}\pi^2 - 6$ | M1A1 | OE |
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9 Evaluate $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x$.
Given that $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x$, prove that, for $n > 1$,
$$I _ { n } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 }$$
By first using the substitution $x = \cos ^ { - 1 } u$, find the value of
$$\int _ { 0 } ^ { 1 } \left( \cos ^ { - 1 } u \right) ^ { 3 } \mathrm {~d} u$$
giving your answer in an exact form.
\hfill \mbox{\textit{CAIE FP1 2016 Q9 [11]}}