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 7

To solve the integral ( \int_1^{\sqrt{2}} \frac{1}{x^2 \sqrt{4 - x^2}} , dx ), we start by applying the substitution:

Let ( x = 2 \cos u ). Thus, ( dx = -2 \sin u , du ).

The limits of integration change as follows:

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

Replacing ( x ) gives us: [ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-2 \sin u}{(2 \cos u)^2 \sqrt{4 - (2 \cos u)^2}} , du ] Simplifying this results in: [ = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-2 \sin u}{4 \cos^2 u \sqrt{4(1 - \cos^2 u)}} , du ] [ = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-2 \sin u}{4 \cos^2 u \cdot 2 \sin u} , du = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{-1}{\cos^2 u} , du ] Recognizing that ( \frac{1}{\cos^2 u} = \sec^2 u ), we have: [ = -\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \sec^2 u , du = -[\tan u]_{\frac{\pi}{4}}^{\frac{\pi}{3}} ] Evaluating this gives: [ = -\left( \tan \frac{\pi}{3} - \tan \frac{\pi}{4} \right) = -\left( \sqrt{3} - 1 \right) = 1 - \sqrt{3} ]