1. **Problem Statement:** Evaluate the integral $$\int_0^{\pi/2} \frac{dx}{4\cos x + 2\sin x}.$$\n\n2. **Formula and Approach:** To solve integrals of the form $$\int \frac{dx}{a\cos x + b\sin x},$$ we use the substitution $$a\cos x + b\sin x = R\cos(x - \alpha),$$ where $$R = \sqrt{a^2 + b^2}$$ and $$\tan \alpha = \frac{b}{a}.$$ This simplifies the denominator and makes the integral easier to evaluate.\n\n3. **Calculate R and \alpha:**\n$$R = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}.$$\n$$\tan \alpha = \frac{2}{4} = \frac{1}{2} \implies \alpha = \tan^{-1}\left(\frac{1}{2}\right).$$\n\n4. **Rewrite the integral:**\n$$\int_0^{\pi/2} \frac{dx}{4\cos x + 2\sin x} = \int_0^{\pi/2} \frac{dx}{2\sqrt{5} \cos(x - \alpha)} = \frac{1}{2\sqrt{5}} \int_0^{\pi/2} \sec(x - \alpha) \, dx.$$\n\n5. **Integrate:**\nThe integral of $$\sec u$$ is $$\ln|\sec u + \tan u| + C.$$\nSo,\n$$\int_0^{\pi/2} \sec(x - \alpha) \, dx = \left[ \ln|\sec(x - \alpha) + \tan(x - \alpha)| \right]_0^{\pi/2}.$$\n\n6. **Evaluate the definite integral:**\nCalculate at upper limit $$x = \pi/2$$:\n$$\sec\left(\frac{\pi}{2} - \alpha\right) = \csc \alpha, \quad \tan\left(\frac{\pi}{2} - \alpha\right) = \cot \alpha.$$\nCalculate at lower limit $$x = 0$$:\n$$\sec(-\alpha) = \sec \alpha, \quad \tan(-\alpha) = -\tan \alpha.$$\n\n7. **Substitute values:**\n$$\int_0^{\pi/2} \sec(x - \alpha) \, dx = \ln|\csc \alpha + \cot \alpha| - \ln|\sec \alpha - \tan \alpha| = \ln \left( \frac{\csc \alpha + \cot \alpha}{\sec \alpha - \tan \alpha} \right).$$\n\n8. **Simplify the expression:**\nUsing identities,\n$$\csc \alpha + \cot \alpha = \frac{1 + \cos \alpha}{\sin \alpha}, \quad \sec \alpha - \tan \alpha = \frac{1 - \sin \alpha}{\cos \alpha}.$$\nSo,\n$$\frac{\csc \alpha + \cot \alpha}{\sec \alpha - \tan \alpha} = \frac{\frac{1 + \cos \alpha}{\sin \alpha}}{\frac{1 - \sin \alpha}{\cos \alpha}} = \frac{(1 + \cos \alpha) \cos \alpha}{\sin \alpha (1 - \sin \alpha)}.$$\n\n9. **Calculate \sin \alpha and \cos \alpha:**\nSince $$\tan \alpha = \frac{1}{2},$$\n$$\sin \alpha = \frac{\tan \alpha}{\sqrt{1 + \tan^2 \alpha}} = \frac{1/2}{\sqrt{1 + (1/2)^2}} = \frac{1/2}{\sqrt{1 + 1/4}} = \frac{1/2}{\sqrt{5/4}} = \frac{1/2}{\frac{\sqrt{5}}{2}} = \frac{1}{\sqrt{5}}.$$\n$$\cos \alpha = \frac{1}{\sqrt{1 + \tan^2 \alpha}} = \frac{1}{\sqrt{1 + (1/2)^2}} = \frac{1}{\sqrt{5/4}} = \frac{2}{\sqrt{5}}.$$\n\n10. **Substitute values:**\n$$\frac{(1 + \cos \alpha) \cos \alpha}{\sin \alpha (1 - \sin \alpha)} = \frac{(1 + \frac{2}{\sqrt{5}}) \cdot \frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}} \cdot (1 - \frac{1}{\sqrt{5}})} = \frac{\left(1 + \frac{2}{\sqrt{5}}\right) \frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}} \left(\frac{\sqrt{5} - 1}{\sqrt{5}}\right)}.$$\nSimplify numerator:\n$$1 + \frac{2}{\sqrt{5}} = \frac{\sqrt{5} + 2}{\sqrt{5}},$$\nso numerator is\n$$\frac{\sqrt{5} + 2}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{2(\sqrt{5} + 2)}{5}.$$\nDenominator is\n$$\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5} - 1}{\sqrt{5}} = \frac{\sqrt{5} - 1}{5}.$$\n\n11. **Divide numerator by denominator:**\n$$\frac{2(\sqrt{5} + 2)/5}{(\sqrt{5} - 1)/5} = \frac{2(\sqrt{5} + 2)}{\sqrt{5} - 1}.$$\n\n12. **Rationalize denominator:**\nMultiply numerator and denominator by $$\sqrt{5} + 1$$:\n$$\frac{2(\sqrt{5} + 2)(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{2(\sqrt{5} + 2)(\sqrt{5} + 1)}{5 - 1} = \frac{2(\sqrt{5} + 2)(\sqrt{5} + 1)}{4}.$$\n\n13. **Expand numerator:**\n$$(\sqrt{5} + 2)(\sqrt{5} + 1) = 5 + \sqrt{5} + 2\sqrt{5} + 2 = 7 + 3\sqrt{5}.$$\nSo the expression is\n$$\frac{2(7 + 3\sqrt{5})}{4} = \frac{7 + 3\sqrt{5}}{2}.$$\n\n14. **Final integral value:**\n$$\int_0^{\pi/2} \frac{dx}{4\cos x + 2\sin x} = \frac{1}{2\sqrt{5}} \ln \left( \frac{7 + 3\sqrt{5}}{2} \right).$$\n\n**Answer:** $$\boxed{\frac{1}{2\sqrt{5}} \ln \left( \frac{7 + 3\sqrt{5}}{2} \right)}.$$