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Nov 2012 p31 q5
1807
(i) By differentiating \(\frac{1}{\cos x}\), show that if \(y = \sec x\) then \(\frac{dy}{dx} = \sec x \tan x\).
(ii) Show that \(\frac{1}{\sec x - \tan x} \equiv \sec x + \tan x\).
(iii) Deduce that \(\frac{1}{(\sec x - \tan x)^2} \equiv 2 \sec^2 x - 1 + 2 \sec x \tan x\).
(iv) Hence show that \(\int_0^{\frac{1}{4}\pi} \frac{1}{(\sec x - \tan x)^2} \, dx = \frac{1}{4}(8\sqrt{2} - \pi)\).
Solution
(i) Let \(y = \sec x = \frac{1}{\cos x}\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{0 \cdot \cos x - (-\sin x) \cdot 1}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x\).
(ii) Multiply numerator and denominator by \(\sec x + \tan x\): \(\frac{1}{\sec x - \tan x} \cdot \frac{\sec x + \tan x}{\sec x + \tan x} = \frac{\sec x + \tan x}{\sec^2 x - \tan^2 x}\). Using \(\sec^2 x - \tan^2 x = 1\), we have \(\sec x + \tan x\).
(iii) Substitute \(\sec x + \tan x\) into \(\frac{1}{(\sec x - \tan x)^2}\): \(\frac{1}{(\sec x - \tan x)^2} = (\sec x + \tan x)^2 = \sec^2 x + 2 \sec x \tan x + \tan^2 x\). Using \(\sec^2 x - \tan^2 x = 1\), we get \(2 \sec^2 x - 1 + 2 \sec x \tan x\).
(iv) Integrate \(\int_0^{\frac{1}{4}\pi} (2 \tan x - x + 2 \sec x) \, dx\). The integral of \(2 \tan x\) is \(-2 \ln |\cos x|\), the integral of \(x\) is \(\frac{x^2}{2}\), and the integral of \(2 \sec x\) is \(2 \ln |\sec x + \tan x|\). Evaluate from 0 to \(\frac{1}{4}\pi\) to obtain \(\frac{1}{4}(8\sqrt{2} - \pi)\).
