Đề bài

Tính các tích phân sau: a) \(\int\limits_0^{\frac{\pi }{2}} {\left( {3\cos x + 2\sin x} \right)dx} \); b) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)dx} \).

Lời giải chi tiết

a) Ta có \(\int\limits_0^{\frac{\pi }{2}} {\left( {3\cos x + 2\sin x} \right)dx}  = 3\int\limits_0^{\frac{\pi }{2}} {\cos xdx}  + 2\int\limits_0^{\frac{\pi }{2}} {\sin xdx}  = 3\left. {\sin x} \right|_0^{\frac{\pi }{2}} - 2\left. {\cos x} \right|_0^{\frac{\pi }{2}} = 3 + 2 = 5\). b) Ta có \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)dx}  = \left. {\left( {\tan x + \cot x} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{4}} = \tan \frac{\pi }{4} + \cot \frac{\pi }{4} - \tan \frac{\pi }{6} - \cot \frac{\pi }{6} = 2 - \frac{1}{{\sqrt 3 }} - \sqrt 3  = 2 - \frac{4}{{\sqrt 3 }}\).