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June 2015 p31 q5
1714
(a) Find \(\int (4 + \tan^2 2x) \, dx\).
(b) Find the exact value of \(\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} \frac{\sin(x + \frac{1}{6}\pi)}{\sin x} \, dx\).
Solution
(a) Use the identity \(\tan^2 2x = \sec^2 2x - 1\) to rewrite the integral as \(\int (4 + \sec^2 2x - 1) \, dx = \int (3 + \sec^2 2x) \, dx\).
The integral of \(3 \, dx\) is \(3x\), and the integral of \(\sec^2 2x \, dx\) is \(\frac{1}{2} \tan 2x\) (using substitution \(u = 2x\), \(du = 2 \, dx\)).
Thus, the integral is \(3x + \frac{1}{2} \tan 2x\).
(b) Use the identity \(\sin x \cos \frac{1}{2}\pi + \cos x \sin \frac{1}{6}\pi\) to simplify the integrand to \(\cos \frac{1}{6}\pi + \frac{\cos x \sin \frac{1}{6}\pi}{\sin x}\).
The integral becomes \(\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} \left( \cos \frac{1}{6}\pi + \frac{\cos x \sin \frac{1}{6}\pi}{\sin x} \right) \, dx\).
Integrate to obtain \(\cos \frac{1}{6}\pi x + \sin \frac{1}{6}\pi \ln(\sin x)\).
Apply the limits and simplify to obtain \(\frac{1}{8} \pi \sqrt{3} - \frac{1}{2} \ln \left( \frac{1}{\sqrt{2}} \right)\).
