C4 June 2005 Q4
4. Use the substitution \(x = \sin\theta\) to find the exact value of
\[\int_0^{\frac{1}{2}} \frac{1}{(1 - x^2)^{\frac{3}{2}}}\,\mathrm{d}x.\]
(7)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{(1 - x^2)^{\frac{3}{2}}}\,\mathrm{d}x = \int \frac{1}{(1 - \sin^2\theta)^{\frac{3}{2}}}\cos\theta\,\mathrm{d}\theta\) Use of \(x = \sin\theta\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \cos\theta\) | M1 |
| \(\displaystyle= \int \frac{1}{\cos^2\theta}\,\mathrm{d}\theta\) | M1 A1 |
| \(\displaystyle= \int \sec^2\theta\,\mathrm{d}\theta = \tan\theta\) | M1 A1 |
| Using the limits 0 and \(\frac{\pi}{6}\) to evaluate integral | M1 |
| \(\Big[\tan\theta\Big]_0^{\frac{\pi}{6}} = \dfrac{1}{\surd 3} \quad \left(= \dfrac{\surd 3}{3}\right)\) cao | A1 |
| (7) | |
| (7 marks) |
Alternative for final M1 A1
| Returning to the variable \(x\) and using the limits 0 and \(\frac{1}{2}\) to evaluate integral | M1 |
| \(\left[\dfrac{x}{\surd(1 - x^2)}\right]_0^{\frac{1}{2}} = \dfrac{1}{\surd 3} \quad \left(= \dfrac{\surd 3}{3}\right)\) cao | A1 |