8. (a) Using the substitution $x = 2 \cos u$, or otherwise, find the exact value of \[ \int_1^{\sqrt{2}} \frac{1}{x^2 \sqrt{4-x^2}} \, dx \] (b) Using your answer to part (a), find the exact volume of the solid of revolution formed. - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 6

We start with the integral:

[ I = \int_1^{\sqrt{2}} \frac{1}{x^2 \sqrt{4-x^2}} , dx ]

Using the substitution $x = 2 \cos u$, we have: [ dx = -2 \sin u , du ]

The limits of integration change:

• When $x=1$, $u=\frac{\pi}{4}$
• When $x=\sqrt{2}$, $u=\frac{\pi}{3}$

Now substituting in the integral gives:
[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-2 \sin u}{(2 \cos u)^2 \sqrt{4 - (2 \cos u)^2}} , du ]

This simplifies to: [ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-2 \sin u}{4 \cos^2 u \sqrt{4(1 - \cos^2 u)}} , du ]

Knowing that $\sqrt{4(1 - \cos^2 u)} = 2 \sin u$ gives us: [ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-2 \sin u}{4 \cos^2 u , 2 \sin u} , du = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-1}{2 \cos^2 u} , du ]

This leads to:
[ I = -\frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \sec^2 u , du ]

Calculating the integral: [ I = -\frac{1}{2} \left[ \tan u \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -\frac{1}{2}\left(\tan\left(\frac{\pi}{3}\right) - \tan\left(\frac{\pi}{4}\right)\right) = -\frac{1}{2}(\sqrt{3} - 1) = \frac{1 - \sqrt{3}}{2} ]