11 Show that \(\int x \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = - \frac { 1 } { 3 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } + c\), where \(c\) is a constant.
Given that \(I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x\), prove that, for \(n \geqslant 2\),
$$( n + 2 ) I _ { n } = ( n - 1 ) I _ { n - 2 }$$
Use the substitution \(x = \sin u\) to show that
$$\int _ { 0 } ^ { 1 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } \pi$$
Find \(I _ { 4 }\).
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Question 11:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\frac{d}{dx}\left(-\frac{1}{3}(1-x^2)^{\frac{3}{2}}+c\right) = -\frac{1}{3} \times \frac{3}{2}(1-x^2)^{\frac{1}{2}} \times (-2x) = x(1-x^2)^{\frac{1}{2}}\) B1
Verifies result
\(I_n = \int_0^1 x^n(1-x^2)^{\frac{1}{2}}\,dx = \int_0^1 x^{n-1} \cdot x(1-x^2)^{\frac{1}{2}}\,dx\) M1
Integrates by parts, correct parts
\(= \left[-x^{n-1}\cdot\frac{1}{3}(1-x^2)^{\frac{3}{2}}\right]_0^1 + \int_0^1(n-1)x^{n-2}\cdot\frac{1}{3}(1-x^2)^{\frac{3}{2}}\,dx\) M1A1
Substitutes limits
\(= \frac{n-1}{3}\int_0^1 x^{n-2}(1-x^2)(1-x^2)^{\frac{1}{2}}\,dx\) M1
Obtains reduction formula
\(= \frac{n-1}{3}I_{n-2} - \frac{n-1}{3}I_n\) \(\Rightarrow (n+2)I_n = (n-1)I_{n-2}\) A1
AG — Part marks: 5
\(x = \sin u \Rightarrow dx = \cos u\,du\); Limits: \(x=0 \Rightarrow u=0\), \(x=1 \Rightarrow u=\frac{\pi}{2}\) M1
Uses substitution correctly
\(\int_0^1(1-x^2)^{\frac{1}{2}}\,dx = \int_0^{\frac{\pi}{2}}\cos^2 u\,du\) A1
Uses double angle formula
\(= \int_0^{\frac{\pi}{2}}\frac{1}{2}(\cos 2u+1)\,du\) M1
Integrates correctly
\(= \frac{1}{2}\left[\frac{\sin 2u}{2}+u\right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}\) M1A1
AG — Part marks: 5
\(\Rightarrow I_2 = \frac{1}{4}\times\frac{\pi}{4} = \frac{\pi}{16} \Rightarrow I_4 = \frac{1}{2}\times\frac{\pi}{16} = \frac{\pi}{32}\) M1A1
Part marks: 2
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# Question 11:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}\left(-\frac{1}{3}(1-x^2)^{\frac{3}{2}}+c\right) = -\frac{1}{3} \times \frac{3}{2}(1-x^2)^{\frac{1}{2}} \times (-2x) = x(1-x^2)^{\frac{1}{2}}$ | B1 | Verifies result |
| $I_n = \int_0^1 x^n(1-x^2)^{\frac{1}{2}}\,dx = \int_0^1 x^{n-1} \cdot x(1-x^2)^{\frac{1}{2}}\,dx$ | M1 | Integrates by parts, correct parts |
| $= \left[-x^{n-1}\cdot\frac{1}{3}(1-x^2)^{\frac{3}{2}}\right]_0^1 + \int_0^1(n-1)x^{n-2}\cdot\frac{1}{3}(1-x^2)^{\frac{3}{2}}\,dx$ | M1A1 | Substitutes limits |
| $= \frac{n-1}{3}\int_0^1 x^{n-2}(1-x^2)(1-x^2)^{\frac{1}{2}}\,dx$ | M1 | Obtains reduction formula |
| $= \frac{n-1}{3}I_{n-2} - \frac{n-1}{3}I_n$ | | |
| $\Rightarrow (n+2)I_n = (n-1)I_{n-2}$ | A1 | AG — **Part marks: 5** |
| $x = \sin u \Rightarrow dx = \cos u\,du$; Limits: $x=0 \Rightarrow u=0$, $x=1 \Rightarrow u=\frac{\pi}{2}$ | M1 | Uses substitution correctly |
| $\int_0^1(1-x^2)^{\frac{1}{2}}\,dx = \int_0^{\frac{\pi}{2}}\cos^2 u\,du$ | A1 | Uses double angle formula |
| $= \int_0^{\frac{\pi}{2}}\frac{1}{2}(\cos 2u+1)\,du$ | M1 | Integrates correctly |
| $= \frac{1}{2}\left[\frac{\sin 2u}{2}+u\right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}$ | M1A1 | AG — **Part marks: 5** |
| $\Rightarrow I_2 = \frac{1}{4}\times\frac{\pi}{4} = \frac{\pi}{16} \Rightarrow I_4 = \frac{1}{2}\times\frac{\pi}{16} = \frac{\pi}{32}$ | M1A1 | **Part marks: 2** |
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11 Show that $\int x \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = - \frac { 1 } { 3 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } + c$, where $c$ is a constant.
Given that $I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x$, prove that, for $n \geqslant 2$,
$$( n + 2 ) I _ { n } = ( n - 1 ) I _ { n - 2 }$$
Use the substitution $x = \sin u$ to show that
$$\int _ { 0 } ^ { 1 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } \pi$$
Find $I _ { 4 }$.
\hfill \mbox{\textit{CAIE FP1 2012 Q11 [13]}}