Tính các tích phân sau:

a) \(\int\limits_0^{2\pi } {\left( {2x + \cos x} \right)dx} \)\( = \int\limits_0^{2\pi } {2xdx} + \int\limits_0^{2\pi } {\cos xdx} \)\( = \left. {{x^2}} \right|_0^{2\pi } + \left. {\sin x} \right|_0^{2\pi }\)\( = 4{\pi ^2}\).

b) \(\int\limits_1^2 {\left( {{3^x} - \frac{3}{x}} \right)dx} \)\( = \int\limits_1^2 {{3^x}dx} - \int\limits_1^2 {\frac{3}{x}dx} \)

\( = \left. {\frac{{{3^x}}}{{\ln 3}}} \right|_1^2 - \left. {3\ln \left| x \right|} \right|_1^2 = \frac{{{3^2}}}{{\ln 3}} - \frac{3}{{\ln 3}} - 3\ln 2 + 3\ln 1 = \frac{6}{{\ln 3}} - 3\ln 2\).

c) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\left( {\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)dx} \)\( = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{1}{{{{\cos }^2}x}}dx} - \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{1}{{{{\sin }^2}x}}dx} \)

\( = \left. {\tan x} \right|_{\frac{\pi }{6}}^{\frac{\pi }{3}} + \left. {\cot x} \right|_{\frac{\pi }{6}}^{\frac{\pi }{3}}\)\( = \tan \frac{\pi }{3} - \tan \frac{\pi }{6} + \cot \frac{\pi }{3} - \cot \frac{\pi }{6}\)\( = \sqrt 3 - \frac{{\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3} - \sqrt 3 = 0\).