Use the substitution $x = a\sec \theta$ to show that $\int_{a\sqrt 2 }^{2a} {\frac{{{\text{d}}x}}{{{x^3}\sqrt {{x^2} - {a^2}} }} = \frac{1}{{24{a^3}}}\left( {3\sqrt 3  + \pi  - 6} \right)}$.

$x = a\sec \theta$

$\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = a\sec \theta \tan \theta$     (A1)

new limits:

$x = a\sqrt 2  \Rightarrow \theta  = \frac{\pi }{4}$ and $x = 2a \Rightarrow \theta  = \frac{\pi }{3}$     (A1)

$\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{a\sec \theta \tan \theta }}{{{a^3}{{\sec }^3}\theta \sqrt {{a^2}{{\sec }^2}\theta  - {a^2}} }}{\text{d}}\theta }$     M1

$= \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}\theta }}{{{a^3}}}{\text{d}}\theta }$     A1

using ${\cos ^2}\theta  = \frac{1}{2}(\cos 2\theta  + 1)$     M1

$\frac{1}{{2{a^3}}}\left[ {\frac{1}{2}\sin 2\theta  + \theta } \right]_{\frac{\pi }{4}}^{\frac{\pi }{3}}$ or equivalent     A1

$= \frac{1}{{4{a^3}}}\left( {\frac{{\sqrt 3 }}{2} + \frac{{2\pi }}{3} - 1 - \frac{\pi }{2}} \right)$ or equivalent     A1

$= \frac{1}{{24{a^3}}}\left( {3\sqrt 3  + \pi  - 6} \right)$     AG

[7 marks]