3. Use the substitution \(x = \sin \theta\) to show that, for \(| x | \leq 1\),
$$\int \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x = \frac { x } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } } + c \text {, where } c \text { is an arbitrary constant. }$$
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Question 3:
Answer Marks
Guidance
Let \(x = \sin\theta\), \(\frac{dx}{d\theta} = \cos\theta\) M1
Substitution stated
\(1 - x^2 = 1 - \sin^2\theta = \cos^2\theta\) M1
Using identity
\((1-x^2)^{3/2} = \cos^3\theta\) Integral becomes \(\int \frac{\cos\theta}{\cos^3\theta} d\theta = \int \sec^2\theta\, d\theta\) A1
Correct simplified integrand
\(= \tan\theta + c\) A1
Integrating correctly
\(\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{x}{\sqrt{1-x^2}} = \frac{x}{(1-x^2)^{1/2}}\) M1 A1
Converting back correctly to given form
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# Question 3:
| Let $x = \sin\theta$, $\frac{dx}{d\theta} = \cos\theta$ | M1 | Substitution stated |
| $1 - x^2 = 1 - \sin^2\theta = \cos^2\theta$ | M1 | Using identity |
| $(1-x^2)^{3/2} = \cos^3\theta$ | | |
| Integral becomes $\int \frac{\cos\theta}{\cos^3\theta} d\theta = \int \sec^2\theta\, d\theta$ | A1 | Correct simplified integrand |
| $= \tan\theta + c$ | A1 | Integrating correctly |
| $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{x}{\sqrt{1-x^2}} = \frac{x}{(1-x^2)^{1/2}}$ | M1 A1 | Converting back correctly to given form |
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3. Use the substitution $x = \sin \theta$ to show that, for $| x | \leq 1$,
$$\int \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x = \frac { x } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } } + c \text {, where } c \text { is an arbitrary constant. }$$
\hfill \mbox{\textit{Edexcel C4 Q3 [6]}}