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2014
P1 Q8
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Maclaurin
14 P1 Q8
2014 H2 Mathematics Paper 1 Question 8
Maclaurin Series
Answers
(i)
sin
−
1
(
x
3
)
+
C
{\sin^{-1} \left(\frac{x}{3}\right) +C}
sin
−
1
(
3
x
)
+
C
(ii)
1
3
+
1
54
x
2
+
1
648
x
4
+
5
34992
x
6
+
…
\frac{1}{3} + \frac{1}{54} x^2 + \frac{1}{648} x^4 + \frac{5}{34992} x^6 + \ldots
3
1
+
54
1
x
2
+
648
1
x
4
+
34992
5
x
6
+
…
(iii)
1
3
x
+
1
162
x
3
+
1
3240
x
5
+
5
244944
x
7
+
…
\frac{1}{3} x + \frac{1}{162} x^3 + \frac{1}{3240} x^5 + \frac{5}{244944} x^7 + \ldots
3
1
x
+
162
1
x
3
+
3240
1
x
5
+
244944
5
x
7
+
…
Full solutions
(i)
∫
f
(
x
)
d
x
=
∫
1
9
−
x
2
d
x
=
sin
−
1
(
x
3
)
+
C
■
\begin{align*} & \int f(x) \; \mathrm{d}x \\ & = \int \frac{1}{\sqrt{9-x^2}} \; \mathrm{d}x \\ & = \sin^{-1} \left(\frac{x}{3}\right) +C \; \blacksquare \end{align*}
∫
f
(
x
)
d
x
=
∫
9
−
x
2
1
d
x
=
sin
−
1
(
3
x
)
+
C
■
(ii)
f
(
x
)
=
(
9
−
x
2
)
−
1
2
=
9
−
1
2
(
1
−
x
2
9
)
−
1
2
=
1
3
(
1
−
1
2
(
−
x
2
9
)
=
1
3
(
1
−
+
−
1
2
(
−
1
2
−
1
)
2
(
−
x
2
9
)
2
=
1
3
(
1
−
+
+
−
1
2
(
−
1
2
−
1
)
(
−
1
2
−
2
)
3
!
(
−
x
2
9
)
3
=
1
3
(
1
−
+
+
+
…
)
=
1
3
(
1
+
x
2
18
+
3
8
⋅
1
81
x
4
+
5
16
⋅
1
729
x
6
+
…
)
=
1
3
+
1
54
x
2
+
1
648
x
4
+
5
34992
x
6
+
…
■
\begin{align*} & f(x) \\ & = \left( 9-x^2 \right)^{- \frac{1}{2}} \\ & = 9^{- \frac{1}{2}} \left( 1 - \frac{x^2}{9} \right)^{- \frac{1}{2}} \\ & = \frac{1}{3} \Bigg( 1 - \frac{1}{2} \left( - \frac{x^2}{9} \right) \\ & \phantom{= \frac{1}{3} \Bigg( 1 -} + \frac{- \frac{1}{2}\left(- \frac{1}{2}-1\right)}{2} \left( - \frac{x^2}{9} \right)^2 \\ & \phantom{= \frac{1}{3} \Bigg( 1 - +} + \frac{- \frac{1}{2}\left(- \frac{1}{2}-1\right)\left(- \frac{1}{2}-2\right)}{3!} \left( - \frac{x^2}{9} \right)^3 \\ & \phantom{= \frac{1}{3} \Bigg( 1 - + +} +\ldots \Bigg) \\ & = \frac{1}{3} \left( 1 + \frac{x^2}{18} + \frac{3}{8} \cdot \frac{1}{81} x^4 + \frac{5}{16} \cdot \frac{1}{729}x^6 + \ldots \right) \\ &= \frac{1}{3} + \frac{1}{54} x^2 + \frac{1}{648} x^4 + \frac{5}{34992} x^6 + \ldots \; \blacksquare \end{align*}
f
(
x
)
=
(
9
−
x
2
)
−
2
1
=
9
−
2
1
(
1
−
9
x
2
)
−
2
1
=
3
1
(
1
−
2
1
(
−
9
x
2
)
=
3
1
(
1
−
+
2
−
2
1
(
−
2
1
−
1
)
(
−
9
x
2
)
2
=
3
1
(
1
−
+
+
3
!
−
2
1
(
−
2
1
−
1
)
(
−
2
1
−
2
)
(
−
9
x
2
)
3
=
3
1
(
1
−
+
+
+
…
)
=
3
1
(
1
+
18
x
2
+
8
3
⋅
81
1
x
4
+
16
5
⋅
729
1
x
6
+
…
)
=
3
1
+
54
1
x
2
+
648
1
x
4
+
34992
5
x
6
+
…
■
(iii)
sin
−
1
(
1
3
x
)
=
∫
f
(
x
)
d
x
=
∫
(
1
3
+
1
54
x
2
+
1
648
x
4
+
5
34992
x
6
+
…
)
d
x
=
1
3
x
+
1
162
x
3
+
1
3240
x
5
+
5
244944
x
7
+
…
■
\begin{align*} & \sin^{-1} \left({\textstyle \frac{1}{3}x}\right) \\ &= \int f(x) \; \mathrm{d}x \\ &= \int \left( \frac{1}{3} + \frac{1}{54} x^2 + \frac{1}{648} x^4 + \frac{5}{34992} x^6 + \ldots \right) \; \mathrm{d}x \\ &= \frac{1}{3} x + \frac{1}{162} x^3 + \frac{1}{3240} x^5 + \frac{5}{244944} x^7 + \ldots \; \blacksquare \end{align*}
sin
−
1
(
3
1
x
)
=
∫
f
(
x
)
d
x
=
∫
(
3
1
+
54
1
x
2
+
648
1
x
4
+
34992
5
x
6
+
…
)
d
x
=
3
1
x
+
162
1
x
3
+
3240
1
x
5
+
244944
5
x
7
+
…
■
We note that
sin
−
1
(
0
)
=
0
{\sin^{-1} (0) = 0}
sin
−
1
(
0
)
=
0
so the constant of integration
C
=
0
{C=0}
C
=
0
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