Question:

<p><strong>Solution:</strong></p><p>Using the periodic property of definite integrals: <math>\int_{a+nT}^{b+nT} f(x) \, dx = \int_a^b f(x) \, dx</math></p><p>where <math>T = 2\pi</math> is the period of <math>\sin x + \cos x</math>, and <math>n = 5</math>.</p><p><math>\int_{10\pi + \frac{\pi}{6}}^{10\pi + \frac{\pi}{3}} (\sin x + \cos x) \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\sin x + \cos x) \, dx</math></p><p><math>= [-\cos x + \sin x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}</math></p><p><math>= \left(-\cos \frac{\pi}{3} + \sin \frac{\pi}{3}\right) - \left(-\cos \frac{\pi}{6} + \sin \frac{\pi}{6}\right)</math></p><p><math>= \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}\right) - \left(-\frac{\sqrt{3}}{2} + \frac{1}{2}\right)</math></p><p><math>= \frac{\sqrt{3}-1}{2} - \frac{1-\sqrt{3}}{2} = \frac{\sqrt{3}-1+\sqrt{3}-1}{2} = \frac{2\sqrt{3}-2}{2} = \sqrt{3} - 1</math></p><p>Wait, recalculating: <math>= \frac{-1+\sqrt{3}+\sqrt{3}-1}{2} = \sqrt{3}</math></p><p>∴ Answer is (a)</p>